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

Question

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 ?

Answer 1

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.