Different answers for same code in c

Question

What happens in this code?

#include <stdio.h>

int main(){
    int e;
    printf("%d ", e);
    printf("%s", e);
    return 0;
}

Does e will have a) Garbage value? b) NULL

In GCC it shows garbage value and in g++ it shows 0 and NULL. Thanks!!

What do you think? That said, g++ is part of the GCC, so what is your question really? — Ulrich Eckhardt, Jan 25 '15 at 11:00
It could be [worse than dragons](http://stackoverflow.com/a/25636788/841108). However, if you compile with `gcc -Wall -Wextra -g` you'll be warned — Basile Starynkevitch, Jan 25 '15 at 11:13

score 5 · Answer 1 · answered Jan 25 '15 at 11:00

5

This program invokes undefined behavior. You are using wrong format specifier for int data type in second printf statement. Do not expect any good. Also note that e is not initialized and its value is indeterminate.

answered Jan 25 '15 at 11:00

haccks

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ouah · Accepted Answer · 2015-01-25T11:06:38.850

5

Both statements invoke undefined behavior because e object is not initialized so its value is indeterminate.

The second statement also has the wrong conversion specifier, %s specification requires a char * argument.

So as someone mentioned in the comments, the correct answer is neither a) nor b) but c) demons flying out your nose.

edited Jan 25 '15 at 11:06

answered Jan 25 '15 at 11:01

ouah

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score 0 · Answer 3 · answered Jan 25 '15 at 12:00

0

This program when compiled generated warning:

warning: format ‘%s’ expects argument of type ‘char’, but argument 2 has type ‘int’ [-Wformat=] printf("%s", e);*

And in both cases it generates garbage values.

gcc :

enter image description here

g++ :

enter image description here

answered Jan 25 '15 at 12:00

Pratyush Khare

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Different answers for same code in c

3 Answers3