115

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?

3
  • That begs a question, what happens when there are multiple threads. will each thread have separate initialization of static local variable ? Does it work like thread-local ? Commented Jan 21, 2024 at 18:46
  • Sorry, I can't help it, but since the site is all around spreading knowledge 😏: en.wikipedia.org/wiki/Begging_the_question Commented Jan 22, 2024 at 20:20
  • 1
    But good question, and my uneducated guess would be that no, there's a single instance, and perhaps it even carries a risk of a race conditions if it is used in multiple threads. Talks about some of this: stackoverflow.com/questions/22794382/… Commented Jan 22, 2024 at 20:27

8 Answers 8

116

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 
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10 Comments

As a clarification: the static variable is initialized the first time execution hits its declaration, not when the containing function is called. If you just have a static at the start of the function (e.g. in your example) these are the same, but are not necessarily: e.g. if you have 'if (...) { static MyClass x; ... }', then 'x' will not be initialized at ALL during the first execution of that function in the case where the if statement's condition evaluates to false.
But doesn't this lead to a runtime overhead, since each time the static variable is used, the program has to check if it has been previously used, since if not, it has to be initialized? In that case that kind of sucks a bit.
@veio: Yes, the initialization is thread-safe. See that question for more details: stackoverflow.com/questions/23829389/…
@HelloGoodbye: yes, it leads to a runtime overhead. Also see that question: stackoverflow.com/questions/23829389/…
@463035818_is_not_a_number compared to if the static variable had been initialized at the beginning of the program
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72

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 definition 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. ]

[more text below adding more liberties for compiler writers]

6.7 Declaration statement [stmt.dcl]

...

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 constant-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 control 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]

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. ]

2 Comments

This answered my question and doesn't rely on "anecdotal evidence" unlike the accepted answer. I was specifically looking for this mention of exceptions in the constructor of statically-initialized function local static objects: 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 anyone is curious, a recursion leads a GCC-compiled program to terminate with __gnu_cxx::recursive_init_error, and an MSVC-compiled program to freeze.
34

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.

3 Comments

not quite, local statics are allocated and zero-initialized "at program load" (in quotes, because that's not quite right either), and then reinitialized the first time the function they're in is entered.
Looks like that link is now broken, 7 years later.
10

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.

Comments

8

I try to test again code from Adam Pierce and added two more cases: static variable in class and POD type. My compiler is g++ 4.8.1, in Windows OS(MinGW-32). Result is static variable in class is treated same with global variable. Its constructor will be called before enter main function.

  • Conclusion (for g++, Windows environment):
  1. Global variable and static member in class: constructor is called before enter main function (1).
  2. Local static variable: constructor is only called when execution reaches its declaration at first time.
  3. If Local static variable is POD type, then it is also initialized before enter main function (1). Example for POD type: static int number = 10;

(1): The correct state should be: "before any function from the same translation unit is called". However, for simple, as in example below, then it is main function.

#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; static test t; // static member }; test test::t("static in class"); test t("global variable"); void f() { static test t("static variable"); static int num = 10 ; // POD type, init before enter main function 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; } 

result:

static in class created 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 static in class destroyed 

Anybody tested in Linux env ?

1 Comment

how did you conclude that num is initialized before main? Note that if the initializer is something more complicated than an integer literal this is not the case: godbolt.org/z/Gasr7dTY8 obvioulsy in this example, expected side-effects would be off it was initialized before main
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).

Comments

3

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.

5 Comments

This question is about function-scoped statics. At least when they have nontrivial constructors they are initialized on first entry into the function. Or more specifically, when that line is reached.
True -- but the question talks about the space allocated to the variable, and uses simple data types. The space is still allocated in the code segment
I don't see how code segment vs. data segment really matters here. I think we need clarification from the OP. He did say "and initialized, if applicable".
variables are never allocated inside the code segment; this way they wouldn't be write-able.
static variables are allocated space in data segment or in bss segment depending upon whether they are initialized or not.
-2

In the following code it prints Initial = 4 which is the value of static_x as it is implemented in the compiling time.

 int func(int x) { static int static_x = 4; static_x = x; printf ("Address = 0x%x",&static_x ); // prints 0x40a010 return static_x; } int main() { int x = 8; uint32_t *ptr = (uint32_t *)(0x40a010); // static_x location printf ("Initial = %d\n",*ptr); func(x); return 0; } 

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