C ++ static array leading to memory leak?

Question

Say I have something like ...

void foo()
{
    char c[100];
    printf("this function does nothing useful");
}

When foo is called, it creates an array on the stack, and when it goes out of scope, is memory freed automatically? Or is c destroyed, but the memory remains allocated, without the ability to access / get it, except for restarting the computer?

Answer 1

- is memory freed up automatically?

Yes. And destructors will also be called if you're interested. That's why they are in a automatic storage class .

( 100 ( ), 100 * sizeof(T) "".)

Answer 2

, . - :

int var = 100;
char* c = new char[var];

, .

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Answer 3

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Answer 4

.

But memory is available for use by the next function immediately. the stack is only implemented as a pointer, when you execute c [100], it moves the pointer down 100 bytes, so the next requested memory comes after c. When you leave a function, the stack pointer just returns to the previous position before you enter the function. This is a very fast and efficient way to manage memory compared to the new / delete / malloc