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?
ONE TWO TWO ONE
C passes every argument by value. Passing a char * copies the address, so swapping the copies inside a function does not affect the caller; passing the address of the pointer (char **) lets the function change the caller's pointers themselves.
Step 1 — f1(a, b): x and y are local copies of a and b. Swapping x and y leaves a = "ONE" and b = "TWO", so the first printf prints ONE TWO.
Step 2 — f2(&a, &b): x points to a and y points to b. *x = *y and *y = t1 exchange the actual pointers, so now a = "TWO" and b = "ONE"; the second printf prints TWO ONE.
Step 3 — The combined output is ONE TWO followed by TWO ONE (no separator is printed between the two calls).
TWO ONE ONE TWO and TWO ONE TWO ONE wrongly assume f1 swaps the caller's strings; ONE TWO ONE TWO wrongly assumes f2 also has no effect.
ONE TWO TWO ONE option is correct.
Similar Questions
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));
}
Which of the following is correct way to declare a functional pointer in C ?
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 ?
#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]) ( )) ( );
Important Questions from Pointer
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;
}
- The following statement in ‘C’
int (*f())[ ];
declares - What would be the equivalent pointer expression for referring the array element ar[m][n][o]
- Only legal pointer operations:
A. pointer + number $\rightarrow$ pointer
B. pointer - number $\rightarrow$ number
C. pointer + pointer $\rightarrow$ pointer
D. pointer - pointer $\rightarrow$ pointer
E. pointer - pointer $\rightarrow$ number
Choose the most appropriate answer from the options given below: 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.
