Redeclaration of global variable vs local variable

Question

When I compile the code below

#include<stdio.h>

int main()
{
  int a;
  int a = 10;
  printf("a is %d \n",a);
  return 0;
}

I get an error:

test3.c: In function ‘main’:
test3.c:6:5: error: redeclaration of ‘a’ with no linkage
test3.c:5:5: note: previous declaration of ‘a’ was here

But if I make the variable global then it works fine.

#include<stdio.h>

int a;
int a = 10;
int main()
{
  printf("a is %d \n",a);
  return 0;
}

Why is declaring the same global variable twice not an error, but doing that for a local variable is an error?

Answer 1

The other reason I could think of is that un-initialized global variables are stored in the BSS (Block Structured Segment) where are the global variables that are initialized are stored in Data Segment.

I am guessing that there is some kind of a name space resolution and when there is a conflict the variable in the Data segment overrides the one in the Block Structured Segment.

if you were to declare

int a =5 int a = 10

in the global scope (both in the data segment) there would be conflict as expected.

Answer 2

In C, the statement int a; when made at file scope, is a declaration and a tentative definition. You can have as many tentative definitions as you want, as long as they all match each other.

If a definition (with an initializer) appears before the end of the translation unit, the variable will be initialized to that value. Having multiple initialization values is a compiler error.

If the end of the translation unit is reached, and no non-tentative definition was found, the variable will be zero initialized.

The above does not apply for local variables. Here a declaration also serves as a definition, and having more than one leads to an error.