Memory allocation to functions in C++

Question

Hi All !

I am still a C++ newbie. Just came to read that the static member function of a class is not object specific - there is a single copy of the member functions for all the objects.

Now two questions arise in my mind :

1. What is the difference between an ordinary function and a static function "in terms of memory allocation only" ?

2. What if the member function contains some local variables ? In that case the function "should" have a separate copy of that variable - specific to the object invoking the function... How is this problem solved in C++ ?

Thanks !

Answer 1

1. no difference
2. local variables are created and destroyed on the stack per invocation of the function

Answer 2

1. I would find it very unlikely that there would be a difference

2. Seems like you should read a bit about the difference between heap and stack allocations. This gives a good idea about how memory works at a low, but still high level. Sorry i can't be more helpful right now.

edit: too slow :)

Answer 3

What is the difference between an ordinary function and a static function "in terms of memory allocation only" ?

Nothing. A static function is just like a global function except for the scope.

Even for non-static member functions, there's no extra memory required. The member function int C::f(int arg1, int arg2) is just syntactic sugar for something like int C__f(C* this, int arg1, int arg2).

What if the member function contains some local variables ? In that case the function "should" have a separate copy of that variable - specific to the object invoking the function...

There's a copy of the local variables for each invocation of the function (unless they're static). This is why recursion in possible in C++.

How is this problem solved in C++ ?

Function calls are based around "stack frames". A stack frame consists of:

• The arguments to the function (including the implicit this if applicable).
• Any other non-static local variables in the function.
• The "return address", which tells the processor where to resume execution once the function is done.

Whenever a function is called, a stack frame is created. When the function is, the stack frame is destroyed. If a function is called recursively, each level of recursion gets its own stack frame. For example, if you have

int factorial(int n) {
    if (n <= 1)
        return 1;
    else
        return factorial(n - 1) * n;
}

Then when you call factorial(3), a stack frame gets created like so:

------------------------ stack pointer (SP)
n = 3
RA = <in main()>

When the recursive call is made to factorial(2), an additional frame is added to the top of the stack

------------------------ SP
n = 2
RA = <in factorial()>
------------------------
n = 3
RA = <in main()>

Another recursive call is made, to factorial(1).

------------------------ SP
n = 1
RA = <in factorial()>
------------------------
n = 2
RA = <in factorial()>
------------------------
n = 3
RA = <in main()>

This is the base case for the recursion, and the return value of 1 is stored in a register. The function call being complete, the top stack frame is destroyed, and execution continues at saved return address.

------------------------ SP
n = 2                                      
RA = <in factorial()>
------------------------
n = 3
RA = <in main()>

Now, the call to factorial(2) can compute its return value (2), and another stack frame can be destroyed:

------------------------ SP
n = 3
RA = <in main()>

Finally, we can compute the result of the original function call (6), and destroy this stack frame too.

Answer 4

1. There really isn't that much of a difference. Both are stored in memory only once, when a non-static method is called, a pointer to the current object (the this pointer) is pushed onto the stack (or stored in ECX depending on your compiler) in addition to all of the function's parameters. A static function doesn't need an instance of a class, so it is just called like a regular C function.

2. Same way as in C, with the call stack.