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我正在处理应用程序的一个非常低级别的部分，其中性能至关重要。

在调查生成的程序集时，我注意到以下说明：

lea eax,[edx*8+8]

我习惯于在使用内存引用时看到加法（例如 [edx+4]），但这是我第一次看到乘法。

  • 这是否意味着 x86 处理器可以在 lea 指令中执行简单的乘法运算？
  • 这种乘法是否会影响执行指令所需的周期数？
  • 乘法是否仅限于 2 的幂（我假设是这种情况）？

提前致谢。

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3 回答 3

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要扩展我的评论并回答其余问题...

是的，它仅限于二的幂。（特别是 2、4 和 8）所以不需要乘数，因为它只是一个转变。它的重点是从索引变量和指针快速生成地址 - 其中数据类型是简单的 2、4 或 8 字节字。（尽管它也经常被滥用于其他用途。）

至于所需的周期数：根据Agner Fog 的表格，看起来该lea指令在某些机器上是恒定的，而在其他机器上是可变的。

在 Sandy Bridge 上，如果它是“复杂的或相对撕裂的”，则会受到 2 个循环的惩罚。但它并没有说明“复杂”是什么意思……所以我们只能猜测，除非你做一个基准测试。

于 2012-05-09T08:17:54.247 回答
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实际上，这不是特定于lea指令的内容。

This type of addressing is called Scaled Addressing Mode. The multiplication is achieved by a bit shift, which is trivial:

A Left Shift

You could do 'scaled addressing' with a mov too, for example (note that this is not the same operation, the only similarity is the fact that ebx*4 represents an address multiplication):

 mov edx, [esi+4*ebx]

(source: http://www.cs.virginia.edu/~evans/cs216/guides/x86.html#memory)

For a more complete listing, see this Intel document. Table 2-3 shows that a scaling of 2, 4, or 8 is allowed. Nothing else.

Latency (in terms of number of cycles): I don't think this should be affected at all. A shift is a matter of connections, and selecting between three possible shifts is the matter of 1 multiplexer worth of delay.

于 2012-05-09T08:38:46.140 回答
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To expand on your last question:

Is the multiplication limited to powers of 2 (I would assume this is the case)?

Note that you get the result of base + scale * index, so while scale has to be 1, 2, 4 or 8 (the size of x86 integer datatypes), you can get the equivalent of a multiplication by some different constants by using the same register as base and index, e.g.:

lea eax, [eax*4 + eax]   ; multiply by 5

This is used by the compiler to do strength reduction, e.g: for a multiplication by 100, depending on compiler options (target CPU model, optimization options), you may get:

lea    (%edx,%edx,4),%eax   ; eax = orig_edx * 5
lea    (%eax,%eax,4),%eax   ; eax = eax * 5 = orig_edx * 25
shl    $0x2,%eax            ; eax = eax * 4 = orig_edx * 100
于 2012-05-09T21:37:41.903 回答