Question

我第一次尝试使用英特尔C编译器，但我得到了完全错误的答案。我做错了什么？

我有一些代码如下：

#include <stdint.h>
#include <stdio.h>

#define CHUNK_SIZE 12
#define NUM_THREADS 8

#define popcnt __builtin_popcountll
#define BILLION (1000 * 1000 * 1000)
#define UPDATE_ROW_PPROD() \
    update_row_pprod(row_pprod, row, rows, row_sums, mask, mask_popcnt)

typedef __int128 int128_t;

static inline int64_t update_row_pprod
(
    int64_t* row_pprod, int64_t row, int64_t* rows,
    int64_t* row_sums, int64_t mask, int64_t mask_popcnt
)
{
    int64_t temp = 2 * popcnt(rows[row] & mask) - mask_popcnt;

    row_pprod[0] *= temp;
    temp -= 1;
    row_pprod[1] *= temp;
    temp -= row_sums[row];
    row_pprod[2] *= temp;
    temp += 1;
    row_pprod[3] *= temp;

    return row + 1;
}

int main(int argc, char* argv[])
{
    int64_t size = argc - 1, rows[argc - 1];
    int64_t row_sums[argc - 1];
    int128_t permanent = 0, sign = size & 1 ? -1 : 1;

    if (argc == 2)
    {
        printf("%d\n", argv[1][0] == '-' ? -1 : 1);
        return 0;
    }

    for (int64_t row = 0; row < size; row++)
    {
        char positive = argv[row + 1][0] == '+' ? '-' : '+';

        sign *= ',' - positive;
        rows[row] = row_sums[row] = 0;

        for (char* p = &argv[row + 1][1]; *p; p++)
        {
            rows[row] <<= 1;
            rows[row] |= *p == positive;
            row_sums[row] += *p == positive;
        }

        row_sums[row] = 2 * row_sums[row] - size;
    }

    #pragma omp parallel for reduction(+:permanent) num_threads(NUM_THREADS)
    for (int64_t mask = 1; mask < 1LL << (size - 1); mask += 2)
    {
        int64_t mask_popcnt = popcnt(mask);
        int64_t row = 0;
        int128_t row_prod = 1 - 2 * (mask_popcnt & 1);
        int128_t row_prod_high = -row_prod;
        int128_t row_prod_inv = row_prod;
        int128_t row_prod_inv_high = -row_prod;

        for (int64_t chunk = 0; chunk < size / CHUNK_SIZE; chunk++)
        {
            int64_t row_pprod[4] = {1, 1, 1, 1};

            for (int64_t i = 0; i < CHUNK_SIZE; i++)
                row = UPDATE_ROW_PPROD();

            row_prod *= row_pprod[0], row_prod_high *= row_pprod[1];
            row_prod_inv *= row_pprod[3], row_prod_inv_high *= row_pprod[2];
        }

        int64_t row_pprod[4] = {1, 1, 1, 1};

        while (row < size)
            row = UPDATE_ROW_PPROD();

        row_prod *= row_pprod[0], row_prod_high *= row_pprod[1];
        row_prod_inv *= row_pprod[3], row_prod_inv_high *= row_pprod[2];
        permanent += row_prod + row_prod_high + row_prod_inv + row_prod_inv_high;
    }

    permanent *= sign;

    if (permanent < 0)
        printf("-"), permanent *= -1;

    int32_t output[5], print = 0;

    output[0] = permanent % BILLION, permanent /= BILLION;
    output[1] = permanent % BILLION, permanent /= BILLION;
    output[2] = permanent % BILLION, permanent /= BILLION;
    output[3] = permanent % BILLION, permanent /= BILLION;
    output[4] = permanent % BILLION;

    if (output[4])
        printf("%u", output[4]), print = 1;
    if (print)
        printf("%09u", output[3]);
    else if (output[3])
        printf("%u", output[3]), print = 1;
    if (print)
        printf("%09u", output[2]);
    else if (output[2])
        printf("%u", output[2]), print = 1;
    if (print)
        printf("%09u", output[1]);
    else if (output[1])
        printf("%u", output[1]), print = 1;
    if (print)
        printf("%09u\n", output[0]);
    else
        printf("%u\n", output[0]);
}

如果我用

编译它

gcc -Wall -std=c99 -fopenmp  -o permanent permanent.c

然后我可以用

运行它

permanent -+ -+

并获取输出

-2

这是正确的。

如果我使用带有

的英特尔C编译器（17.0.1）进行编译

icc -std=c99 -qopenmp -Wall permanent.c

然后再做

a.out -+ -+

我得到了

11910984139051480114196905982

如评论中所述，如果删除-qopenmp，则icc会生成一个正确运行的版本，尽管只在一个核心上运行。

Answer 1

看起来像英特尔在__int128类型变量上处理OpenMP减少的错误。这很容易复制：

#include <stdio.h>
#include <inttypes.h>

int main() {
    __int128 sum = 0;
    #pragma omp parallel for reduction(+:sum)
    for (int i = 0; i < 0; i++) {
        sum += 1;
    }
    printf("%" PRIX64 " %" PRIX64 "\n", (uint64_t)sum, (uint64_t)(sum >> 64));
}

使用-fopenmp输出：

14000000000 78778300

在没有gcc -fopenmp的情况下正确输出：

0 0

您可以使用简单的自定义缩减声明解决此问题（并保留__int128类型）：

#pragma omp declare reduction(add128: int128_t: omp_out = omp_out + omp_in) initializer(omp_priv = 0)
[...]
#pragma omp parallel for reduction(add128:permanent) num_threads(NUM_THREADS)

据我所知，标准不限制缩减列表元素的类型。至少编译器应该告诉你它是否不受支持。

Answer 2

感谢测试用例，这绝对是还原变量私有副本初始化中的编译器错误。您可以自由地报告错误，但我们会尝试针对相应的编译器开发人员提交内部错误。

所以，另一个（不是很优雅）的解决方法可能是显式初始化私有副本，即替换

#pragma omp parallel for reduction(+:permanent) num_threads(NUM_THREADS)

与

#pragma omp parallel reduction(+:permanent) num_threads(NUM_THREADS)
{
  permanent = 0; // that should be done by compiler actually
  #pragma omp for
  ...
}

使用openmp时，英特尔C编译器输出错误

2 个答案: