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, Jul 13 '14 at 09:01
@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, Jul 13 '14 at 09:04
@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, Jul 13 '14 at 09:04
@mafso: And I too slow. Will add it to the question... Thanks — barak manos, Jul 13 '14 at 09:06
Pretty sure that's the same question. Edit: Maybe not. It's similar. — Joseph Mansfield, Jul 13 '14 at 09:08
The compiler does not need an invalid value, a flag 'initialized' will do it, even for an object with a constructor. — , Jul 13 '14 at 09:10
@Mansuro It does, but it's not the same question. The answer was in the question of that question. — Joseph Mansfield, Jul 13 '14 at 09:15
@DieterLücking: Thanks. I actually feel kinda stupid asking this question in the first place. — barak manos, Jul 13 '14 at 10:09

score 2 · Accepted Answer · answered Jul 13 '14 at 10:05

2

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.

answered Jul 13 '14 at 10:05

Baptiste Wicht

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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). – barak manos Jul 13 '14 at 10:19

score 1 · Answer 2 · answered Jul 13 '14 at 10:19

1

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.

answered Jul 13 '14 at 10:19

George Kourtis

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Initialize a static local variable to a value unknown during compilation

2 Answers2