GATE CS 2018 Q.39 (CS Q.29): Pointers, Arrays and Strings

Consider the following C program:

#include<stdio.h> 
 
void fun1(char *s1, char *s2){ 
 char *tmp; 
 tmp = s1; 
 s1 = s2; 
 s2 = tmp; 
} 
void fun2(char **s1, char **s2){ 
 char *tmp; 
 tmp = *s1; 
 *s1 = *s2; 
 *s2 = tmp; 
} 
int main(){ 
 char *str1 = "Hi", *str2 = "Bye"; 
 fun1(str1, str2);   printf("%s %s ", str1, str2); 
 fun2(&str1, &str2); printf("%s %s", str1, str2); 
 return 0; 
}

The output of the program above is

Hi Bye Bye Hi

Hi Bye Hi Bye

Bye Hi Hi Bye

Bye Hi Bye Hi

Official answer and solution

Official answer: A
From: Official GATE 2018 answer key

Solution

Contrast pass-by-value with pass-by-reference when manipulating pointer variables in C.

Trace fun1(str1, str2):

str1 and str2 are pointer values passed by value. fun1 receives copies of these pointers in local parameters s1 and s2.
The swap operation inside fun1 only modifies the local copies s1 and s2, leaving the original str1 and str2 in main unaffected.
Thus, printf("%s %s ", str1, str2) prints: Hi Bye .

Trace fun2(&str1, &str2):

The addresses &str1 and &str2 are passed to fun2, which receives them as char **.
Inside fun2, *s1 and *s2 directly dereference the variables str1 and str2 in main, swapping the pointers themselves.
After fun2, str1 points to "Bye" and str2 points to "Hi".
Thus, printf("%s %s", str1, str2) prints: Bye Hi.

Combining both outputs gives Hi Bye Bye Hi.

Final answer

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2018 Computer Science and Information Technology (CS)
Question: 39 of 65, printed as CS Q.29
Section: Computer Science and Information Technology

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