How does a compiler optimise this factorial function so well?

Question

So I have been having a look at some of the magic that is O3 in GCC (well actually I\'m compiling using Clang but it\'s the same with GCC and I\'m guessing a large

Answer 1

If you compile with gcc -O3 -fdump-tree-all, you can see that the first dump in which the recursion has been turned into a loop is foo.c.035t.tailr1. This means the same optimisation that handles other tail calls also handles this slightly extended case. Recursion in the form of n * foo(...) or n + foo(...) is not that hard to handle manually (see below), and since it's possible to describe exactly how, the compiler can perform that optimisation automatically.

The optimisation of main is much simpler: inlining can turn this into 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 * 1, and if all the operands of a multiplication are constants, then the multiplication can be performed at compile time.

Update: Here's how you can manually remove the recursion from foo, which can be done automatically. I'm not saying this is the method used by GCC, but it's one realistic possibility.

First, create a helper function. It behaves exactly as foo(n), except that its results are multiplied by an extra parameter f.

int foo(int n)
{
    return foo_helper(n, 1);
}

int foo_helper(int n, int f)
{
    if (n == 0) return f * 1;
    return f * n * foo(n-1);
}

Then, turn recursive calls of foo into recursive calls of foo_helper, and rely on the factor parameter to get rid of the multiplication.

int foo(int n)
{
    return foo_helper(n, 1);
}

int foo_helper(int n, int f)
{
    if (n == 0) return f;
    return foo_helper(n-1, f * n);
}

Turn this into a loop:

int foo(int n)
{
    return foo_helper(n, 1);
}

int foo_helper(int n, int f)
{
restart:
    if (n == 0) return f;
    {
        int newn = n-1;
        int newf = f * n;
        n = newn;
        f = newf;
        goto restart;
    }
}

Finally, inline foo_helper:

int foo(int n)
{
    int f = 1;
restart:
    if (n == 0) return f;
    {
        int newn = n-1;
        int newf = f * n;
        n = newn;
        f = newf;
        goto restart;
    }
}

(Naturally, this is not the most sensible way to manually write the function.)