带有循环迭代器和 omp simd 的私有子句

Question

当我尝试这个时

#pragma omp simd private(i)
for(i=0; i<n; i++)

使用 GCC 5.2.1 时出现错误

error: iteration variable ‘i’ should not be private

我对 firstprivate 和 lastprivate 也有同样的错误。我从这段代码中发现了这一点

int i;
#pragma omp parallel for simd reduction(+:sum) lastprivate(i)
for(i=0; i<(n/8)*8; i++) sum += a[i];
for(; i<n; i++) sum += a[i];

然而，这工作正常

int i,j;
#pragma omp parallel for simd reduction(+:sum) lastprivate(j)
for(i=0; i<(n/8)*8; i++) {sum += a[i]; j=i;}
for(; j<n; j++) sum += a[j];

但这似乎是一个愚蠢的 hack。

GCC 也给我警告

warning: ‘i’ may be used uninitialized in this function

with parallel for simd 但仅使用 parallel for.

时没有警告

为什么我不能用 omp simd 或 omp for simd 将并行循环迭代器显式声明为 private？反正不是隐含的private吗？

来自 Hristo Iliev 的回答中推荐的 OpenMP 4.0 规范第 14.1.2 节。

The loop iteration variable in the associated for-loop of a simd construct with just one associated for-loop may be listed in a linear clause with a constant-linear-step that is the increment of the associated for-loop.

The loop iteration variables in the associated for-loops of a simd construct with multiple associated for-loops may be listed in a lastprivate clause.

这个有效

int i;
#pragma omp parallel for simd reduction(+:sum) linear(i)
for(i=0; i<(n/8)*8; i++) sum += a[i];
for(; i<n; i++) sum += a[i];

但似乎也不需要 linear(i) 所以迭代器默认为 linear(i).

所以我本来可以做到的

 #pragma omp parallel for simd reduction(+:sum)

看来迭代器默认有效 lastprivate omp simd 和 omp for simd.

我仍然收到警告

‘i’ may be used uninitialized in this function

我应该选择第 14.1.1 节

Certain variables and objects have predetermined data-sharing attributes as follows:

The loop iteration variable in the associated for-loop of a simd construct with just one associated for-loop is linear with a constant-linear-step that is the increment of the associated for-loop.

这是一个工作示例，它给出带有 -fopenmp 的警告，但没有 -fopenmp

则没有警告

‘i’ may be used uninitialized in this function

代码

#include <stdio.h>
//#define N 101                                                                                                       
float foo(float *a, int n)
{
    float sum = 0;
    int i;
    #pragma omp parallel for simd reduction(+:sum)
    for(i=0; i<(n/8)*8; i++) sum += a[i];
    for(; i<n; i++) sum += a[i];
    return sum;
}

int main(void) {
    const int N = 101;
    float a[N]; for(int i=0; i<N; i++) a[i] = i;
    printf("%f\n", foo(a,N));
}

Answer 1

不，simd 构造中的循环迭代变量具有单个关联的 for 循环具有预定的数据共享 class linear 而不是 private（OpenMP 4.0, §2.14.1.1). linear 提供的功能是 private 的超集，因此将 i 声明为 (last)private 是降级，因此会出现错误。

带有循环迭代器和 omp simd 的私有子句

private clause with loop iterators and omp simd

c

gcc

openmp