使用带阵列限制？ [英] Using restrict with arrays?

问题描述

有没有办法告诉C99编译，只有这样，我会定的数组访问是通过使用myArray的[指数]
说是这样的：

Is there a way to tell a C99 compiler that the only way I am going to access given array is by using myarray[index] ? Say something like this:

int heavy_calcualtions(float* restrict range1, float* restrict range2)
{
    float __I promise I won't alias this__ tmpvalues[1000] = {0};

    ....
    heavy calculations using range1, range2 and tmpvalues;
    ....
}

通过使用限制我答应，我不会别名范围1和范围，但我怎么做同样的事情阵列我的函数内声明？

By using restrict I promised that I won't alias range1 and range2 but how do I do the same thing for array declared inside my function ?

推荐答案

为什么不能你下面？您没有访问通过该变量， tmpvalues​​ 相关联的数据，因此它是有效的在code的计算密集型部分使用限制指针。

Why can't you do the following? You are not accessing the data associated with tmpvalues via that variable, so it is valid to use a restrict pointer in the compute-intensive portion of the code.

#include <stdio.h>
#include <stdlib.h>

int heavy_calcs(int n, float* restrict range1, float* restrict range2)
{
    if (n>1000) return 1;
    float tmpvalues[1000] = {0};
    {
        float * restrict ptv = tmpvalues;
        for (int i=0; i<n; i++) {
            ptv[i] = range1[i] + range2[i];
        }
    }
    return 0;
}

int main(int argc, char * argv[])
{
    int n = (argc>1) ? atoi(argv[1]) : 1000;
    float * r1 = (float*)malloc(n*sizeof(float));
    float * r2 = (float*)malloc(n*sizeof(float));
    int rc = heavy_calcs(n,r1,r2);
    free(r1);
    free(r2);
    return rc;
}

我跑这通过英特尔编译器15和它没有任何麻烦，矢量化循环。当然，这个循环是微不足道的相比，要什么，我认为你的是，所以你的里程可能会有所不同。

I ran this through the Intel 15 compiler and it had no trouble vectorizing the loop. Granted, this loop is trivial to compared to what I assume yours is, so your mileage may vary.

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