C Aptitude Part V

Note : All the programs are tested under Turbo C/C++ compilers.
It is assumed that,
Programs run under DOS environment,
The underlying machine is an x86 system,
Program is compiled using Turbo C/C++ compiler.
The program output may depend on the information based on this assumptions
(for example sizeof(int) == 2 may be assumed).
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Predict the output or error(s) for the following:
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66) main()
{
int y;
scanf("%d",&y); // input given is 2000
if( (y%4==0 && y%100 != 0) || y%100 == 0 )
printf("%d is a leap year");
else
printf("%d is not a leap year");
}
Answer:
2000 is a leap year
Explanation:
An ordinary program to check if leap year or not.
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67) #define max 5
#define int arr1[max]
main()
{
typedef char arr2[max];
arr1 list={0,1,2,3,4};
arr2 name="name";
printf("%d %s",list[0],name);
}
Answer:
Compiler error (in the line arr1 list = {0,1,2,3,4})
Explanation:
arr2 is declared of type array of size 5 of characters. So it can be used to
declare the variable name of the type arr2. But it is not the case of arr1.
Hence an error.
Rule of Thumb:
#defines are used for textual replacement whereas typedefs are used for
declaring new types.
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68) int i=10;
main()
{
extern int i;
{
int i=20;
{
const volatile unsigned i=30;
printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}
Answer:
30,20,10
Explanation:
'{' introduces new block and thus new scope. In the innermost block i is
declared as,
const volatile unsigned
which is a valid declaration. i is assumed of type int. So printf prints 30. In
the next block, i has value 20 and so printf prints 20. In the outermost
block, i is declared as extern, so no storage space is allocated for it. After
compilation is over the linker resolves it to global variable i (since it is the
only variable visible there). So it prints i's value as 10.
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69) main()
{
int *j;
{
int i=10;
j=&i;
}
printf("%d",*j);
}
Answer:
10
Explanation:
The variable i is a block level variable and the visibility is inside that
block only. But the lifetime of i is lifetime of the function so it lives upto
the exit of main function. Since the i is still allocated space, *j prints the
value stored in i since j points i.
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70) main()
{
int i=-1;
-i;
printf("i = %d, -i = %d \n",i,-i);
}
Answer:
i = -1, -i = 1
Explanation:
-i is executed and this execution doesn't affect the value of i. In printf first
you just print the value of i. After that the value of the expression -i = -(-1)
is printed.
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71) #include<stdio.h>
main()
{
const int i=4;
float j;
j = ++i;
printf("%d %f", i,++j);
}
Answer:
Compiler error
Explanation:
i is a constant. you cannot change the value of constant
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72) #include<stdio.h>
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d..%d",*p,*q);
}
Answer:
garbagevalue..1
Explanation:
p=&a[2][2][2] you declare only two 2D arrays. but you are trying to
access the third 2D(which you are not declared) it will print garbage
values. *q=***a starting address of a is assigned integer pointer. now q is
pointing to starting address of a.if you print *q meAnswer:it will print first
element of 3D array.
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73) #include<stdio.h>
main()
{
register i=5;
char j[]= "hello";
printf("%s %d",j,i);
}
Answer:
hello 5
Explanation:
if you declare i as register compiler will treat it as ordinary integer and it
will take integer value. i value may be stored either in register or in
memory.
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74) main()
{
int i=5,j=6,z;
printf("%d",i+++j);
}
Answer:
11
Explanation:
the expression i+++j is treated as (i++ + j)
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76) struct aaa{
struct aaa *prev;
int i;
struct aaa *next;
};
main()
{
struct aaa abc,def,ghi,jkl;
int x=100;
abc.i=0;abc.prev=&jkl;
abc.next=&def;
def.i=1;def.prev=&abc;def.next=&ghi;
ghi.i=2;ghi.prev=&def;
ghi.next=&jkl;
jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;
x=abc.next->next->prev->next->i;
printf("%d",x);
}
Answer:
2
Explanation:
above all statements form a double circular linked list;
abc.next->next->prev->next->i
this one points to "ghi" node the value of at particular node is 2.
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77) struct point
{
int x;
int y;
};
struct point origin,*pp;
main()
{
pp=&origin;
printf("origin is(%d%d)\n",(*pp).x,(*pp).y);
printf("origin is (%d%d)\n",pp->x,pp->y);
}
Answer:
origin is(0,0)
origin is(0,0)
Explanation:
pp is a pointer to structure. we can access the elements of the structure
either with arrow mark or with indirection operator.
Note:
Since structure point is globally declared x & y are initialized as zeroes
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78) main()
{
int i=_l_abc(10);
printf("%d\n",--i);
}
int _l_abc(int i)
{
return(i++);
}
Answer:
9
Explanation:
return(i++) it will first return i and then increments. i.e. 10 will be
returned.
