Static variable inside of a function in C

Question

There are two issues here, lifetime and scope.

The scope of variable is where the variable name can be seen. Here, x is visible only inside function foo().

The lifetime of a variable is the period over which it exists. If x were defined without the keyword static, the lifetime would be from the entry into foo() to the return from foo(); so it would be re-initialized to 5 on every call.

The keyword static acts to extend the lifetime of a variable to the lifetime of the programme; e.g. initialization occurs once and once only and then the variable retains its value - whatever it has come to be - over all future calls to foo().

Output: 6 7

Reason: static variable is initialised only once (unlike auto variable) and further definition of static variable would be bypassed during runtime. And if it is not initialised manually, it is initialised by value 0 automatically. So,

void foo() {
    static int x = 5; // assigns value of 5 only once
    x++;
    printf("%d", x);
}

int main() {
    foo(); // x = 6
    foo(); // x = 7
    return 0;
}

That is the same as having the following program:

static int x = 5;

void foo()
{
    x++;
    printf("%d", x);
}

int main()
{
     foo();
     foo();
     return 0;
}

All that the static keyword does in that program is it tells the compiler (essentially) 'hey, I have a variable here that I don't want anyone else accessing, don't tell anyone else it exists'.

Inside a method, the static keyword tells the compiler the same as above, but also, 'don't tell anyone that this exists outside of this function, it should only be accessible inside this function'.

I hope this helps

6 7

compiler arranges that static variable initialization does not happen each time the function is entered

Output: 6,7

Reason

The declaration of x is inside foo but the x=5 initialization takes place outside of foo!

What we need to understand here is that

static int x = 5;

is not the same as

static int x;
x = 5;

Other answers have used the important words here, scope and lifetime, and pointed out that the scope of x is from the point of its declaration in the function foo to the end of the function foo. For example I checked by moving the declaration to the end of the function, and that makes x undeclared at the x++; statement.

So the static int x (scope) part of the statement actually applies where you read it, somewhere INSIDE the function and only from there onwards, not above it inside the function.

However the x = 5 (lifetime) part of the statement is initialization of the variable and happening OUTSIDE of the function as part of the program loading. Variable x is born with a value of 5 when the program loads.

I read this in one of the comments: "Also, this doesn't address the really confusing part, which is the fact that the initializer is skipped on subsequent calls." It is skipped on all calls. Initialization of the variable is outside of the function code proper.

The value of 5 is theoretically set regardless of whether or not foo is called at all, although a compiler might optimize the function away if you don't call it anywhere. The value of 5 should be in the variable before foo is ever called.

Inside of foo, the statement static int x = 5; is unlikely to be generating any code at all.

I found the address x uses when I put a function foo into a program of mine, and then (correctly) guessed that the same location would be used if I ran the program again. The partial screen capture below shows that x has the value 5 even before the first call to foo.

A static variable inside a function has a lifespan as long as your program runs. It won't be allocated every time your function is called and deallocated when your function returns.

Let's just read the Wikipedia article on Static Variables...

Static local variables: variables declared as static inside a function are statically allocated while having the same scope as automatic local variables. Hence whatever values the function puts into its static local variables during one call will still be present when the function is called again.