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).
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Predict the output or error(s) for the following:
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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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