使用NEON内在函数除以浮点数 [英] Divide by floating-point number using NEON intrinsics

问题描述

我当时正在处理四个像素的图像，这在Android应用程序的armv7上.

I'm processing an image by four pixels at the time, this on a armv7 for an Android application.

我想将一个float32x4_t向量除以另一个向量，但是其中的数字从大约0.7到3.85不等，在我看来，除法的唯一方法是使用右移，但这是2^n的数字.

I want to divide a float32x4_t vector by another vector but the numbers in it are varying from circa 0.7 to 3.85, and it seems to me that the only way to divide is using right shift but that is for a number which is 2^n.

此外，我是新手，因此欢迎您提供任何建设性的帮助或意见.

Also, I'm new in this, so any constructive help or comment is welcomed.

示例:

如何使用NEON内部函数执行这些操作?

How can I perform these operations with NEON intrinsics?

float32x4_t a = {25.3,34.1,11.0,25.1};
float32x4_t b = {1.2,3.5,2.5,2.0};
//    somthing like this
float32x4 resultado = a/b; // {21.08,9.74,4.4,12.55}

推荐答案

NEON指令集没有浮点除法.

The NEON instruction set does not have a floating-point divide.

如果您先验知道您的值缩放比例不佳，并且您不需要正确的舍入(如果您要进行图像处理，几乎可以肯定是这种情况)，那么您可以使用倒数估算，细化步骤，然后乘以而不是除法:

If you know a priori that your values are not poorly scaled, and you do not require correct rounding (this is almost certainly the case if you're doing image processing), then you can use a reciprocal estimate, refinement step, and multiply instead of a divide:

// get an initial estimate of 1/b.
float32x4_t reciprocal = vrecpeq_f32(b);

// use a couple Newton-Raphson steps to refine the estimate.  Depending on your
// application's accuracy requirements, you may be able to get away with only
// one refinement (instead of the two used here).  Be sure to test!
reciprocal = vmulq_f32(vrecpsq_f32(b, reciprocal), reciprocal);
reciprocal = vmulq_f32(vrecpsq_f32(b, reciprocal), reciprocal);

// and finally, compute a/b = a*(1/b)
float32x4_t result = vmulq_f32(a,reciprocal);

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