Static variables exist outside of the function, in terms of their memory at least (not scope), right? But one thing that always concerned me, is what happens when I call the function a second time. For instance:
f(){
static char buffer[256*256];
stuff(buffer);
}
When I call this function a second time, wouldn't it technically be declaring the variable 'buffer' a second time? Or does it work differently with static variables (as opposed to normal ones) once everything is compiled?
... I sometimes wish there was a chart or something of what a c++ compiler usually turns code into (minus optimizations) so I wouldn't have to bother you fine folks with little questions like this, aha. Thank you in advance!
edit: I know it works like this, I just want to know why though. It's probably something mind numbingly simple...
Static storage duration objects in function scope.
These objects are created on first use.
Then destroyed in reverse order of creation (with other static storage duration objects).
#include <iostream>
class X
{
public:
X(int x): m(x) {std::cout << "X: " << m << " created\n"; }
~X() {std::cout << "X: " << m << " destroyed\n";}
private:
int m;
};
static X x1(1);
int test()
{
std::cout << "Test: Start\n";
static X x3(3);
std::cout << "Test: Finished\n";
return 5;
}
int main()
{
std::cout << "Main: Start\n";
X x2(2);
test();
X x4(4);
std::cout << "Main: Finished\n";
}
Now Try it: (comments added). SSDO => Static Storage Duration object.
g++ X.cpp
./a.out
X: 1 created // SSDO file scope.
Main: Start
X: 2 created
Test: Start
X: 3 created // SSDO created on first use (Notice not destroyed)
Test: Finished
Test: Start // Notice not created here.
Test: Finished
X: 4 created
Main: Finished
X: 4 destroyed
X: 2 destroyed // Main now really finished. after destroying local variables.
X: 3 destroyed // Destroy SSDO in reverse order of creation. (3 - 1)
X: 1 destroyed