Discussion
What is the output of following program?
#include <stdio.h>
void swap(char *str1, char *str2)
{
char *temp = str1;
str1 = str2;
str2 = temp;
}
int main()
{
char *str1 = "Geeks";
char *str2 = "Quiz";
swap(str1, str2);
printf("str1 is %s, str2 is %s", str1, str2);
return 0;
}
Answer: B
Correct answer: str1 is Geeks, str2 is Quiz
swap receives copies of the two pointers and exchanges the copies; the originals in main are untouched. To swap the caller’s pointers the function would have to take char ** and be called as swap(&str1, &str2).
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