Which of the following is correct way to declare a functional pointer in C ?
int (*func) (int, int);
A pointer to a function is declared by putting the * and the name inside parentheses, followed by the parameter list: return_type (*name)(parameter types). The parentheses are needed because the function-call operator () binds more tightly than the * declarator.
Step 1 — int (*func) (int, int); declares func as a pointer to a function that takes two int arguments and returns an int; it can be assigned, for example, func = add; and called as func(2, 3).
Step 2 — int *func (int, int); without parentheses declares a function named func that returns a pointer to int, not a pointer to a function.
Step 3 — int (func*) (int, int); and int *func* (int, int); place the * after the identifier, which is not valid declarator syntax in C and does not compile.
int (*func) (int, int); option is correct.
What is the output of the following program ?
(Assume that the appropriate preprocessor directives are included and there is no syntax error)
main ( )
{ char S[ ]="ABCDEFGH";
printf ("%C", * (& S[3]));
printf ("%s", S+4);
printf ("%u", S);
/* Base address of S is 1000 */
}
Which of the following statements about pointers in C are TRUE.
(A) Pointers can be used to access array elements
(B) Pointers can store the address of another pointer
(C) Pointers are automatically deferenced in expression
(D) Pointers cannot be used to access structure members
Choose the correct answer from the options given below :
What will be the output of the following C code?
#include<stdio.h>
void main()
{
int arr[5] = {10, 20, 30, 40, 50};
int *p = (int*) (&arr + 1);
printf("%d%d", *(arr + 1), *(p - 1));
}
Consider following C program :
#include<stdio.h>
int main()
{
int x[]={2, 4, 6, 8, 10};
int a, b=0, *y=x+4;
for(a=0; a<5; a++)
{
b=b+(*y-a)-*(y-a);
}
printf("%d\n", b);
return 0;
}
What will be the output of the above C program ?
Consider the following code :
#include<stdio.h>
void f1(char *x, char *y) {
char *t1; t1=x; x=y; y=t1;}
void f2(char **x, char **y) {
char *t1; t1=*x; *x=*y; *y=t1;}
int main()
{
char *a="ONE", *b="TWO";
f1(a, b); printf("%s %s", a, b);
f2(&a, &b); printf("%s %s", a, b);
return 0;
}
What will be the output of the above code?
#include<stdio.h>
void main()
{
int arr[]={1, 2, 3, 4, 5};
int *p=arr;
printf("%d", *p++);
printf("%d", *(p+1));
}
Find the output of the above code?
What does the following expression means ?
char *(*(* a[N]) ( )) ( );
What is the output of the following C programming language code?
#include<stdio.h>
int main()
{
int x = 10;
int *p = &x;
*p = *p + 5;
printf("%d %d", x, *p);
return 0;
}
Consider the following ANSI-C program.
#include <stdio.h>
int main(){
int *ptr, a, b, c;
a=5; b=11; c=20;
ptr=&a; *ptr=c; ptr=&c;
a=*(&b); c=*ptr-a;
printf("%d",c);
return(0);
}
The output of this program is ____________. (answer in integer)
Note: Assume that the program compiles and runs successfully.