C Aptitude Part VII

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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126)
int aaa() {printf(―Hi‖);}
int bbb(){printf(―hello‖);}
iny ccc(){printf(―bye‖);}
main()
{
int ( * ptr[3]) ();
ptr[0] = aaa;
54ptr[1] = bbb;
ptr[2] =ccc;
ptr[2]();
}
Answer:
bye
Explanation:
int (* ptr[3])() says that ptr is an array of pointers to functions that takes
no arguments and returns the type int. By the assignment ptr[0] = aaa; it
means that the first function pointer in the array is initialized with the
address of the function aaa. Similarly, the other two array elements also
get initialized with the addresses of the functions bbb and ccc. Since ptr[2]
contains the address of the function ccc, the call to the function ptr[2]() is
same as calling ccc(). So it results in printing "bye".
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127)
main()
{
int i=5;
printf(―%d‖,i=++i ==6);
}
Answer:
1
Explanation:
The expression can be treated as i = (++i==6), because == is of higher
precedence than = operator. In the inner expression, ++i is equal to 6
yielding true(1). Hence the result.
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128)
main()
{
char p[ ]="%d\n";
p[1] = 'c';
printf(p,65);
}
Answer:
A
Explanation:
Due to the assignment p[1] =  ̳c‘ the string becomes, ―%c\n‖. Since this
string becomes the format string for printf and ASCII value of 65 is  ̳A‘,
the same gets printed.
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129)
void ( * abc( int, void ( *def) () ) ) ();
Answer::
abc is a ptr to a function which takes 2 parameters .(a). an integer
variable.(b).
a ptrto a funtion which returns void. the return type of the
function is void.
Explanation:
Apply the clock-wise rule to find the result.
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130) main()
{
while (strcmp(―some‖,‖some\0‖))
printf(―Strings are not equal\n‖);
}
Answer:
No output
Explanation:
Ending the string constant with \0 explicitly makes no difference. So
―some‖ and ―some\0‖ are equivalent. So, strcmp returns 0 (false) hence
breaking out of the while loop.
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131) main()
{
char str1[] = { ̳s‘,‘o‘,‘m‘,‘e‘};
char str2[] = { ̳s‘,‘o‘,‘m‘,‘e‘,‘\0‘};
while (strcmp(str1,str2))
printf(―Strings are not equal\n‖);
}
Answer:
―Strings are not equal‖
―Strings are not equal‖
....
Explanation:
If a string constant is initialized explicitly with characters,  ̳\0‘ is not
appended automatically to the string. Since str1 doesn‘t have null
termination, it treats whatever the values that are in the following positions
as part of the string until it randomly reaches a  ̳\0‘. So str1 and str2 are
not the same, hence the result.
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132)
main()
{
int i = 3;
for (;i++=0;) printf(―%d‖,i);
}
Answer:
Compiler Error: Lvalue required.
Explanation:
As we know that increment operators return rvalues and hence it
cannot appear on the left hand side of an assignment operation.
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133) void main()
{
int *mptr, *cptr;
mptr = (int*)malloc(sizeof(int));
printf(―%d‖,*mptr);
int *cptr = (int*)calloc(sizeof(int),1);
printf(―%d‖,*cptr);
}
Answer:
garbage-value 0
Explanation:
The memory space allocated by malloc is uninitialized, whereas calloc
returns the allocated memory space initialized to zeros.
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134) void main()
{
static int i;
while(i<=10)
(i>2)?i++:i--;
printf(―%d‖, i);
}
Answer:
32767
Explanation:
Since i is static it is initialized to 0. Inside the while loop the conditional
operator evaluates to false, executing i--. This continues till the integer
value rotates to positive value (32767). The while condition becomes false
and hence, comes out of the while loop, printing the i value.
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135) main()
{
int i=10,j=20;
j = i, j?(i,j)?i:j:j;
printf("%d %d",i,j);
}
Answer:
10 10
Explanation:
The Ternary operator ( ? : ) is equivalent for if-then-else statement. So the
question can be written as:
if(i,j)
{
if(i,j)
j = i;
else
j = j;
}
else
j = j;
136)
1. const char *a;
2. char* const a;
3. char const *a;
-Differentiate the above declarations.
