有序/无序比较是什么意思? [英] What does ordered / unordered comparison mean?

问题描述

看上证所运营商

CMPORDPS - ordered compare packed singles
CMPUNORDPS - unordered compare packed singles

有序和无序是什么意思?我在x86指令集中寻找了等效指令，但它似乎只有无序(FUCOM).

What do ordered and unordered mean? I looked for equivalent instructions in the x86 instruction set, and it only seems to have unordered (FUCOM).

推荐答案

有序比较检查两个操作数是否都不为NaN.相反，无序比较检查两个操作数是否为NaN.

An ordered comparison checks if neither operand is NaN. Conversely, an unordered comparison checks if either operand is a NaN.

此页面提供了更多有关此的信息:

This page gives some more information on this:

• http://csapp.cs.cmu.edu/public/waside/waside-sse.pdf (第5节)

• http://csapp.cs.cmu.edu/public/waside/waside-sse.pdf (section 5)

这里的想法是与NaN的比较是不确定的. (无法确定结果).因此，有序/无序比较会检查情况是否如此.

The idea here is that comparisons with NaN are indeterminate. (can't decide the result) So an ordered/unordered comparison checks if this is (or isn't) the case.

double a = 0.;
double b = 0.;

__m128d x = _mm_set1_pd(a / b);     //  NaN
__m128d y = _mm_set1_pd(1.0);       //  1.0
__m128d z = _mm_set1_pd(1.0);       //  1.0

__m128d c0 = _mm_cmpord_pd(x,y);    //  NaN vs. 1.0
__m128d c1 = _mm_cmpunord_pd(x,y);  //  NaN vs. 1.0
__m128d c2 = _mm_cmpord_pd(y,z);    //  1.0 vs. 1.0
__m128d c3 = _mm_cmpunord_pd(y,z);  //  1.0 vs. 1.0
__m128d c4 = _mm_cmpord_pd(x,x);    //  NaN vs. NaN
__m128d c5 = _mm_cmpunord_pd(x,x);  //  NaN vs. NaN

cout << _mm_castpd_si128(c0).m128i_i64[0] << endl;
cout << _mm_castpd_si128(c1).m128i_i64[0] << endl;
cout << _mm_castpd_si128(c2).m128i_i64[0] << endl;
cout << _mm_castpd_si128(c3).m128i_i64[0] << endl;
cout << _mm_castpd_si128(c4).m128i_i64[0] << endl;
cout << _mm_castpd_si128(c5).m128i_i64[0] << endl;

结果:

0
-1
-1
0
0
-1

• NaN和1.0的有序比较得到false.
• NaN和1.0的无序比较给出true.
和1.0的
• 有序比较给出true.
和1.0的
• 无序比较给出false.
• NaN和Nan的有序比较得到false.
• NaN和NaN的无序比较给出true.
• Ordered comparison of NaN and 1.0 gives false.
• Unordered comparison of NaN and 1.0 gives true.
• Ordered comparison of 1.0 and 1.0 gives true.
• Unordered comparison of 1.0 and 1.0 gives false.
• Ordered comparison of NaN and Nan gives false.
• Unordered comparison of NaN and NaN gives true.

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