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79) main()
{
char *p;
int *q;
long *r;
p=q=r=0;
p++;
q++;
r++;
printf("%p...%p...%p",p,q,r);
}
Answer:
0001...0002...0004
Explanation:
++ operator when applied to pointers increments address according to
their corresponding data-types.
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80) main()
{
char c=' ',x,convert(z);
getc(c);
37if((c>='a') && (c<='z'))
x=convert(c);
printf("%c",x);
}
convert(z)
{
return z-32;
}
Answer:
Compiler error
Explanation:
declaration of convert and format of getc() are wrong.
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81) main(int argc, char **argv)
{
printf("enter the character");
getchar();
sum(argv[1],argv[2]);
}
sum(num1,num2)
int num1,num2;
{
return num1+num2;
}
Answer:
Compiler error.
Explanation:
argv[1] & argv[2] are strings. They are passed to the function sum without
converting it to integer values.
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82) # include <stdio.h>
int one_d[]={1,2,3};
main()
{
int *ptr;
ptr=one_d;
ptr+=3;
printf("%d",*ptr);
}
Answer:
garbage value
Explanation:
ptr pointer is pointing to out of the array range of one_d.
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83) # include<stdio.h>
aaa() {printf("hi");
}
bbb(){
printf("hello");
}
ccc(){
printf("bye");
}
main()
{
int (*ptr[3])();
ptr[0]=aaa;
ptr[1]=bbb;
ptr[2]=ccc;
ptr[2]();
}
Answer:
bye
Explanation:
ptr is array of pointers to functions of return type int.ptr[0] is assigned to
address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc
respectively. ptr[2]() is in effect of writing ccc(), since ptr[2] points to ccc.
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85) #include<stdio.h>
main()
{
FILE *ptr;
char i;
ptr=fopen("zzz.c","r");
while((i=fgetch(ptr))!=EOF)
printf("%c",i);
}
Answer:
contents of zzz.c followed by an infinite loop
Explanation:
The condition is checked against EOF, it should be checked against
NULL.
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86) main()
{
int i =0;j=0;
if(i && j++)
printf("%d..%d",i++,j);
printf("%d..%d,i,j);
}
Answer:
390..0
Explanation:
The value of i is 0. Since this information is enough to determine the truth
value of the boolean expression. So the statement following the if
statement is not executed. The values of i and j remain unchanged and get
printed.
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87) main()
{
int i;
i = abc();
printf("%d",i);
}
abc()
{
_AX = 1000;
}
Answer:
1000
Explanation:
Normally the return value from the function is through the information
from the accumulator. Here _AH is the pseudo global variable denoting
the accumulator. Hence, the value of the accumulator is set 1000 so the
function returns value 1000.
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88) int i;
main(){
int t;
for ( t=4;scanf("%d",&i)-t;printf("%d\n",i))
printf("%d--",t--);
}
// If the inputs are 0,1,2,3 find the o/p
Answer:
4--0
3--1
2--2
Explanation:
Let us assume some x= scanf("%d",&i)-t the values during execution
will be,
t
i
x
4
0
-4
3
1
-2
2
2
0
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89) main(){
int a= 0;int b = 20;char x =1;char y =10;
40if(a,b,x,y)
printf("hello");
}
Answer:
hello
Explanation:
The comma operator has associativity from left to right. Only the
rightmost value is returned and the other values are evaluated and ignored.
Thus the value of last variable y is returned to check in if. Since it is a non
zero value if becomes true so, "hello" will be printed.
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90) main(){
unsigned int i;
for(i=1;i>-2;i--)
printf("c aptitude");
}
Explanation:
i is an unsigned integer. It is compared with a signed value. Since the both
types doesn't match, signed is promoted to unsigned value. The unsigned
equivalent of -2 is a huge value so condition becomes false and control
comes out of the loop.
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C Aptitude Part IV

Note : All the programs are tested under Turbo C/C++ compilers.
It is assumed that,
Programs run under DOS environment,
The underlying machine is an x86 system,
Program is compiled using Turbo C/C++ compiler.
The program output may depend on the information based on this assumptions
(for example sizeof(int) == 2 may be assumed).
---------------------------------------------------------------------------------------------------------------------------------------
Predict the output or error(s) for the following:
---------------------------------------------------------------------------------------------------------------------------------------  
46) main()
{
show();
}
void show()
{
printf("I'm the greatest");
}
Answer:
Compier error: Type mismatch in redeclaration of show.
Explanation:
When the compiler sees the function show it doesn't know anything about
it. So the default return type (ie, int) is assumed. But when compiler sees
the actual definition of show mismatch occurs since it is declared as void.
