How can I get the size of a C function

Question

In C++, how can I find the size of a function?

Assume I have a function:

void f()
{
/*do something*/
}

... By "size of f," I mean the length of the code that will /*do something*/ beginning at the location specified by a reference to f.

Answer 1

You just cannot. It's possible that it's not even a well-defined idea. Consider the following code as an example:

int f(int a) {
    return (3*a)+1;
}

int g(int a) {
    return (2*a)+1;
}

I doubt that happens in practise for this example because it would not be an optimization, but the compiler is allowed to insert a block of code that computes a+1 then returns and jump to that block from each of f and g's entry points (after doing a multiplication in each case). What is the size of f then? Should the size of the shared block be included? How about half that size? It just does not make sense in C++ to assert that the function f has a size.

A "pointer to f" may also be more than just the location of the function f. It undoubtedly gives

Answer 2

You just cannot. It's possible that it's not even a well-defined idea. Consider the following code as an example:

int f(int a) {
    return (3*a)+1;
}

int g(int a) {
    return (2*a)+1;
}

I doubt that happens in practise for this example because it would not be an optimization, but the compiler is allowed to insert a block of code that computes a+1 then returns and jump to that block from each of f and g's entry points (after doing a multiplication in each case). What is the size of f then? Should the size of the shared block be included? How about half that size? It just does not make sense in C++ to assert that the function f has a size.

A "pointer to f" may also be more than just the location of the function f. It undoubtedly gives