Answer:
1. 'const' applies to char * rather than 'a' ( pointer to a constant char )
*a='F'
: illegal
a="Hi"
: legal
2. 'const' applies to 'a' rather than to the value of a (constant pointer to
char )
*a='F'
a="Hi"
: legal
: illegal
3. Same as 1.
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137)
main()
{
int i=5,j=10;
i=i&=j&&10;
printf("%d %d",i,j);
}
Answer:
1 10
Explanation:
The expression can be written as i=(i&=(j&&10)); The inner expression
(j&&10) evaluates to 1 because j==10. i is 5. i = 5&1 is 1. Hence the
result.
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138)
main()
{
int i=4,j=7;
j = j || i++ && printf("YOU CAN");
printf("%d %d", i, j);
}
Answer:
4 1
Explanation:
The boolean expression needs to be evaluated only till the truth value of
the expression is not known. j is not equal to zero itself means that the
expression‘s truth value is 1. Because it is followed by || and true ||
(anything) => true where (anything) will not be evaluated. So the
remaining expression is not evaluated and so the value of i remains the
same.
Similarly when && operator is involved in an expression, when any of the
operands become false, the whole expression‘s truth value becomes false
and hence the remaining expression will not be evaluated.
false && (anything) => false where (anything) will not be evaluated.
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139)
main()
{
register int a=2;
printf("Address of a = %d",&a);
printf("Value of a = %d",a);
}
Answer:
Compier Error: '&' on register variable
Rule to Remember:
& (address of ) operator cannot be applied on register variables.
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140)
main()
{
float i=1.5;
switch(i)
{
case 1: printf("1");
case 2: printf("2");
default : printf("0");
}
}
Answer:
Compiler Error: switch expression not integral
Explanation:
Switch statements can be applied only to integral types.
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141)
main()
{
extern i;
printf("%d\n",i);
{
int i=20;
printf("%d\n",i);
}
}
Answer:
Linker Error : Unresolved external symbol i
Explanation:
The identifier i is available in the inner block and so using extern has no
use in resolving it.
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142)
main()
{
int a=2,*f1,*f2;
f1=f2=&a;
*f2+=*f2+=a+=2.5;
printf("\n%d %d %d",a,*f1,*f2);
}
Answer:
16 16 16
Explanation:
f1 and f2 both refer to the same memory location a. So changes through f1
and f2 ultimately affects only the value of a.
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143)
main()
{
char *p="GOOD";
char a[ ]="GOOD";
printf("\n sizeof(p) = %d, sizeof(*p) = %d, strlen(p) = %d", sizeof(p),
sizeof(*p), strlen(p));
printf("\n sizeof(a) = %d, strlen(a) = %d", sizeof(a), strlen(a));
}
Answer:
sizeof(p) = 2, sizeof(*p) = 1, strlen(p) = 4
sizeof(a) = 5, strlen(a) = 4
Explanation:
sizeof(p) => sizeof(char*) => 2
sizeof(*p) => sizeof(char) => 1
Similarly,
sizeof(a) => size of the character array => 5
When sizeof operator is applied to an array it returns the sizeof the array
and it is not the same as the sizeof the pointer variable. Here the sizeof(a)
where a is the character array and the size of the array is 5 because the
space necessary for the terminating NULL character should also be taken
into account.
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144)
#define DIM( array, type) sizeof(array)/sizeof(type)
main()
{
int arr[10];
printf(―The dimension of the array is %d‖, DIM(arr, int));
}
Answer:
10
Explanation:
The size of integer array of 10 elements is 10 * sizeof(int). The macro
expands to sizeof(arr)/sizeof(int) => 10 * sizeof(int) / sizeof(int) => 10.
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145) int DIM(int array[])
{
return sizeof(array)/sizeof(int );
}
main()
{
int arr[10];
printf(―The dimension of the array is %d‖, DIM(arr));
}
Answer:
1
Explanation:
Arrays cannot be passed to functions as arguments and only the pointers
can be passed. So the argument is equivalent to int * array (this is one of
the very few places where [] and * usage are equivalent). The return
statement becomes, sizeof(int *)/ sizeof(int) that happens to be equal in
this case.