Hence the error.
The solutions are as follows:
1. declare void show() in main() .
2. define show() before main().
3. declare extern void show() before the use of show().
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47) main( )
{
int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
printf(―%u %u %u %d \n‖,a,*a,**a,***a);
printf(―%u %u %u %d \n‖,a+1,*a+1,**a+1,***a+1);
}
Answer:
100, 100, 100, 2
114, 104, 102, 3
Explanation:
The given array is a 3-D one. It can also be viewed as a 1-D array.
2 4
7 8 3
4 2
2 2
3
3
4
100 102 104 106 108 110 112 114 116 118 120 122
thus, for the first printf statement a, *a, **a give address of first element .
since the indirection ***a gives the value. Hence, the first line of the
output.
for the second printf a+1 increases in the third dimension thus points to
value at 114, *a+1 increments in second dimension thus points to 104, **a
+1 increments the first dimension thus points to 102 and ***a+1 first gets
the value at first location and then increments it by 1. Hence, the output.
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48) main( )
{
int a[ ] = {10,20,30,40,50},j,*p;
for(j=0; j<5; j++)
{
printf(―%d‖ ,*a);
a++;
}
p = a;
for(j=0; j<5; j++)
{
printf(―%d ‖ ,*p);
p++;
}
}
Answer:
Compiler error: lvalue required.
Explanation:
Error is in line with statement a++. The operand must be an lvalue and
may be of any of scalar type for the any operator, array name only when
subscripted is an lvalue. Simply array name is a non-modifiable lvalue.
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49) main( )
{
static int a[ ] = {0,1,2,3,4};
int *p[ ] = {a,a+1,a+2,a+3,a+4};
int **ptr = p;
ptr++;
printf(―\n %d %d %d‖, ptr-p, *ptr-a, **ptr);
*ptr++;
printf(―\n %d %d %d‖, ptr-p, *ptr-a, **ptr);
*++ptr;
printf(―\n %d %d %d‖, ptr-p, *ptr-a, **ptr);
++*ptr;
printf(―\n %d %d %d‖, ptr-p, *ptr-a, **ptr);
}
Answer:
111
222
333
344
Explanation:
Let us consider the array and the two pointers with some address
a
260
100
1
102
2
104
3
106
4
108
p
100
102
104
106
108
1000 1002 1004 1006 1008
ptr
1000
2000
After execution of the instruction ptr++ value in ptr becomes 1002, if
scaling factor for integer is 2 bytes. Now ptr – p is value in ptr – starting
location of array p, (1002 – 1000) / (scaling factor) = 1, *ptr – a = value at
address pointed by ptr – starting value of array a, 1002 has a value 102 so
the value is (102 – 100)/(scaling factor) = 1, **ptr is the value stored in
the location pointed by the pointer of ptr = value pointed by value pointed
by 1002 = value pointed by 102 = 1. Hence the output of the firs printf is
1, 1, 1.
After execution of *ptr++ increments value of the value in ptr by scaling
factor, so it becomes1004. Hence, the outputs for the second printf are ptr
– p = 2, *ptr – a = 2, **ptr = 2.
After execution of *++ptr increments value of the value in ptr by scaling
factor, so it becomes1004. Hence, the outputs for the third printf are ptr –
p = 3, *ptr – a = 3, **ptr = 3.
After execution of ++*ptr value in ptr remains the same, the value pointed
by the value is incremented by the scaling factor. So the value in array p at
location 1006 changes from 106 10 108,. Hence, the outputs for the fourth
printf are ptr – p = 1006 – 1000 = 3, *ptr – a = 108 – 100 = 4, **ptr = 4.
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50) main( )
{
char *q;
int j;
for (j=0; j<3; j++) scanf(―%s‖ ,(q+j));
for (j=0; j<3; j++) printf(―%c‖ ,*(q+j));
for (j=0; j<3; j++) printf(―%s‖ ,(q+j));
}
Explanation:
Here we have only one pointer to type char and since we take input in the
same pointer thus we keep writing over in the same location, each time
shifting the pointer value by 1. Suppose the inputs are MOUSE, TRACK
and VIRTUAL. Then for the first input suppose the pointer starts at
location 100 then the input one is stored as
M
O
U
S
E
\0
When the second input is given the pointer is incremented as j value
becomes 1, so the input is filled in memory starting from 101.
M
T
R
A
C
K
\0
The third input starts filling from the location 102
27M
T
V
I
R
T
U
A
L
\0
This is the final value stored .
The first printf prints the values at the position q, q+1 and q+2 = M T V
The second printf prints three strings starting from locations q, q+1, q+2
i.e MTVIRTUAL, TVIRTUAL and VIRTUAL.
