Question

I am starting again with c++ and was thinking about the scope of variables. If I have a variable inside a function and then I return that variable will the variable not be ‘dead’ when it’s returned because the scope it was in has ended?

I have tried this with a function returning a string and it did work. Can anyone explain this? Or at least point me to some place that can explain this to me please.

Thanks

Answer 1

When the function terminates, the following steps happen:

• The function’s return value is copied into the placeholder that was put on the stack for this purpose.

• Everything after the stack frame pointer is popped off. This destroys all local variables and arguments.

• The return value is popped off the stack and is assigned as the value of the function. If the value of the function isn’t assigned to anything, no assignment takes place, and the value is lost.

• The address of the next instruction to execute is popped off the stack, and the CPU resumes execution at that instruction.

The stack and the heap