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146) main()
{
static int a[3][3]={1,2,3,4,5,6,7,8,9};
int i,j;
static *p[]={a,a+1,a+2};
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
printf("%d\t%d\t%d\t%d\n",*(*(p+i)+j),
*(*(j+p)+i),*(*(i+p)+j),*(*(p+j)+i));
}
}
Answer:
1
2
3
4
1
4
7
2
1
2
3
4
1
4
7
2
615
6
7
8
9
5
8
3
6
9
5
6
7
8
9
5
8
3
6
9
Explanation:
*(*(p+i)+j) is equivalent to p[i][j].
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147)
main()
{
void swap();
int x=10,y=8;
swap(&x,&y);
printf("x=%d y=%d",x,y);
}
void swap(int *a, int *b)
{
*a ^= *b, *b ^= *a, *a ^= *b;
}
Answer:
x=10 y=8
Explanation:
Using ^ like this is a way to swap two variables without using a temporary
variable and that too in a single statement.
Inside main(), void swap(); means that swap is a function that may take
any number of arguments (not no arguments) and returns nothing. So this
doesn‘t issue a compiler error by the call swap(&x,&y); that has two
arguments.
This convention is historically due to pre-ANSI style (referred to as
Kernighan and Ritchie style) style of function declaration. In that style, the
swap function will be defined as follows,
void swap()
int *a, int *b
{
*a ^= *b, *b ^= *a, *a ^= *b;
}
where the arguments follow the (). So naturally the declaration for swap
will look like, void swap() which means the swap can take any number of
arguments.
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148)
main()
{
int i = 257;
int *iPtr = &i;
printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer:
1 1
Explanation:
The integer value 257 is stored in the memory as, 00000001 00000001, so
the individual bytes are taken by casting it to char * and get printed.
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149)
main()
{
int i = 258;
int *iPtr = &i;
printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer:
2 1
Explanation:
The integer value 257 can be represented in binary as, 00000001
00000001. Remember that the INTEL machines are  ̳small-endian‘
machines. Small-endian means that the lower order bytes are stored in the
higher memory addresses and the higher order bytes are stored in lower
addresses. The integer value 258 is stored in memory as: 00000001
00000010.
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150)
main()
{
int i=300;
char *ptr = &i;
*++ptr=2;
printf("%d",i);
}
Answer:
556
Explanation:
The integer value 300 in binary notation is: 00000001 00101100. It is
stored in memory (small-endian) as: 00101100 00000001. Result of the
expression *++ptr = 2 makes the memory representation as: 00101100
00000010. So the integer corresponding to it is 00000010 00101100 =>
556.
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151)
#include <stdio.h>
main()
{
char * str = "hello";
char * ptr = str;
char least = 127;
while (*ptr++)
least = (*ptr<least ) ?*ptr :least;
printf("%d",least);
}
Answer:
0
Explanation:
After  ̳ptr‘ reaches the end of the string the value pointed by  ̳str‘ is  ̳\0‘.
So the value of  ̳str‘ is less than that of  ̳least‘. So the value of  ̳least‘
finally is 0.
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152) Declare an array of N pointers to functions returning pointers to functions
returning pointers to characters?
Answer:
(char*(*)( )) (*ptr[N])( );
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153) main()
{
struct student
{
char name[30];
struct date dob;
}stud;
struct date
{
int day,month,year;
};
scanf("%s%d%d%d",
stud.rollno,
&student.dob.month,
&student.dob.year);
&student.dob.day,
}
Answer:
Compiler Error: Undefined structure date
Explanation:
Inside the struct definition of  ̳student‘ the member of type struct date is
given. The compiler doesn‘t have the definition of date structure (forward
reference is not allowed in C in this case) so it issues an error.
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154)
main()
{
struct date;
struct student
{
char name[30];
struct date dob;
}stud;
struct date
{
int day,month,year;};
scanf("%s%d%d%d", stud.rollno, &student.dob.day, &student.dob.month,
&student.dob.year);
}
Answer:
Compiler Error: Undefined structure date
Explanation:
Only declaration of struct date is available inside the structure definition
of  ̳student‘ but to have a variable of type struct date the definition of the
structure is required.