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51) main( )
{
void *vp;
char ch =  ̳g‘, *cp = ―goofy‖;
int j = 20;
vp = &ch;
printf(―%c‖, *(char *)vp);
vp = &j;
printf(―%d‖,*(int *)vp);
vp = cp;
printf(―%s‖,(char *)vp + 3);
}
Answer:
g20fy
Explanation:
Since a void pointer is used it can be type casted to any other type pointer.
vp = &ch stores address of char ch and the next statement prints the value
stored in vp after type casting it to the proper data type pointer. the output
is  ̳g‘. Similarly the output from second printf is  ̳20‘. The third printf
statement type casts it to print the string from the 4 th value hence the
output is  ̳fy‘.
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52) main ( )
{
static char *s[ ] = {―black‖, ―white‖, ―yellow‖, ―violet‖};
char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;
p = ptr;
**++p;
printf(―%s‖,*--*++p + 3);
}
Answer:
ck
Explanation:
In this problem we have an array of char pointers pointing to start of 4
strings. Then we have ptr which is a pointer to a pointer of type char and a
variable p which is a pointer to a pointer to a pointer of type char. p hold
the initial value of ptr, i.e. p = s+3. The next statement increment value in
p by 1 , thus now value of p = s+2. In the printf statement the expression
is evaluated *++p causes gets value s+1 then the pre decrement is
executed and we get s+1 – 1 = s . the indirection operator now gets the
28value from the array of s and adds 3 to the starting address. The string is
printed starting from this position. Thus, the output is  ̳ck‘.
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53)
main()
{
int i, n;
char *x = ―girl‖;
n = strlen(x);
*x = x[n];
for(i=0; i<n; ++i)
{
printf(―%s\n‖,x);
x++;
}
}
Answer:
(blank space)
irl
rl
l
Explanation:
Here a string (a pointer to char) is initialized with a value ―girl‖. The
strlen function returns the length of the string, thus n has a value 4. The
next statement assigns value at the nth location ( ̳\0‘) to the first location.
Now the string becomes ―\0irl‖ . Now the printf statement prints the string
after each iteration it increments it starting position. Loop starts from 0 to
4. The first time x[0] =  ̳\0‘ hence it prints nothing and pointer value is
incremented. The second time it prints from x[1] i.e ―irl‖ and the third
time it prints ―rl‖ and the last time it prints ―l‖ and the loop terminates.
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54) int i,j;
for(i=0;i<=10;i++)
{
j+=5;
assert(i<5);
}
Answer:
Runtime error: Abnormal program termination.
assert failed (i<5), <file name>,<line number>
Explanation:
asserts are used during debugging to make sure that certain conditions are
satisfied. If assertion fails, the program will terminate reporting the same.
After debugging use,
#undef NDEBUG
and this will disable all the assertions from the source code. Assertion
is a good debugging tool to make use of.
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55) main()
{
int i=-1;
+i;
printf("i = %d, +i = %d \n",i,+i);
}
Answer:
i = -1, +i = -1
Explanation:
Unary + is the only dummy operator in C. Where-ever it comes you can
just ignore it just because it has no effect in the expressions (hence the
name dummy operator).
56) What are the files which are automatically opened when a C file is executed?
Answer:
stdin, stdout, stderr (standard input,standard output,standard error).
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57) what will be the position of the file marker?
a: fseek(ptr,0,SEEK_SET);
b: fseek(ptr,0,SEEK_CUR);
Answer :
a: The SEEK_SET sets the file position marker to the starting of the file.
b: The SEEK_CUR sets the file position marker to the current position
of the file.
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58) main()
{
char name[10],s[12];
scanf(" \"%[^\"]\"",s);
}
How scanf will execute?
Answer:
First it checks for the leading white space and discards it.Then it matches
with a quotation mark and then it reads all character upto another
quotation mark.
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59) What is the problem with the following code segment?
while ((fgets(receiving array,50,file_ptr)) != EOF)
;
Answer & Explanation:
fgets returns a pointer. So the correct end of file check is checking for !=
NULL.
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60) main()
{
main();
}
Answer:
Runtime error : Stack overflow.
Explanation:
main function calls itself again and again. Each time the function is called
its return address is stored in the call stack. Since there is no condition to
terminate the function call, the call stack overflows at runtime. So it
terminates the program and results in an error.
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61) main()
{
char *cptr,c;
void *vptr,v;
c=10; v=0;
cptr=&c; vptr=&v;
printf("%c%v",c,v);
}
Answer:
Compiler error (at line number 4): size of v is Unknown.
Explanation:
You can create a variable of type void * but not of type void, since void is
an empty type. In the second line you are creating variable vptr of type
void * and v of type void hence an error.
