Double definition of variable

Question

As we know extern declaration of variables can be initialized.

#include<stdio.h>

extern int a=5; //will work as both definition and declaration.
int main()
{

}

Why this programme is compiling and running with no error.

extern int a=5; //will work as both definition and declaration.
int main()
{

}

int a; // isn't it second definition??

Probably has to do with http://stackoverflow.com/questions/3095861/about-tentative-definition — Kos, Apr 29 '13 at 17:30
Thanks @Kos, this thread is very informative. I didn't find it in my search.:( — user2333014, Apr 29 '13 at 17:39
Also related: http://stackoverflow.com/questions/1490693/tentative-definitions-in-c99-and-linking — Pascal Cuoq, Apr 29 '13 at 18:00

score 4 · Answer 1 · answered Apr 29 '13 at 17:30

C has a concept of a "tentative definition". Most definitions without initializers are "tentative", so an arbitrary number of them are allowed, as long as they don't conflict with each other. At the end of the translation unit, the definition of the variable becomes basically the composite of those (e.g., if one definition says the variable is const and another says it's volatile, the variable will end up as both const and volatile.

A definition with an initializer is never tentative, so only one of them is allowed.

Which declaration specifiers can differ between tentative definitions? cv-qualifiers, storage specifiers (extern, static?), what else? — Kos, Apr 29 '13 at 17:41

Double definition of variable

1 Answers1