查看编译器的输出来帮助您做出决定。我在x86-64上使用 gcc (gcc) 4.2.1 20070719 [FreeBSD]

也可以在godbolt上看到编译器的在线输出。

What you see is the compiler does use a sarl (arithmetic right-shift) instruction in both cases, so it does recognize the similarity between the two expressions. If you use the divide, the compiler also needs to adjust for negative numbers. To do that it shifts the sign bit down to the lowest order bit, and adds that to the result. This fixes the off-by-one issue when shifting negative numbers, compared to what a divide would do. Since the divide case does 2 shifts, while the explicit shift case only does one, we can now explain some of the performance differences measured by other answers here.

C代码与汇编输出:

对于除法，你的输入是

int div2signed(int a) {
  return a / 2;
}

这个编译成

    movl    %edi, %eax
    shrl    $31, %eax            # (unsigned)x >> 31
    addl    %edi, %eax           # tmp = x + (x<0)
    sarl    %eax                 # (x + 0 or 1) >> 1  arithmetic right shift
    ret

shift也是一样

int shr2signed(int a) {
  return a >> 1;
}

输出:

    sarl    %edi
    movl    %edi, %eax
    ret

其他isa即使不能做得更好，也能同样有效地做到这一点。例如GCC For AArch64使用:

        add     w0, w0, w0, lsr 31      // x += (unsigned)x>>31
        asr     w0, w0, 1               // x >>= 1
        ret

使用最能描述您要做的事情的操作。

如果你将数字作为一个比特序列来处理，请使用bitshift。 如果你把它当作一个数值，使用除法。

请注意，它们并不完全相等。对于负整数，它们可以给出不同的结果。例如:

-5 / 2  = -2
-5 >> 1 = -3

(ideone)

一般右移分为:

q = i >> n; is the same as: q = i / 2**n;

这有时被用来加快程序的速度，但以清晰度为代价。我觉得你不应该这么做。编译器足够智能，可以自动执行加速。这意味着，以清晰度为代价，你不会获得任何好处。

看看《实用c++编程》的这一页。

X = X / 2;是合适的代码使用..但是一个操作取决于你自己的程序，你想要产生怎样的输出。

Mod 2, test for = 1。不知道c中的语法，但这可能是最快的。

我们有很多理由支持使用x = x / 2;以下是一些例子:

it expresses your intent more clearly (assuming you're not dealing with bit twiddling register bits or something) the compiler will reduce this to a shift operation anyway even if the compiler didn't reduce it and chose a slower operation than the shift, the likelihood that this ends up affecting your program's performance in a measurable way is itself vanishingly small (and if it does affect it measurably, then you have an actual reason to use a shift) if the division is going to be part of a larger expression, you're more likely to get the precedence right if you use the division operator: x = x / 2 + 5; x = x >> 1 + 5; // not the same as above signed arithmetic might complicate things even more than the precedence problem mentioned above to reiterate - the compiler will already do this for you anyway. In fact, it'll convert division by a constant to a series of shifts, adds, and multiplies for all sorts of numbers, not just powers of two. See this question for links to even more information about this.

简而言之，当你真正想要进行乘法或除法运算时，编写移位代码并没有什么好处，除了可能会增加引入错误的可能性。自从编译器不够聪明到在适当的时候优化这类事情到移位以来，已经过去了。