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62) main()
{
char *str1="abcd";
char str2[]="abcd";
printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));
}
Answer:
2 5 5
Explanation:
In first sizeof, str1 is a character pointer so it gives you the size of the
pointer variable. In second sizeof the name str2 indicates the name of the
array whose size is 5 (including the '\0' termination character). The third
sizeof is similar to the second one.
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63) main()
{
char not;
not=!2;
printf("%d",not);
}
Answer:
0
Explanation:
! is a logical operator. In C the value 0 is considered to be the boolean
value FALSE, and any non-zero value is considered to be the boolean
value TRUE. Here 2 is a non-zero value so TRUE. !TRUE is FALSE (0)
so it prints 0.
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64) #define FALSE -1
#define TRUE 1
#define NULL 0
main() {
if(NULL)
puts("NULL");
else if(FALSE)
puts("TRUE");
else
puts("FALSE");
}
Answer:
TRUE
Explanation:
The input program to the compiler after processing by the preprocessor is,
main(){
if(0)
puts("NULL");
else if(-1)
puts("TRUE");
else
puts("FALSE");
}
Preprocessor doesn't replace the values given inside the double quotes.
The check by if condition is boolean value false so it goes to else. In
second if -1 is boolean value true hence "TRUE" is printed.
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65) main()
{
int k=1;
printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");
}
Answer:
1==1 is TRUE
Explanation:
When two strings are placed together (or separated by white-space) they
are concatenated (this is called as "stringization" operation). So the string
32is as if it is given as "%d==1 is %s". The conditional operator( ?: )
evaluates to "TRUE".
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C Aptitude Part III

Note : All the programs are tested under Turbo C/C++ compilers.
It is assumed that,
Programs run under DOS environment,
The underlying machine is an x86 system,
Program is compiled using Turbo C/C++ compiler.
The program output may depend on the information based on this assumptions
(for example sizeof(int) == 2 may be assumed).
---------------------------------------------------------------------------------------------------------------------------------------
Predict the output or error(s) for the following:
---------------------------------------------------------------------------------------------------------------------------------------

27) main()
{
clrscr();
}
clrscr();
Answer:
No output/error
Explanation:
The first clrscr() occurs inside a function. So it becomes a function
call. In the second clrscr(); is a function declaration (because it is not
inside any function).
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28) enum colors {BLACK,BLUE,GREEN}
main()
{
printf("%d..%d..%d",BLACK,BLUE,GREEN);
return(1);
}
Answer:
0..1..2
Explanation:
enum assigns numbers starting from 0, if not explicitly defined.
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29) void main()
{
char far *farther,*farthest;
printf("%d..%d",sizeof(farther),sizeof(farthest));
}
Answer:
4..2
Explanation:
the second pointer is of char type and not a far pointer
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30) main()
{
int i=400,j=300;
printf("%d..%d");
19}
Answer:
400..300
Explanation:
printf takes the values of the first two assignments of the program. Any
number of printf's may be given. All of them take only the first two
values. If more number of assignments given in the program,then printf
will take garbage values.
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31) main()
{
char *p;
p="Hello";
printf("%c\n",*&*p);
}
Answer:
H
Explanation:
* is a dereference operator & is a reference operator. They can be
applied any number of times provided it is meaningful. Here p points to
the first character in the string "Hello". *p dereferences it and so its value
is H. Again & references it to an address and * dereferences it to the value
H.

32) main()
{
int i=1;
while (i<=5)
{
printf("%d",i);
if (i>2)
goto here;
i++;
}
}
fun()
{
here:
printf("PP");
}
Answer:
Compiler error: Undefined label 'here' in function main
Explanation:
Labels have functions scope, in other words The scope of the labels is
limited to functions . The label 'here' is available in function fun() Hence it
is not visible in function main.
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33) main()
{
static char names[5][20]={"pascal","ada","cobol","fortran","perl"};
int i;
char *t;
t=names[3];
names[3]=names[4];
names[4]=t;
for (i=0;i<=4;i++)
printf("%s",names[i]);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
Array names are pointer constants. So it cannot be modified.
---------------------------------------------------------------------------------------------------------------------------------------
34) void main()
{
int i=5;
printf("%d",i++ + ++i);
}
Answer:
Output Cannot be predicted exactly.
Explanation:
Side effects are involved in the evaluation of i
---------------------------------------------------------------------------------------------------------------------------------------
35) void main()
{
int i=5;
printf("%d",i+++++i);
}
Answer:
Compiler Error
Explanation:
The expression i+++++i is parsed as i ++ ++ + i which is an illegal
combination of operators.
