how/when is a static variable in a function declared within C language?

Question

My question is in the foo() function, the sa variable seems to be declared and initialized there, however since it is static is it ignored by the compiler after the first time? How come it is not initialized back to the value 10 even if it's static?

#include <stdio.h>

void foo()
{
int a = 10;
static int sa = 10;

a += 5;
sa += 5;

printf("a = %d, sa = %d\n", a, sa);
}


int main()
{
int i;

for (i = 0; i < 10; ++i)
    foo();
}

This prints:

a = 15, sa = 15
a = 15, sa = 20
a = 15, sa = 25
a = 15, sa = 30
a = 15, sa = 35
a = 15, sa = 40
a = 15, sa = 45
a = 15, sa = 50
a = 15, sa = 55
a = 15, sa = 60

Please reread what the static keyword means

Ed Heal
– Ed Heal

2015-11-05 20:52:27 +00:00
Commented Nov 5, 2015 at 20:52 — Ed Heal
– Ed Heal, Commented Nov 5, 2015 at 20:52

Scott Hunter · Accepted Answer · 2015-11-05 20:52:55Z

2

In C, static means that it persists between calls, so it is the opposite of what you suggest: the initialization only occurs on the first call.

answered Nov 5, 2015 at 20:52

Scott Hunter

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Comments

Sourav Ghosh · Accepted Answer · 2015-11-05 21:02:54Z

Quoting C11 standard, chapter §6.2.4, Storage durations of objects

An object whose identifier is declared without the storage-class specifier _Thread_local, and either with external or internal linkage or with the storage-class specifier static, has static storage duration. Its lifetime is the entire execution of the program and its stored value is initialized only once, prior to program startup.

So, sa is not being initialized on every call. It is initialized only once and holds the last stored value.

That said, regarding the retention of the value, quoting the same spec,

[..] An object exists, has a constant address, and retains its last-stored value throughout its lifetime. [...]

The reason of putting it inside a function is to limit the scope of the variable to that function scope itself.

Fiddling Bits · Accepted Answer · 2015-11-05 22:05:54Z

1

Your program would work identically if you declared sa globally, though its scope would be different:

int sa = 10;

void foo()
{
    int a = 10;

    a += 5;
    sa += 5;

    printf("a = %d, sa = %d\n", a, sa);
}

The reason you might want to declare sa within foo is to limit its access.

edited Nov 5, 2015 at 22:05

answered Nov 5, 2015 at 21:01

Fiddling Bits

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Comments

mksteve · Accepted Answer · 2015-11-05 21:05:17Z

0

static can declare a function which is only visible within the compilation unit.

This use of static variables within a function, a) reserves a piece of memory which will hold the single value (irrespective of threads), and b) Initialize it only once. In C++11 this is guaranteed to be thread safe, although I don't think there is such a guarantee in C.

So there is a single piece of memory reserved. Until the function is called, it will not be setup

answered Nov 5, 2015 at 21:05

mksteve

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Comments

John Bode · Accepted Answer · 2015-11-05 21:55:02Z

0

Objects with static storage duration are allocated and initialized once at program startup, and that storage is held until the program terminates.

Thus, the sa variable is allocated and initialized when the program starts, not when foo is first executed, and it persists between calls to foo.

The sa variable is only visible by name within the foo function, however.

answered Nov 5, 2015 at 21:55

John Bode

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