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155) There were 10 records stored in ―somefile.dat‖ but the following program printed 11 names. What went wrong?
void main()
{
struct student
{
char name[30], rollno[6];
}stud;
FILE *fp = fopen(―somefile.dat‖,‖r‖);
while(!feof(fp))
{
fread(&stud, sizeof(stud), 1 , fp);
puts(stud.name);
}
}
Explanation:
fread reads 10 records and prints the names successfully. It will
return EOF only when fread tries to read another record and fails
reading EOF (and returning EOF). So it prints the last record
again. After this only the condition feof(fp) becomes false, hence
comes out of the while loop.
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156) Is there any difference between the two declarations,
1. int foo(int *arr[]) and
2. int foo(int *arr[2])
Answer:
No
Explanation:
Functions can only pass pointers and not arrays. The numbers that are
allowed inside the [] is just for more readability. So there is no difference
between the two declarations.
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157) What is the subtle error in the following code segment?
void fun(int n, int arr[])
{
int *p=0;
int i=0;
while(i++<n)
p = &arr[i];
*p = 0;
}
Answer & Explanation:
If the body of the loop never executes p is assigned no address. So
p remains NULL where *p =0 may result in problem (may rise to
runtime error ―NULL pointer assignment‖ and terminate the
program).
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158) What is wrong with the following code?
int *foo()
{
int *s = malloc(sizeof(int)100);
assert(s != NULL);
return s;
}
Answer & Explanation:
assert macro should be used for debugging and finding out bugs. The
check s != NULL is for error/exception handling and for that assert
shouldn‘t be used. A plain if and the corresponding remedy statement has
to be given.
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159) What is the hidden bug with the following statement?
assert(val++ != 0);
Answer & Explanation:
Assert macro is used for debugging and removed in release version. In
assert, the experssion involves side-effects. So the behavior of the code
becomes different in case of debug version and the release version thus
leading to a subtle bug.
Rule to Remember:
Don‟t use expressions that have side-effects in assert statements.
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160) void main()
{
int *i = 0x400; // i points to the address 400
*i = 0;
// set the value of memory location pointed by i;
}
Answer:
Undefined behavior
Explanation:
The second statement results in undefined behavior because it points to
some location whose value may not be available for modification. This
66type of pointer in which the non-availability of the implementation of the
referenced location is known as 'incomplete type'.
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C Aptitude Part VI

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:
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 91) In the following pgm add a stmt in the function fun such that the address of
'a' gets stored in 'j'.
main(){
int * j;
void fun(int **);
fun(&j);
}
void fun(int **k) {
int a =0;
/* add a stmt here*/
}
Answer:
*k = &a
Explanation:
The argument of the function is a pointer to a pointer.
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92) What are the following notations of defining functions known as?
i.
int abc(int a,float b)
{
/* some code */
}
ii. int abc(a,b)
int a; float b;
{
/* some code*/
}
Answer:
i. ANSI C notation
ii. Kernighan & Ritche notation
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93) main()
{
char *p;
p="%d\n";
p++;
p++;
printf(p-2,300);
}
Answer:
300
Explanation:
The pointer points to % since it is incremented twice and again
decremented by 2, it points to '%d\n' and 300 is printed.
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94) main(){
char a[100];
a[0]='a';a[1]]='b';a[2]='c';a[4]='d';
abc(a);
}
abc(char a[]){
a++;
printf("%c",*a);
a++;
printf("%c",*a);
}
Explanation:
The base address is modified only in function and as a result a points to 'b'
then after incrementing to 'c' so bc will be printed.
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95) func(a,b)
int a,b;
{
return( a= (a==b) );
}
main()
{
int process(),func();
printf("The value of process is %d !\n ",process(func,3,6));
}
process(pf,val1,val2)
int (*pf) ();
int val1,val2;
{
return((*pf) (val1,val2));
}
Answer:
The value if process is 0 !
Explanation:
The function 'process' has 3 parameters - 1, a pointer to another function 2
and 3, integers. When this function is invoked from main, the following
substitutions for formal parameters take place: func for pf, 3 for val1 and 6
for val2. This function returns the result of the operation performed by the
function 'func'. The function func has two integer parameters. The formal
parameters are substituted as 3 for a and 6 for b. since 3 is not equal to 6,
a==b returns 0. therefore the function returns 0 which in turn is returned
by the function 'process'.
