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)

让你的意图更清楚……例如，如果你想除法，使用x / 2，并让编译器将其优化为shift运算符(或其他任何运算符)。

今天的处理器不会让这些优化对程序的性能产生任何影响。

就CPU而言，位移运算比除法运算快。 然而，编译器知道这一点，并将适当地优化到它可以的程度， 因此，您可以以最有意义的方式编写代码，并且知道您的代码是有意义的 有效地运行。但是请记住，由于前面指出的原因，在某些情况下，无符号整型可以比整型更好地优化。 如果你不需要符号算术，那么不要包含符号位。

这个问题的答案取决于你工作的环境。

If you're working on an 8-bit microcontroller or anything without hardware support for multiplication, bit shifting is expected and commonplace, and while the compiler will almost certainly turn x /= 2 into x >>= 1, the presence of a division symbol will raise more eyebrows in that environment than using a shift to effect a division. If you're working in a performance-critical environment or section of code, or your code could be compiled with compiler optimization off, x >>= 1 with a comment explaining its reasoning is probably best just for clarity of purpose. If you're not under one of the above conditions, make your code more readable by simply using x /= 2. Better to save the next programmer who happens to look at your code the 10 second double-take on your shift operation than to needlessly prove you knew the shift was more efficient sans compiler optimization.

所有这些假设都是无符号整数。简单的移位可能不是你想要的符号。此外，DanielH提出了一个关于在ActionScript等特定语言中使用x *= 0.5的好观点。

x / 2更清晰，x >> 1快不了多少(根据一个微基准测试，对于Java JVM快30%左右)。正如其他人所注意到的，对于负数，舍入略有不同，所以当您想处理负数时必须考虑这一点。一些编译器可能会自动将x / 2转换为x >> 1，如果他们知道这个数字不可能是负数(即使我无法验证这一点)。

即使x / 2可能不使用(慢)除法CPU指令，因为一些捷径是可能的，但它仍然比x >> 1慢。

(This is a C / C++ question, other programming languages have more operators. For Java there is also the unsigned right shift, x >>> 1, which is again different. It allows to correctly calculate the mean (average) value of two values, so that (a + b) >>> 1 will return the mean value even for very large values of a and b. This is required for example for binary search if the array indices can get very large. There was a bug in many versions of binary search, because they used (a + b) / 2 to calculate the average. This doesn't work correctly. The correct solution is to use (a + b) >>> 1 instead.)