如何将单精度浮点数的XMM寄存器转换为整数? [英] How can I convert an XMM register of single-precision floats to integers?

问题描述

我在XMM寄存器中有一堆打包的浮点数(使用SSE内在函数):

I have a bunch of packed floats inside an XMM register (using SSE intrinsics):

__m128 xmm = _mm_set_ps(4.0f, 3.0f, 2.0f, 1.0f);

我想一次将所有这些转换为整数.我发现了一个内在函数，它可以满足我的要求( _mm_cvtps_pi16())，但是它会生成4x16位的 short 而不是成熟的 int .名为 _mm_cvtps_pi32()的内部函数会产生 int ，但仅适用于 xmm 中的两个较低值.我可以使用它，提取值，四处移动并再次使用它，但是有没有更简单的方法?为什么没有直接的32位压缩浮点-> 32位整数指令?两者肯定都适合XMM寄存器的同一空间吗?

I'd like to convert all of these to integers in one go. I found an intrinsic, that does what I want (_mm_cvtps_pi16()), but it yields 4x16-bit short instead of full-blown int. An intrinsic called _mm_cvtps_pi32() yields int, but only for the two lower values in xmm. I can use it, extract the values, move things around and use it again, but is there a simpler way? Why wouldn't there be a straightforward 32bit packed float -> 32bit integer instruction? Surely both fit in the same space of an XMM register?

好的，我现在看到 _mm_cvtps_pi32()返回__m64而不是__m128，这意味着它在MMX风格的MM ...寄存器上运行.那可以解释为什么它只返回两个整数，但是现在我在想:

Okay, I see now that _mm_cvtps_pi32() returns __m64 instead of __m128, which means it operates on a MMX-style MM... register. That would explain why it returns just two ints, but now I'm wondering:

• 为x64编译时会遇到麻烦吗?据说那里不支持__m64 ...
• 为什么在SSE推出时他们没有扩展此指令?

谢谢！

推荐答案

根据

According to this documentation: __m128d _mm_cvtps_epi32(__m128d a) generates a cvtps2dq instruction, which does what you want.

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