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

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。

使用除法(/)，假设这样更清楚。编译器将相应地进行优化。

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

我想说有几件事需要考虑。

Bitshift should be faster, as no special computation is really needed to shift the bits, however as pointed out, there are potential issues with negative numbers. If you are ensured to have positive numbers, and are looking for speed then I would recommend bitshift. The division operator is very easy for humans to read. So if you are looking for code readability, you could use this. Note that the field of compiler optimization has come a long way, so making code easy to read and understand is good practice. Depending on the underlying hardware, operations may have different speeds. Amdal's law is to make the common case fast. So you may have hardware that can perform different operations faster than others. For example, multiplying by 0.5 may be faster than dividing by 2. (Granted you may need to take the floor of the multiplication if you wish to enforce integer division).

如果您追求的是纯粹的性能，我建议您创建一些可以执行数百万次操作的测试。对执行进行多次采样(您的样本量)，以确定哪一个在统计上最适合您的操作系统/硬件/编译器/代码。

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。