iOS:在 64 位中使用 vDSP 功能 [英] iOS: using vDSP functions in 64bit

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本文介绍了iOS:在 64 位中使用 vDSP 功能的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

在 iOS 中,我有一个函数调用依赖于 vDSP_vgathr,Apple 加速框架:

In iOS, I have a function call that depends on vDSP_vgathr, a member of Apple's Accelerate framework:

int pix = image.size.height * image.size.width;
float * array = (float *)malloc(sizeof(float) * pix * 4);
float * colorPlane = (float *)malloc(sizeof(float) * pix);
float * y_r = (float *)malloc(sizeof(float) * pix)
int * malloc(sizeof(int) * pix)

vDSP_vgathr(colorPlane, (const vDSP_Length *)idx, 1, y_r, 1, (int)(rate*pix));

我看过一个SO问题与我得到的错误有关,EXC_I386_GPFLT,以及 64 位系统和非规范指针.解决方案(和另一个) 我已经看到建议包括 tgmath.h,但是当使用(成功编译)时,它什么也不做,我仍然得到 EXC_I386_GPFLT.

I've seen an SO question that relating to the error I get, EXC_I386_GPFLT, and 64 bit systems and non-canonical pointers. The solution (and another one) I've seen suggest including tgmath.h, but when used (with a successful compile), it does nothing and I still get EXC_I386_GPFLT.

我尝试了许多其他潜在的解决方案,包括类型转换各种对象和不同的函数调用.我怎样才能使这个函数调用适用于 64 位架构?

I've tried a host of other potential solutions including typecasting various objects and different function calls. How can I do I make this function call work with the 64 bit architecture?

推荐答案

vDSP_vgathr 的文档说它被定义为

The docs for vDSP_vgathr say it's defined as

void vDSP_vgathr (
   float *__vDSP_A,
   vDSP_Length *__vDSP_B,
   vDSP_Stride __vDSP_J,
   float *__vDSP_C,
   vDSP_Stride __vDSP_K,
   vDSP_Length __vDSP_N
);

您正在将一个 int 传递给该函数,并且 vDSP_Length 被定义为 unsigned long. 因此,将一个 unsigned long 传递给:

You're passing an int into that function and vDSP_Length is defined as unsigned long. So, pass an unsigned long in:

unsigned long * idxSend = (unsigned long *)malloc(sizeof(unsigned long) * N);
for (i=0; i<N; i++) idxSend[i] = (unsigned long)idx[i];
vDSP_vgathr(..., idxSend, ...);

这就解决了你的问题.

这篇关于iOS:在 64 位中使用 vDSP 功能的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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