Why is the static storage behaving differently in the two examples?-CodePudding

So here is an example:

#include <stdio.h>

int main(void) {
    static int i=0;
    printf("%d",i);
    static int i=0;
    printf("%d",i);
    return 0;
}

This gives me an error:
error: redefinition of 'i'

Now here is another Example :

#include <stdio.h>
void increment(void) {
    static unsigned int counter = 0;
    counter  ;
    printf("%d ", counter);
}

int main(void) {
    for (int i = 0; i < 5; i  ) {
        increment();
    }
    return 0;
}

This gives the output :
1 2 3 4 5

Why does this happen ?
In the second example by calling the function aren't we redeclaring it? And shouldn't the output be 1 1 1 1 1 ?

CodePudding user response：

In the second example by calling the function aren't we redeclaring it?

No, we aren't: We are declaring i in a different scope; specifically, in the body of a different function. You can only define a variable once within a given scope (ignoring sub-scopes); but you can define a variable of the same name in different scopes:

int i; // global scope

void foo()
{
    int i; // scope - body of foo
    {
        int i; // a sub-scope of the body of foo
    }
}

int main()
{
    int i; // a different scope - body of main, not body of foo
}

and the "closest" definition to a command is the one which will be relevant; it will "shadow" the other variables of the same name which might otherwise be usable in that scope.

Of course, it's not a good idea to define many variables with the same name which may shadow each other - it's confusing and their names will likely not be meaningful. People often use a one-letter variable for a loop counter: i, j, k etc. - but then you would tend avoid using i for something more long-lasting.

1 2 3 4 5... Why does this happen ?

Because within the increment() function, the counter is a static unsigned int, meaning it has static storage duration - it is only initialized once, and maintains its value between invocations of the function.