---------------------------------------------------------------------------------------------------------------------------------------
36) #include<stdio.h>
main()
{
int i=1,j=2;
switch(i)
{
case 1: printf("GOOD");
break;
case j: printf("BAD");
break;
}
}
Answer:
Compiler Error: Constant expression required in function main.
Explanation:
The case statement can have only constant expressions (this implies that
we cannot use variable names directly so an error).
Note:
Enumerated types can be used in case statements.
---------------------------------------------------------------------------------------------------------------------------------------
37) main()
{
int i;
printf("%d",scanf("%d",&i)); // value 10 is given as input here
}
Answer:
1
Explanation:
Scanf returns number of items successfully read and not 1/0. Here 10 is
given as input which should have been scanned successfully. So number
of items read is 1.
---------------------------------------------------------------------------------------------------------------------------------------
38) #define f(g,g2) g##g2
main()
{
int var12=100;
printf("%d",f(var,12));
}
Answer:
100
---------------------------------------------------------------------------------------------------------------------------------------
39) main()
{
int i=0;
for(;i++;printf("%d",i)) ;
printf("%d",i);
}
Answer:
1
Explanation:
before entering into the for loop the checking condition is "evaluated".
Here it evaluates to 0 (false) and comes out of the loop, and i is
incremented (note the semicolon after the for loop).
---------------------------------------------------------------------------------------------------------------------------------------
40) #include<stdio.h>
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}
Answer:
M
Explanation:
p is pointing to character '\n'.str1 is pointing to character 'a' ++*p
meAnswer:"p is pointing to '\n' and that is incremented by one." the ASCII
value of '\n' is 10. then it is incremented to 11. the value of ++*p is 11.
++*str1 meAnswer:"str1 is pointing to 'a' that is incremented by 1 and it
becomes 'b'. ASCII value of 'b' is 98. both 11 and 98 is added and result is
subtracted from 32.
i.e. (11+98-32)=77("M");
---------------------------------------------------------------------------------------------------------------------------------------
41) #include<stdio.h>
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s=malloc(sizeof(struct xx));
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
Initialization should not be done for structure members inside the structure
declaration
---------------------------------------------------------------------------------------------------------------------------------------
42) #include<stdio.h>
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}
Answer:
Compiler Error
Explanation:
in the end of nested structure yy a member have to be declared.
---------------------------------------------------------------------------------------------------------------------------------------
43) main()
{
extern int i;
i=20;
printf("%d",sizeof(i));
}
Answer:
Linker error: undefined symbol '_i'.
Explanation:
extern declaration specifies that the variable i is defined somewhere else.
The compiler passes the external variable to be resolved by the linker. So
compiler doesn't find an error. During linking the linker searches for the
definition of i. Since it is not found the linker flags an error.
---------------------------------------------------------------------------------------------------------------------------------------
44) main()
{
printf("%d", out);
}
int out=100;
Answer:
Compiler error: undefined symbol out in function main.
Explanation:
The rule is that a variable is available for use from the point of declaration.
Even though a is a global variable, it is not available for main. Hence an
error.
---------------------------------------------------------------------------------------------------------------------------------------
45) main()
{
extern out;
printf("%d", out);
}
int out=100;
Answer:
24100
Explanation:
This is the correct way of writing the previous program.
---------------------------------------------------------------------------------------------------------------------------------------

C Aptitude Part II

Note : All the programs are tested under Turbo C/C++ compilers.
It is assumed that,
Programs run under DOS environment,
The underlying machine is an x86 system,
Program is compiled using Turbo C/C++ compiler.
The program output may depend on the information based on this assumptions
(for example sizeof(int) == 2 may be assumed).
---------------------------------------------------------------------------------------------------------------------------------------
Predict the output or error(s) for the following:
---------------------------------------------------------------------------------------------------------------------------------------

11. main()
{
char string[]="Hello World";
display(string);
}
void display(char *string)
{
printf("%s",string);
}
Answer:
Compiler Error : Type mismatch in redeclaration of function display
Explanation :
In third line, when the function display is encountered, the compiler
doesn't know anything about the function display. It assumes the arguments and
return types to be integers, (which is the default type). When it sees the actual
function display, the arguments and type contradicts with what it has assumed
previously. Hence a compile time error occurs.
--------------------------------------------------------------------------------------------------------------------------------------
12. main()
{
int c=- -2;
printf("c=%d",c);
}
Answer:
c=2;
Explanation:
Here unary minus (or negation) operator is used twice. Same maths
rules applies, ie. minus * minus= plus.
