11

Consider the following two programs that perform the same computations in two different ways:

// v1.c
#include <stdio.h>
#include <math.h>
int main(void) {
   int i, j;
   int nbr_values = 8192;
   int n_iter = 100000;
   float x;
   for (j = 0; j < nbr_values; j++) {
      x = 1;
      for (i = 0; i < n_iter; i++)
         x = sin(x);
   }
   printf("%f\n", x);
   return 0;
}

and

// v2.c
#include <stdio.h>
#include <math.h>
int main(void) {
   int i, j;
   int nbr_values = 8192;
   int n_iter = 100000;
   float x[nbr_values];
   for (i = 0; i < nbr_values; ++i) {
      x[i] = 1;
   }
   for (i = 0; i < n_iter; i++) {
      for (j = 0; j < nbr_values; ++j) {
         x[j] = sin(x[j]);
      }
   }
   printf("%f\n", x[0]);
   return 0;
}

When I compile them using gcc 4.7.2 with -O3 -ffast-math and run on a Sandy Bridge box, the second program is twice as fast as the first one.

Why is that?

One suspect is the data dependency between successive iterations of the i loop in v1. However, I don't quite see what the full explanation might be.

(Question inspired by Why is my python/numpy example faster than pure C implementation?)

EDIT:

Here is the generated assembly for v1:

        movl    $8192, %ebp
        pushq   %rbx
LCFI1:
        subq    $8, %rsp
LCFI2:
        .align 4
L2:
        movl    $100000, %ebx
        movss   LC0(%rip), %xmm0
        jmp     L5
        .align 4
L3:
        call    _sinf
L5:
        subl    $1, %ebx
        jne     L3
        subl    $1, %ebp
        .p2align 4,,2
        jne     L2

and for v2:

        movl    $100000, %r14d
        .align 4
L8:
        xorl    %ebx, %ebx
        .align 4
L9:
        movss   (%r12,%rbx), %xmm0
        call    _sinf
        movss   %xmm0, (%r12,%rbx)
        addq    $4, %rbx
        cmpq    $32768, %rbx
        jne     L9
        subl    $1, %r14d
        jne     L8
4

2 回答 2

15

Ignore the loop structure all together, and only think about the sequence of calls to sin. v1 does the following:

x <-- sin(x)
x <-- sin(x)
x <-- sin(x)
...

that is, each computation of sin( ) cannot begin until the result of the previous call is available; it must wait for the entirety of the previous computation. This means that for N calls to sin, the total time required is 819200000 times the latency of a single sin evaluation.

In v2, by contrast, you do the following:

x[0] <-- sin(x[0])
x[1] <-- sin(x[1])
x[2] <-- sin(x[2])
...

notice that each call to sin does not depend on the previous call. Effectively, the calls to sin are all independent, and the processor can begin on each as soon as the necessary register and ALU resources are available (without waiting for the previous computation to be completed). Thus, the time required is a function of the throughput of the sin function, not the latency, and so v2 can finish in significantly less time.

I should also note that DeadMG is right that v1 and v2 are formally equivalent, and in a perfect world the compiler would optimize both of them into a single chain of 100000 sin evaluations (or simply evaluate the result at compile time). Sadly, we live in an imperfect world.

于 2013-01-22T21:51:39.483 回答
0

In the first example, it runs 100000 loops of sin, 8192 times.

In the second example, it runs 8192 loops of sin, 100000 times.

Other than that and storing the result differently, I don't see any difference.

However, what does make a difference is that the input is being changed for each loop in the second case. So I suspect what happens is that the sin value, at certain times in the loop, gets much easier to calculate. And that can make a big difference. Calculating sin is not entirely trivial, and it's a series calculation that loops until the exit condition is hit.

于 2013-01-22T22:02:17.040 回答