CODEWITHSUNDEEP
c
Armia Wagdy
Armia Wagdy

Reputation: 599

What happens if i declared another variable with the same name in another file?

I have declared int x in file one and by mistake I declared another variable of type char with the same name x in file two, and I wait the compiler or the linker to give me an error, but there are no errors displayed. and when I use the debugger I see that int x is converted to char x , is that true?! and what actually happens here?!

Show this modification on my code:

File One

#include <stdio.h>

int x = 50;  /** declare global variable called
                           x **/
int main()
{
    print();
    printf("      global in file one = %d",x);  /** Modification is just here **/
    return 0;
}

File Two

char x;   

void print(void)
{

    x = 100;
    printf("global in file two = %d ",x);
    return;

}

My expected results are = global in file two = 100 global in file one = 50

but The results are : global in file two = 100 global in file one = 100

When I use the debugger I see that int x is converted to char x , is that true?! and what actually happens here?

Upvotes: 5

Views: 155

Answers (1)

Carl Norum
Carl Norum

Reputation: 224884

You're in troublesome territory here. Technically your program causes undefined behaviour. char x is a tentative definition, since it doesn't have an initializer. That means the linker unifies it with the int x in the other file at link time. It's a bit weird looking, since your two declarations have different types, but it appears to have successfully linked in your case. Anyway, you have only one x, and luck is making it work the way you're seeing it (and little-endian architecture, probably).

If you want the two variables to be independent, make them static, and they'll be restricted to the scope of their respective translation units.

Upvotes: 6

Related Questions