Note:
However you cannot give like --2. Because -- operator can only be
applied to variables as a decrement operator (eg., i--). { “ 2 is a constant and not a
variable. ” }
--------------------------------------------------------------------------------------------------------------------------------------
13. #define int char
main()
{
int i=65;
printf("sizeof(i)=%d",sizeof(i));
}
Answer:
sizeof(i)=1
Explanation:
Since the #define replaces the string int by the macro char
--------------------------------------------------------------------------------------------------------------------------------------
14. main()
{
int i=10;
i=!i>14; // Here Not of I is compared with 14,that’s how result is executed
Printf ("i=%d",i);
}
Answer:
14i=0
Explanation:
In the expression !i>14 , NOT (!) operator has more precedence than „
>‟ symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is
false (zero).
--------------------------------------------------------------------------------------------------------------------------------------
15. #include<stdio.h>
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}
Answer:
77
Explanation:
p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is
pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10, which
is then incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to 'a'
that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.
Now performing (11 + 98 – 32), we get 77("M");
So we get the output 77 :: "M" (Ascii is 77).
--------------------------------------------------------------------------------------------------------------------------------------
16. #include<stdio.h>
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d----%d",*p,*q);
}
Answer:
SomeGarbageValue---1
Explanation:
p=&a[2][2][2] you declare only two 2D arrays, but you are trying to
access the third 2D(which you are not declared) it will print garbage values. *q=***a
starting address of a is assigned integer pointer. Now q is pointing to starting address of a.
If you print *q, it will print first element of 3D array.
--------------------------------------------------------------------------------------------------------------------------------------
17. #include<stdio.h>
main()
{
struct xx
{
int x=3; // canot initialized variable at deecleration
char name[]="hello";
};
struct xx *s;
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
You should not initialize variables in declaration
--------------------------------------------------------------------------------------------------------------------------------------
18. #include<stdio.h>
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}
Answer:
Compiler Error
Explanation:
The structure yy is nested within structure xx. Hence, the elements
are of yy are to be accessed through the instance of structure xx, which needs an
instance of yy to be known. If the instance is created after defining the structure the
compiler will not know about the instance relative to xx. Hence for nested structure
yy you have to declare member.
--------------------------------------------------------------------------------------------------------------------------------------
19. main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}
Answer:
16hai
Explanation:
\n - newline
\b - backspace
\r - linefeed
--------------------------------------------------------------------------------------------------------------------------------------
20. main()
{
int i=5;
printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
}
Answer:
45545
Explanation:
The arguments in a function call are pushed into the stack from left to
right. The evaluation is by popping out from the stack. and the evaluation is from
right to left, hence the result.
--------------------------------------------------------------------------------------------------------------------------------------
21. #define square(x) x*x
main()
{
int i;
i = 64/square(4);
printf("%d",i);
}
Answer:
64
Explanation:
the macro call square(4) will substituted by 4*4 so the expression
becomes i = 64/4*4 . Since / and * has equal priority the expression will be evaluated
as (64/4)*4 i.e. 16*4 = 64
--------------------------------------------------------------------------------------------------------------------------------------
22. main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s %s",p,p1);
}
Answer:
ibj!gsjfoet
Explanation:
++*p++ will be parse in the given order
*p that is value at the location currently pointed by p will be taken
++*p the retrieved value will be incremented
when ; is encountered the location will be incremented that is p++ will be executed
Hence, in the while loop initial value pointed by p is  ̳h‘, which is changed to  ̳i‘ by
executing ++*p and pointer moves to point,  ̳a‘ which is similarly changed to  ̳b‘ and so
on. Similarly blank space is converted to  ̳!‘. Thus, we obtain value in p becomes
―ibj!gsjfoet‖ and since p reaches  ̳\0‘ and p1 points to p thus p1doesnot print anything.
--------------------------------------------------------------------------------------------------------------------------------------
23. #include <stdio.h>
#define a 10
main()
{
#define a 50
printf("%d",a);
}
Answer:
50
Explanation:
The preprocessor directives can be redefined anywhere in the program. So
the most recently assigned value will be taken.
--------------------------------------------------------------------------------------------------------------------------------------
24. #define clrscr() 100
main()
{
clrscr();
printf("%d\n",clrscr());
}
Answer:
100
Explanation:
Preprocessor executes as a seperate pass before the execution of the
compiler. So textual replacement of clrscr() to 100 occurs.The input program to
compiler looks like this :
main()
{
100;
printf("%d\n",100);
}
Note:
100; is an executable statement but with no action. So it doesn't give
any problem
--------------------------------------------------------------------------------------------------------------------------------------
25. main()
{
printf("%p",main);
}
Answer:
Some address will be printed.
Explanation:
Function names are just addresses (just like array names are addresses).
main() is also a function. So the address of function main will be printed. %p in printf
specifies that the argument is an address. They are printed as hexadecimal numbers.

C Aptitude Part I

Note : All the programs are tested under Turbo C/C++ compilers.
