Redeclaration of global variable vs local variable

问题

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?

回答1:

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.

回答2:

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.

回答3:

You can't have two global variables with the same name in C program. C might allow multiple definitions in the same file scope through the tentative definition rule, but in any case all definitions will refer to the same variable.

Local Variable

In C, multiple local variables are not "merged" into one.

All the local variables with the same name will be referring to the different int-sized piece of memory.

 #include<stdio.h>

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

So when assigning the memory to the redeclaration of the same variable it gives an Error.

Global Variable

In C, multiple global variables are "merged" into one. So you have indeed just one global variable, declared multiple times. This goes back to a time when extern wasn't needed (or possibly didn't exist - not quite sure) in C.

In other words, all global variables with the same name will be converted to be one variable - so your

#include<stdio.h>

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

will be referring to the same int-sized piece of memory.

来源：https://stackoverflow.com/questions/21275992/redeclaration-of-global-variable-vs-local-variable