Initialize a static local variable to a value unknown during compilation

Question

In the C language, in order to initialize a static local variable to a value unknown during compilation, I would normally do something like this (for example):

void func() { static int var = INVALID_VALUE; if (var == INVALID_VALUE) var = some_other_func(); ... }

In the C++ language, I can simply do:

void func() { static int i = some_other_func(); ... }

The only way (that I can think of) for a C++ compiler to resolve it properly, is by replacing this code with a mechanism similar to the C example above.

But how would the compiler determine a "proper" invalid value? Or is there another way which I haven't taken into consideration?

Thanks

Clarification:

INVALID_VALUE is a value which function some_other_func never returns.

It is used in order to ensure that this function is never invoked more than once.

I'm a bit confused. What does INVALID_VALUE represent? And what's the point of it? — Joseph Mansfield
– Joseph Mansfield, Commented Jul 13, 2014 at 9:01
@JosephMansfield: I think, OP means some value some_other_func never returns to check if var has been assigned a return value of that function. — mafso
– mafso, Commented Jul 13, 2014 at 9:04
@JosephMansfield: A value which I consider invalid, and that I know some_other_func will never return (for example, if it was a pointer, then I would normally use NULL). — barak manos
– barak manos, Commented Jul 13, 2014 at 9:04
The compiler does not need an invalid value, a flag 'initialized' will do it, even for an object with a constructor. — user2249683
– user2249683, Commented Jul 13, 2014 at 9:10

Baptiste Wicht · Accepted Answer · 2014-07-13 10:05:18Z

The compiler will not generate code to do it based on its value but on a thread safe flag that ensure that the code is only executed once.

Something like that:

void func() { static int i; static bool i_initialized; if (!i_initialized) { i = some_other_func(); i_initialized = true; } }

Except that generally it is not a bool but a thread safe way of testing it.

Thank you. It's pretty much the same answer as the one given by @Dieter Lücking in one of the comments above (which makes me feel kinda stupid asking this question in the first place).

George Kourtis · Accepted Answer · 2014-07-13 10:19:45Z

According to code seen by disassembling and debugging the g++ compiled code, there is a hidden variable that is initialized to 0 and when the initialization is run it is set to 1. So the next time the initialization code isn't executed.

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