难道一个C / C ++编译器通过优化电源的二值恒师为转移？ [英] Does a c/c++ compiler optimize constant divisions by power-of-two value into shifts?

问题描述

问题说明了一切。有谁知道，如果下面...

Question says it all. Does anyone know if the following...

size_t div(size_t value) {
    const size_t x = 64;
    return value / x;
}

...被优化成？

...is optimized into?

size_t div(size_t value) {
    return value >> 6;
}

不要编译器这样做呢？ （我的兴趣在于GCC）。有没有在那里它的情况和其他地方不？

Do compilers do this? (My interest lies in GCC). Are there situations where it does and others where it doesn't?

我真的很想知道，因为我每次写，可以这样进行优化，我花费了一些精力想知道是否第二的precious空话被浪费在做了分工，其中的转变就足够一个部门。

I would really like to know, because every time I write a division that could be optimized like this I spend some mental energy wondering about whether precious nothings of a second is wasted doing a division where a shift would suffice.

推荐答案

即使 G ++ -O0 （是的， -O0 ！），这种情况发生。你的函数编译为：

Even with g++ -O0 (yes, -O0!), this happens. Your function compiles down to:

_Z3divm:
.LFB952:
        pushq   %rbp
.LCFI0:
        movq    %rsp, %rbp
.LCFI1:
        movq    %rdi, -24(%rbp)
        movq    $64, -8(%rbp)
        movq    -24(%rbp), %rax
        shrq    $6, %rax
        leave
        ret

请注意在 SHRQ $ 6 ，这是由6位右移。

Note the shrq $6, which is a right shift by 6 places.

使用 -O1 ，不必要的垃圾被删除：

With -O1, the unnecessary junk is removed:

_Z3divm:
.LFB1023:
        movq    %rdi, %rax
        shrq    $6, %rax
        ret

结果在G ++ 4.3.3，64。

Results on g++ 4.3.3, x64.

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