When do function-level static variables get allocated/initialized?

Question

I'm quite confident that globally declared variables get allocated (and initialized, if applicable) at program start time.

int globalgarbage;
unsigned int anumber = 42;

But what about static ones defined within a function?

void doSomething()
{
  static bool globalish = true;
  // ...
}

When is the space for globalish allocated? I'm guessing when the program starts. But does it get initialized then too? Or is it initialized when doSomething() is first called?

Answer 1

Static variables are allocated inside a code segment -- they are part of the executable image, and so are mapped in already initialized.

Static variables within function scope are treated the same, the scoping is purely a language level construct.

For this reason you are guaranteed that a static variable will be initialized to 0 (unless you specify something else) rather than an undefined value.

There are some other facets to initialization you can take advantage off -- for example shared segments allow different instances of your executable running at once to access the same static variables.

In C++ (globally scoped) static objects have their constructors called as part of the program start up, under the control of the C runtime library. Under Visual C++ at least the order that objects are initialized in can be controlled by the init_seg pragma.

Answer 2

I was curious about this so I wrote the following test program and compiled it with g++ version 4.1.2.

include <iostream>
#include <string>

using namespace std;

class test
{
public:
        test(const char *name)
                : _name(name)
        {
                cout << _name << " created" << endl;
        }

        ~test()
        {
                cout << _name << " destroyed" << endl;
        }

        string _name;
};

test t("global variable");

void f()
{
        static test t("static variable");

        test t2("Local variable");

        cout << "Function executed" << endl;
}


int main()
{
        test t("local to main");

        cout << "Program start" << endl;

        f();

        cout << "Program end" << endl;
        return 0;
}

The results were not what I expected. The constructor for the static object was not called until the first time the function was called. Here is the output:

global variable created
local to main created
Program start
static variable created
Local variable created
Function executed
Local variable destroyed
Program end
local to main destroyed
static variable destroyed
global variable destroyed

Answer 3

Adam Pierce:

The results were not what I expected. The constructor for the static object was not called until the first time the function was called.

They were not what you expected but this how C++ works. Function-level statics with nontrivial constructors are initialized the first time that line of code is executed.

Answer 4

The memory for all static variables is allocated at program load. But local static variables are created and initialized the first time they are used, not at program start up. There's some good reading about that, and statics in general, here. In general I think some of these issues depend on the implementation, especially if you want to know where in memory this stuff will be located.

Answer 5

The compiler will allocate static variable(s) defined in a function foo at program load, however the compiler will also add some additional instructions (machine code) to your function foo so that the first time it is invoked this additional code will initialize the static variable (e.g. invoking the constructor, if applicable).

@Adam: This behind the scenes injection of code by the compiler is the reason for the result you saw.

Answer 6

Or is it initialized when doSomething() is first called?

Yes, it is. This, among other things, lets you initialize globally-accessed data structures when it is appropriate, for example inside try/catch blocks. E.g. instead of

int foo = init(); // bad if init() throws something

int main() { try { ... } catch(...){ ... } }

you can write

int& foo() { static int myfoo = init(); return myfoo; }

and use it inside the try/catch block. On the first call, the variable will be initialized. Then, on the first and next calls, its value will be returned (by reference).

Answer 7

Some relevant verbiage from C++ Standard:

3.6.2 Initialization of non-local objects [basic.start.init]

1 The storage for objects with static storage duration (basic.stc.static) shall be zero-initialized (dcl.init) before any other initialization takes place. Objects of POD types (basic.types) with static storage duration initialized with constant expressions (expr.const) shall be initialized before any dynamic initialization takes place. Objects of namespace scope with static storage duration defined in the same translation unit and dynamically initialized shall be initialized in the order in which their defini- tion appears in the translation unit. [Note: dcl.init.aggr describes the order in which aggregate members are initialized. The initialization of local static objects is described in stmt.dcl. ]

6.7 Declaration statement [stmt.dcl]

<...snip...>

4 The zero-initialization (dcl.init) of all local objects with static storage duration (basic.stc.static) is performed before any other initialization takes place. A local object of POD type (basic.types) with static storage duration initialized with con- stant-expressions is initialized before its block is first entered. An implementation is permitted to perform early initialization of other local objects with static storage duration under the same conditions that an implementation is permitted to statically initialize an object with static storage duration in namespace scope (basic.start.init). Otherwise such an object is initialized the first time control passes through its declaration; such an object is considered initialized upon the completion of its initialization. If the initialization exits by throwing an exception, the initialization is not complete, so it will be tried again the next time control enters the declaration. If con- trol re-enters the declaration (recursively) while the object is being initialized, the behavior is undefined. [Example: int foo(int i) { static int s = foo(2*i); // recursive call - undefined return i+1; } --end example]

---

2) The transfer from the condition of a switch statement to a case la- bel is considered a jump in this respect.

5 The destructor for a local object with static storage duration will be executed if and only if the variable was constructed. [Note: basic.start.term describes the order in which local objects with static storage duration are destroyed. ]