GATE CSE 2018 | Question: 29

Question

#include<stdio.h>
void fun1(char* s1, char* s2){
    char* temp;
    temp = s1;
    s1 = s2;
    s2 = temp;
}
void fun2(char** s1, char** s2){
    char* temp;
    temp = *s1;
    *s1 = *s2;
    *s2 = temp;
}
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:

• A. $\text{Hi Bye Bye Hi}$
• B. $\text{Hi Bye Hi Bye}$
• C. $\text{Bye Hi Hi Bye}$
• D. $\text{Bye Hi Bye Hi}$

Comments

@arjun sir why? , i do not know the reason :(

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Call by value and reference, a visualization of the problem

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https://www.youtube.com/watch?v=d6BriTtMB6s

Answer 1

func1(char* s1, char* s2){
    char* temp;
    temp = s1;
    s1 = s2;
    s2 = temp;
}

Everything is local here. So, once function completes its execution all modification go in vain.

func2(char** s1, char** s2){
    char* temp
    temp = *s1
    *s1 = *s2
    *s2 = temp
}

This will retain modification and swap pointers of string.
So output would be Hi Bye Bye Hi

Correct Answer: $A$

Comments

Thank you so much

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void fun2(char* s1, char* s2)

change as above. still getting same output.

https://ideone.com/WOdFCN

Also run this code in local CodeBlocks it gives warning(no error). But same output.

“incompatible pointer type” and cast warning.

@ShaikMasthan

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replied at my answer

Answer 2

The first call to the function 'func1(str1,str2);' is Call by Value. Hence, any change in the formal parameters are NOT reflected in actual parameters. Hence, str1 points at "hi" and str2 points at "bye".

The second call to the function 'func2(&str1,&str2);' is Call by Reference. Hence, any change in the formal parameters are reflected in actual parameters. Hence, str1 now points at "bye" and str2 points at "hi".

Hence, answer is hi byebye hi ....

Comments

Both are call by value only just that in second case the value is a pointer to pointer and using ** the pointed value gets changed.

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I understand Arjun sir that C doesn't have Call by Reference unlike C++, but it is a way to implement call by reference like mechanism in C, hence mentioned it likewise and thanks for noting this sir :)

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Sir, so will the answer: be "hi bye bye hi" ? Someone posted a screenshot of some compiler giving output as
"hi bye hi bye"

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@Chandrashis Mazumd 1  no ThatsWrong 

Answer 3

hi bye bye hi

Answer 4

The first call to the function ‘func1(str1, str2);’ is call by value.
Hence, any change in the formal parameters are NOT reflected in actual parameters.
Hence, str1 points at “hi” and str2 points at “bye”.
The second call to the function ‘func2(&str1, &str2);’ is call by reference.
Hence, any change in formal parameters are reflected in actual parameters.
Hence, str1 now points at “bye” and str2 points at “hi”.

Hence answer is “hi bye bye hi”.

Answer 5

The correct answer is (A).

i.e. Hi Bye Bye Hi

In short, the fun1 function attempts to swap the pointers s1 and s2, but since C uses pass-by-value, the swap is local to fun1 and doesn't affect the pointers in main. Thus, it performs a "fake swap" and has no effect on the output.

In the second function fun2, pointers to pointers are used as arguments (char**). This allows the function to modify the original pointers str1 and str2 from main by dereferencing them. Therefore, when fun2 swaps the values pointed to by s1 and s2, it directly affects str1 and str2 in main. This results in a successful swap of the strings "Hi" and "Bye", which is reflected in the output.

Answer 6

Str1 and Str2 are Constant Pointers
and Base Address of the Constant Pointers Cannot get Modified
Str1 and Str2 will remains the same