Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Why does GCC generate a faster program than Clang in this recursive Fibonacci code?

This is the code that I tested:

#include <iostream>
#include <chrono>
using namespace std;

#define CHRONO_NOW                  chrono::high_resolution_clock::now()
#define CHRONO_DURATION(first,last) chrono::duration_cast<chrono::duration<double>>(last-first).count()

int fib(int n) {
    if (n<2) return n;
    return fib(n-1) + fib(n-2);
}

int main() {
    auto t0 = CHRONO_NOW;
    cout << fib(45) << endl;
    cout << CHRONO_DURATION(t0, CHRONO_NOW) << endl;
    return 0;
}

Of course, there are much faster ways of calculating Fibonacci numbers, but this is a good little stress test that focuses on recursive function calls. There's nothing else to the code, other than the use of chrono for measuring time.

First I ran the test a couple of times in Xcode on OS X (so that's clang), using -O3 optimization. It took about 9 seconds to run.

Then, I compiled the same code with gcc (g++) on Ubuntu (using -O3 again), and that version only took about 6.3 seconds to run! Also, I was running Ubuntu inside VirtualBox on my mac, which could only affect the performance negatively, if at all.

So there you go:

  • Clang on OS X -> ~9 secs
  • gcc on Ubuntu in VirtualBox -> ~6.3 secs.

I know that these are completely different compilers so they do stuff differently, but all the tests I've seen featuring gcc and clang only showed much less of a difference, and in some cases, the difference was the other way around (clang being faster).

So is there any logical explanation why gcc beats clang by miles in this particular example?

like image 450
notadam Avatar asked Mar 21 '15 18:03

notadam


1 Answers

GCC 4.9.2 in compiler explorer really does loop-unrolling and inlines a lot of function calls while Clang 3.5.1 calls fib twice each iteration without even tail call optimization like below

fib(int):                                # @fib(int)
        push    rbp
        push    rbx
        push    rax
        mov     ebx, edi
        cmp     ebx, 2
        jge     .LBB0_1
        mov     eax, ebx
        jmp     .LBB0_3
.LBB0_1:
        lea     edi, dword ptr [rbx - 1]
        call    fib(int)       # fib(ebx - 1)
        mov     ebp, eax
        add     ebx, -2
        mov     edi, ebx
        call    fib(int)       # fib(ebx - 2)
        add     eax, ebp
.LBB0_3:
        add     rsp, 8
        pop     rbx
        pop     rbp
        ret

The GCC version is more than 10 times longer, with only a single fib call and 20+ labels for inlining the call, which also means that the last call has been tail-optimized into a jmp or GCC has converted some of the recursion into iteration (since it allocates a big array for storing intermediate values)

I've also brought ICC into perspective, and surprisingly it has 10 call instructions inside fib, and it also inlines fib calls 9 times inside main, but it doesn't convert the recursive code to iterative

Here's the compiler outputs for comparison

Note that you can modify the code like this to make the output easier to read

int fib(int n) {
    if (n<2) return n;
    int t = fib(n-1);
    return t + fib(n-2);
}

Now compiler explorer will highlight which source code line an instruction in the assembly output corresponds to with distinct colors, and you'll easily see how the two calls are made. The line return t + fib(n-2) is compiled by GCC to

.L3:
        mov     eax, DWORD PTR [rsp+112]  # n, %sfp
        add     edx, DWORD PTR [rsp+64]   # D.35656, %sfp
        add     DWORD PTR [rsp+60], edx   # %sfp, D.35656
        sub     DWORD PTR [rsp+104], 2    # %sfp,
like image 116
phuclv Avatar answered Sep 19 '22 08:09

phuclv