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96) void main()
{
static int i=5;
if(--i){
main();
printf("%d ",i);
}
}
Answer:
0 0 0 0
Explanation:
The variable "I" is declared as static, hence memory for I will be allocated
for only once, as it encounters the statement. The function main() will be called
recursively unless I becomes equal to 0, and since main() is recursively called, so
the value of static I ie., 0 will be printed every time the control is returned.
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97) void main()
{
int k=ret(sizeof(float));
printf("\n here value is %d",++k);
}
int ret(int ret)
{
ret += 2.5;
return(ret);
}
Answer:
Here value is 7
Explanation:
The int ret(int ret), ie., the function name and the argument name can be
the same.
Firstly, the function ret() is called in which the sizeof(float) ie., 4 is
passed, after the first expression the value in ret will be 6, as ret is integer hence
the value stored in ret will have implicit type conversion from float to int. The ret
is returned in main() it is printed after and preincrement.
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98) void main()
{
char a[]="12345\0";
int i=strlen(a);
printf("here in 3 %d\n",++i);
}
Answer:
here in 3 6
Explanation:
The char array 'a' will hold the initialized string, whose length will be
counted from 0 till the null character. Hence the 'I' will hold the value equal to 5,
after the pre-increment in the printf statement, the 6 will be printed.
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99) void main()
{
unsigned giveit=-1;
int gotit;
printf("%u ",++giveit);
printf("%u \n",gotit=--giveit);
}
Answer:
0 65535
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100) void main()
{
int i;
char a[]="\0";
if(printf("%s\n",a))
printf("Ok here \n");
else
printf("Forget it\n");
}
Answer:
Ok here
Explanation:
Printf will return how many characters does it print. Hence printing
a null character returns 1 which makes the if statement true, thus
"Ok here" is printed.
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101) void main()
{
void *v;
int integer=2;
int *i=&integer;
v=i;
printf("%d",(int*)*v);
}
Answer:
Compiler Error. We cannot apply indirection on type void*.
Explanation:
Void pointer is a generic pointer type. No pointer arithmetic can be
done on it. Void pointers are normally used for,
1. Passing generic pointers to functions and returning such pointers.
2. As a intermediate pointer type.
3. Used when the exact pointer type will be known at a later point of
time.
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102) void main()
{
int i=i++,j=j++,k=k++;
printf(―%d%d%d‖,i,j,k);
}
Answer:
Garbage values.
Explanation:
An identifier is available to use in program code from the point of its
declaration.
So expressions such as i = i++ are valid statements. The i, j and k are
automatic variables and so they contain some garbage value. Garbage in is
garbage out (GIGO).
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103) void main()
{
static int i=i++, j=j++, k=k++;
printf(―i = %d j = %d k = %d‖, i, j, k);
}
Answer:
i = 1 j = 1 k = 1
Explanation:
Since static variables are initialized to zero by default.
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104) void main()
{
while(1){
if(printf("%d",printf("%d")))
break;
else
continue;
}
}
Answer:
Garbage values
Explanation:
The inner printf executes first to print some garbage value. The printf
returns no of characters printed and this value also cannot be predicted.
Still the outer printf prints something and so returns a non-zero value. So
it encounters the break statement and comes out of the while statement.
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104)
main()
{
unsigned int i=10;
while(i-->=0)
printf("%u ",i);
}
Answer:
10 9 8 7 6 5 4 3 2 1 0 65535 65534.....
Explanation:
Since i is an unsigned integer it can never become negative. So the
expression i-- >=0 will always be true, leading to an infinite loop.
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105) #include<conio.h>
main()
{
int x,y=2,z,a;
if(x=y%2) z=2;
a=2;
printf("%d %d ",z,x);
}
Answer:
Garbage-value 0
Explanation:
The value of y%2 is 0. This value is assigned to x. The condition reduces
to if (x) or in other words if(0) and so z goes uninitialized.
Thumb Rule: Check all control paths to write bug free code.
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106) main()
{
int a[10];
printf("%d",*a+1-*a+3);
}
Answer:
4
Explanation:
*a and -*a cancels out. The result is as simple as 1 + 3 = 4 !
