c - 浮点精度的细微差别

我在 NVIDIA 的 CUDA SDK 示例中找到了这段代码。

void computeGold( float* reference, float* idata, const unsigned int len)
{
    reference[0] = 0;
    double total_sum = 0;
    unsigned int i;
    for( i = 1; i < len; ++i)
    {
        total_sum += idata[i-1];
        reference[i] = idata[i-1] + reference[i-1];
    }
    // Here it should be okay to use != because we have integer values
    // in a range where float can be exactly represented
    if (total_sum != reference[i-1])
        printf("Warning: exceeding single-precision accuracy.  Scan will be inaccurate.\n");
}
//(C) Nvidia Corp

有人可以告诉我一个打印警告的案例，最重要的是，为什么。

最佳答案

该函数在编写时考虑了一定范围的输入数据。如果未满足该输入数据期望，则会打印警告:

#include <stdio.h>
#define COUNT_OF(x) (sizeof(x)/sizeof(0[(x)]))

void computeGold( float* reference, float* idata, const unsigned int len)
{
    double total_sum = 0;
    unsigned int i;

    reference[0] = 0;

    for( i = 1; i < len; ++i)
    {
        total_sum += idata[i-1];
        reference[i] = idata[i-1] + reference[i-1];
    }
    // Here it should be okay to use != because we have integer values
    // in a range where float can be exactly represented
    if (total_sum != reference[i-1])
        printf("Warning: exceeding single-precision accuracy.  Scan will be inaccurate.\n");
}
//(C) Nvidia Corp


float data[] = {
    1.0,
    2.0,
    3.0,
    4.0,
    5.0
};

float data2[] = {
    123456.0,
    999999.0,
    999999.0,
    999999.0,
    999999.0,
    999999.0,
    999999.0,
    999999.0,
    999999.0,
    999999.0,
    999999.0,
    999999.0,
    999999.0,
    999999.0,
    999999.0,
    999999.0,
    999999.0,
    999999.0,
    123456.0
};

float ref[COUNT_OF(data2)] = {0.0};

int main()
{
    computeGold( ref, data, COUNT_OF(data));
    computeGold( ref, data2, COUNT_OF(data2));
    return 0;
}

关于c - 浮点精度的细微差别，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/3048043/