C program

Question

why this function not do swapping

#include <stdio.h>

void swap(char *str1, char *str2) {
    char *temp = str1;
    str1 = str2;
    str2 = temp;
}

int main() {
    char *str1 = "Gate";
    char *str2 = "Overflow";
    swap(str1, str2);
    printf("str1 is %s, str2 is %s", str1, str2);
    return 0;
}

Output :

str1 is Gate, str2 is Overflow

Answer 1

In this function whatever the exchange you are doing, is local only since str1, str2 and temp char pointers are local to swap() function.That is the reason your swap function is not working.

Comments

No call by reference in C. It can be simulated as follows:

#include<stdio.h>
void swap(char **str1, char **str2)
{
char *temp = *str1;
*str1 = *str2;
*str2 = temp;
}
int main()
{
char *str1 = "Gate";
char *str2 = "Overflow";
swap(&str1, &str2);
printf("str1 is %s, str2 is %s", str1, str2);
return 0;
}

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Sir.. even now the strings are not swapped..

https://ideone.com/qR0S23

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Sorry, the pointer usage was not correct. See now..