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107) #define prod(a,b) a*b
main()
{
int x=3,y=4;
printf("%d",prod(x+2,y-1));
}
Answer:
10
Explanation:
The macro expands and evaluates to as:
x+2*y-1 => x+(2*y)-1 => 10
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108) main()
{
unsigned int i=65000;
while(i++!=0);
printf("%d",i);
}
Answer:
1
Explanation:
Note the semicolon after the while statement. When the value of i
becomes 0 it comes out of while loop. Due to post-increment on i the
value of i while printing is 1.
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109)
main()
{
int i=0;
while(+(+i--)!=0)
i-=i++;
printf("%d",i);
}
Answer:
-1
Explanation:
Unary + is the only dummy operator in C. So it has no effect on the
expression and now the while loop is,
while(i--!=0) which is false
and so breaks out of while loop. The value –1 is printed due to the post-
decrement operator.
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113)
main()
{
float f=5,g=10;
enum{i=10,j=20,k=50};
printf("%d\n",++k);
printf("%f\n",f<<2);
printf("%lf\n",f%g);
printf("%lf\n",fmod(f,g));
}
Answer:
Line no 5: Error: Lvalue required
Line no 6: Cannot apply leftshift to float
Line no 7: Cannot apply mod to float
Explanation:
Enumeration constants cannot be modified, so you cannot apply ++.
Bit-wise operators and % operators cannot be applied on float values.
fmod() is to find the modulus values for floats as % operator is for ints.
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110)
main()
{
int i=10;
void pascal f(int,int,int);
f(i++,i++,i++);
printf(" %d",i);
}
void pascal f(integer :i,integer:j,integer :k)
{
write(i,j,k);
}
Answer:
Compiler error: unknown type integer
Compiler error: undeclared function write
Explanation:
Pascal keyword doesn‘t mean that pascal code can be used. It means that
the function follows Pascal argument passing mechanism in calling the functions.
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111)
void pascal f(int i,int j,int k)
{
printf(―%d %d %d‖,i, j, k);
}
void cdecl f(int i,int j,int k)
{
printf(―%d %d %d‖,i, j, k);
}
main()
48{
int i=10;
f(i++,i++,i++);
printf(" %d\n",i);
i=10;
f(i++,i++,i++);
printf(" %d",i);
}
Answer:
10 11 12 13
12 11 10 13
Explanation:
Pascal argument passing mechanism forces the arguments to be called
from left to right. cdecl is the normal C argument passing mechanism where the
arguments are passed from right to left.
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112)What is the output of the program given below
main()
{
signed char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer
-128
Explanation
Notice the semicolon at the end of the for loop. THe initial value of
the i is set to 0. The inner loop executes to increment the value
from 0 to 127 (the positive range of char) and then it rotates to the
negative value of -128. The condition in the for loop fails and so
comes out of the for loop. It prints the current value of i that is -128.
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113) main()
{
unsigned char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer
infinite loop
Explanation
The difference between the previous question and this one is that the char
is declared to be unsigned. So the i++ can never yield negative value and i>=0
never becomes false so that it can come out of the for loop.
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114) main()
{
char i=0;
for(;i>=0;i++) ;
printf("%d\n",i);
}
Answer:
Behavior is implementation dependent.
Explanation:
The detail if the char is signed/unsigned by default is
implementation dependent. If the implementation treats the char to be
signed by default the program will print –128 and terminate. On the other
hand if it considers char to be unsigned by default, it goes to infinite loop.
Rule:
You can write programs that have implementation dependent
behavior. But dont write programs that depend on such behavior.
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115) Is the following statement a declaration/definition. Find what does it mean?
int (*x)[10];
Answer
Definition.
x is a pointer to array of(size 10) integers.
Apply clock-wise rule to find the meaning of this definition.
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116). What is the output for the program given below
typedef enum errorType{warning, error, exception,}error;
main()
{
error g1;
g1=1;
printf("%d",g1);
}
Answer
Compiler error: Multiple declaration for error
Explanation
The name error is used in the two meanings. One means that it is a
enumerator constant with value 1. The another use is that it is a type name
(due to typedef) for enum errorType. Given a situation the compiler
cannot distinguish the meaning of error to know in what sense the error is
used:
error g1;
g1=error;
// which error it refers in each case?
