- 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.
- 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.
- 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.
- 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.
- 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;
- 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' : legal
a="Hi" :
illegal
3. Same as 1.
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.
- 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.
- main(){
register int a=2;
printf("Address of a = %d",&;
printf("Value of a
= %d",;
}
Answer:
Compier Error: '&' on
register variable
Rule to Remember:
& (address of
) operator cannot be applied on register variables.
- 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.
- 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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