从__m128转换为__m128i会导致错误的值 [英] Converting from __m128 to __m128i results in wrong value

问题描述

我需要将浮点向量（__m128）转换为整数向量（__m128i），并且我正在使用 _mm_cvtps_epi32 ，但未获得预期值。这是一个非常简单的示例：

I need to convert a float vector (__m128) to an integer vector (__m128i), and I am using _mm_cvtps_epi32, but I am not getting the expected value. Here is a very simple example:

__m128 test = _mm_set1_ps(4.5f);
__m128i test_i = _mm_cvtps_epi32(test);

我得到的调试器输出：

(lldb) po test
 ([0] = 4.5, [1] = 4.5, [2] = 4.5, [3] = 4.5)
(lldb) po test_i
 ([0] = 17179869188, [1] = 17179869188)
(lldb) 

如您所见，结果整数是.. 17179869188？从4.5开始？为什么只有两个值？ _mm_cvtps_epi32 应该将4个压缩的32位浮点值转换为4个压缩的32位整数。

As you can see, the resulting integer is.. 17179869188? From 4.5? And why are there only two values? _mm_cvtps_epi32 should convert 4 packed 32-bit floating-point values to 4 packed 32-bit integers.

推荐答案

调试器在此示例中，将 __ m128i 值解释为两个64位整数，而不是您期望的四个32位整数。实际的转换是正确的。

Debugger, in this example, interprets the __m128i value as two 64-bit integers, as opposed to four 32-bit ones expected by you. The actual conversion is correct.

在您的代码中，您需要明确指定如何解释SIMD值，例如： test_i.m128i_i32 [0]

In your code you need to explicitly specify how to interpret the SIMD value, for example: test_i.m128i_i32[0]

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