Static variable inside of a function in C

What will be printed out? 6 6 or 6 7? And why?

void foo()
{
    static int x = 5;
    x++;
    printf("%d", x);
}

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

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.

Vadiklk,

Why ...? Reason is that static variable is initialized only once, and maintains its value throughout the program. means, you can use static variable between function calls. also it can be used to count "how many times a function is called"

main()
{
   static int var = 5;
   printf("%d ",var--);
   if(var)
      main();
} 

and answer is 5 4 3 2 1 and not 5 5 5 5 5 5 .... (infinite loop) as you are expecting. again, reason is static variable is initialized once, when next time main() is called it will not be initialize to 5 because it is already initialized in the program.So we can change the value but can not reinitialized. Thats how static variable works.

or you can consider as per storage: static variables are stored on Data Section of a program and variables which are stored in Data Section are initialized once. and before initialization they are kept in BSS section.

In turn Auto(local) variables are stored on Stack and all the variables on stack reinitialized all time when function is called as new FAR(function activation record) is created for that.

okay for more understanding, do the above example without "static" and let you know what will be the output. That make you to understand the difference between these two.

Thanks Javed

The output will be 6 7. A static variable (whether inside a function or not) is initialized exactly once, before any function in that translation unit executes. After that, it retains its value until modified.

You will get 6 7 printed as, as is easily tested, and here's the reason: When foo is first called, the static variable x is initialized to 5. Then it is incremented to 6 and printed.

Now for the next call to foo. The program skips the static variable initialization, and instead uses the value 6 which was assigned to x the last time around. The execution proceeds as normal, giving you the value 7.

6 7

x is a global variable that is visible only from foo(). 5 is its initial value, as stored in the .data section of the code. Any subsequent modification overwrite previous value. There is no assignment code generated in the function body.

6 and 7 Because static variable intialise only once, So 5++ becomes 6 at 1st call 6++ becomes 7 at 2nd call Note-when 2nd call occurs it takes x value is 6 instead of 5 because x is static variable.

In C++11 at least, when the expression used to initialize a local static variable is not a 'constexpr' (cannot be evaluated by the compiler), then initialization must happen during the first call to the function. The simplest example is to directly use a parameter to intialize the local static variable. Thus the compiler must emit code to guess whether the call is the first one or not, which in turn requires a local boolean variable. I've compiled such example and checked this is true by seeing the assembly code. The example can be like this:

void f( int p )
{
  static const int first_p = p ;
  cout << "first p == " << p << endl ;
}

void main()
{
   f(1); f(2); f(3);
}

of course, when the expresion is 'constexpr', then this is not required and the variable can be initialized on program load by using a value stored by the compiler in the output assembly code.