It is assumed that,
Programs run under DOS environment,
 The underlying machine is an x86 system,
 Program is compiled using Turbo C/C++ compiler.
The program output may depend on the information based on this assumptions
(for example sizeof(int) == 2 may be assumed).
---------------------------------------------------------------------------------------------------------------------------------------
Predict the output or error(s) for the following:
---------------------------------------------------------------------------------------------------------------------------------------
1. void main()
{
int const * p=5;
printf("%d",++(*p));
}
Answer:
Compiler error: Cannot modify a constant value.
Explanation:
p is a pointer to a "constant integer". But we tried to change the value
of the "constant integer".
--------------------------------------------------------------------------------------------------------------------------------------
2. main()
{
char s[ ]="man";
int i;
for(i=0;s[ i ];i++)
printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);
}
Answer:
mmmm
aaaa
nnnn
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea.
Generally array name is the base address for that array. Here s is the base address. i is the
10index number/displacement from the base address. So, indirecting it with * is same as
s[i]. i[s] may be surprising. But in the case of C it is same as s[i].
--------------------------------------------------------------------------------------------------------------------------------------
3. main()
{
float me = 1.1;
double you = 1.1;
if(me==you)
printf("I love U");
else
printf("I hate U");
}
Answer:
I hate U
Explanation:
For floating point numbers (float, double, long double) the values
cannot be predicted exactly. Depending on the number of bytes, the precession with
of the value represented varies. Float takes 4 bytes and long double takes 10 bytes.
So float stores 0.9 with less precision than long double.
Rule of Thumb:
Never compare or at-least be cautious when using floating point
numbers with relational operators (== , >, <, <=, >=,!= ) .
--------------------------------------------------------------------------------------------------------------------------------------
4. main()
{
static int var = 5;
printf("%d ",var--);
if(var)
main();
}
Answer:
5 4 3 2 1
Explanation:
When static storage class is given, it is initialized once. The change in the
value of a static variable is retained even between the function calls. Main is also treated
like any other ordinary function, which can be called recursively.
--------------------------------------------------------------------------------------------------------------------------------------
5. main()
{
int c[ ]={2.8,3.4,4,6.7,5};
int j,*p=c,*q=c;
for(j=0;j<5;j++) {
printf(" %d ",*c);
++q; }
for(j=0;j<5;j++){
printf(" %d ",*p);
++p; }
11}
Answer:
2 2 2 2 2 2 3 4 6 5
Explanation:
Initially pointer c is assigned to both p and q. In the first loop, since only q
is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is
incremented. So the values 2 3 4 6 5 will be printed.
--------------------------------------------------------------------------------------------------------------------------------------
6. main()
{
extern int i;
i=20;
printf("%d",i);
}
Answer:
Linker Error : Undefined symbol '_i'
Explanation:
extern storage class in the following declaration,
extern int i;
specifies to the compiler that the memory for i is allocated in some other program
and that address will be given to the current program at the time of linking. But
linker finds that no other variable of name i is available in any other program with
memory space allocated for it. Hence a linker error has occurred .
--------------------------------------------------------------------------------------------------------------------------------------
7. main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}
Answer:
0 0 1 3 1
Explanation :
Logical operations always give a result of 1 or 0 . And also the logical
AND (&&) operator has higher priority over the logical OR (||) operator. So the
expression „i++ && j++ && k++‟ is executed first. The result of this expression is 0
(-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR
operator always gives 1 except for „0 || 0‟ combination- for which it gives 0). So the
value of m is 1. The values of other variables are also incremented by 1.
--------------------------------------------------------------------------------------------------------------------------------------
8. main()
{
char *p;
printf("%d %d ",sizeof(*p),sizeof(p));
12}
Answer:
1 2
Explanation:
The sizeof() operator gives the number of bytes taken by its operand.
P is a character pointer, which needs one byte for storing its value (a character).
Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of
the character pointer sizeof(p) gives 2.
--------------------------------------------------------------------------------------------------------------------------------------
9. main()
{
int i=3;
switch(i)
{
default:printf("zero");
case 1: printf("one");
break;
case 2:printf("two");
break;
case 3: printf("three");
break;
}
}
Answer :
three
Explanation :
The default case can be placed anywhere inside the loop. It is executed
only when all other cases doesn't match.
--------------------------------------------------------------------------------------------------------------------------------------
10. main()
{
printf("%x",-1<<4);
}
Answer:
fff0
Explanation :
-1 is internally represented as all 1's. When left shifted four times the
least significant 4 bits are filled with 0's.The %x format specifier specifies that the
integer value be printed as a hexadecimal value.
--------------------------------------------------------------------------------------------------------------------------------------