When the compiler can distinguish between usages then it will not
issue error (in pure technical terms, names can only be overloaded in
different namespaces).
Note: the extra comma in the declaration,
enum errorType{warning, error, exception,}
is not an error. An extra comma is valid and is provided just for
programmer‘s convenience.
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117)
typedef struct error{int warning, error, exception;}error;
main()
{
error g1;
g1.error =1;
printf("%d",g1.error);
}
Answer
1
Explanation
The three usages of name errors can be distinguishable by the compiler at
any instance, so valid (they are in different namespaces).
Typedef struct error{int warning, error, exception;}error;
This error can be used only by preceding the error by struct kayword as in:
struct error someError;
typedef struct error{int warning, error, exception;}error;
This can be used only after . (dot) or -> (arrow) operator preceded by the variable
name as in :
g1.error =1;
printf("%d",g1.error);
typedef struct error{int warning, error, exception;}error;
This can be used to define variables without using the preceding struct keyword
as in:
error g1;
Since the compiler can perfectly distinguish between these three usages, it is
perfectly legal and valid.
Note
This code is given here to just explain the concept behind. In real
programming don‘t use such overloading of names. It reduces the readability of
the code. Possible doesn‘t mean that we should use it!
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118)
#ifdef something
int some=0;
#endif
main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}
Answer:
Compiler error : undefined symbol some
Explanation:
This is a very simple example for conditional compilation. The
name something is not already known to the compiler making the
declaration
int some = 0;
effectively removed from the source code.
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119)
#if something == 0
int some=0;
#endif
main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}
Answer
0 0
Explanation
This code is to show that preprocessor expressions are not the
same as the ordinary expressions. If a name is not known the
preprocessor treats it to be equal to zero.
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120). What is the output for the following program
main()
{
int arr2D[3][3];
printf("%d\n", ((arr2D==* arr2D)&&(* arr2D == arr2D[0])) );
}
Answer
1
Explanation
This is due to the close relation between the arrays and pointers. N
dimensional arrays are made up of (N-1) dimensional arrays.
52arr2D is made up of a 3 single arrays that contains 3 integers each .
arr2D
arr2D[1]
arr2D[2]
arr2D[3]
The name arr2D refers to the beginning of all the 3 arrays. *arr2D
refers to the start of the first 1D array (of 3 integers) that is the
same address as arr2D. So the expression (arr2D == *arr2D) is true
(1).
Similarly, *arr2D is nothing but *(arr2D + 0), adding a zero
doesn‘t change the value/meaning. Again arr2D[0] is the another
way of telling *(arr2D + 0). So the expression (*(arr2D + 0) ==
arr2D[0]) is true (1).
Since both parts of the expression evaluates to true the result is
true(1) and the same is printed.
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121) void main()
{
if(~0 == (unsigned int)-1)
printf(―You can answer this if you know how values are represented in
memory‖);
}
Answer
You can answer this if you know how values are represented in
memory
Explanation
~ (tilde operator or bit-wise negation operator) operates on 0 to
produce all ones to fill the space for an integer. –1 is represented in
unsigned value as all 1‘s and so both are equal.
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122) int swap(int *a,int *b)
{
*a=*a+*b;*b=*a-*b;*a=*a-*b;
}
main()
{
int x=10,y=20;
swap(&x,&y);
printf("x= %d y = %d\n",x,y);
}
Answer
x = 20 y = 10
Explanation
This is one way of swapping two values. Simple checking will help
understand this.
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123) main()
{
char *p = ―ayqm‖;
printf(―%c‖,++*(p++));
}
Answer:
b
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124) main()
{
int i=5;
printf("%d",++i++);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
++i yields an rvalue. For postfix ++ to operate an lvalue is
required.
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125)
main()
{
char *p = ―ayqm‖;
char c;
c = ++*p++;
printf(―%c‖,c);
}
Answer:
b
Explanation:
There is no difference between the expression ++*(p++) and
++*p++. Parenthesis just works as a visual clue for the reader to
see which expression is first evaluated.
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