Question

我想知道如何在 C语言

中生成精确到最多6点小数且在（-999到+999）范围内的随机浮点数位

我尝试过：

unsigned int i, num;
int divide;
FILE *f;

f=fopen("/dev/urandom", "r");
if (!f) {
        fprintf(stderr, "Failed open file\n");
        exit(1);
    }
for(i=0; i< w*w; i++)
{
    fread(&num, sizeof(unsigned int), 1, f);
    fread(&divide, sizeof(int), 1, f);
    h[i] = ((float)num)/((float)divide);

}
fclose(f);

但这会产生超过6个小数点。

有没有办法获得浮动没有。例如。 34.123456000000000000000，即在小数点后6位，应截断该值的其余部分。

@sas @Havenard @Lundin标记为重复的问题仍然无法解决问题。代码：

    #include <stdio.h>
#include <stdlib.h>
#include<math.h>
void fill(float* h, int w)
{
    unsigned int i, num;
    int divide;
    FILE *f;

    f=fopen("/dev/urandom", "r");
    if (!f) 
    {
            fprintf(stderr, "Failed open file\n");
            exit(1);
        }
    for(i=0; i< w*w; i++)
    {
        fread(&num, sizeof(unsigned int), 1, f);
        fread(&divide, sizeof(int), 1, f);
        h[i] = ((float)num)/((float)divide);
                h[i] = trunc(100*h[i])/100;   // trial for 2 place decimal
    }
    fclose(f);
} 

void getrand(int n )
{
        const int mybatch = 1;
    float mat1_size = sizeof(float) * n * n;
        float* mat1 = (float*) malloc(mat1_size);
    fill(mat1, n);
        for(int i=0; i<n; i++)
    {
            for(int j=0; j<n; j++)
        {   
            fprintf(stdout,"%12.10f    ",mat1[i*n+j]);

        }
            fprintf(stdout,"\n");
        }

        fprintf(stdout,"\n\n");
}
int main(int argc, char *argv[])
{
    if(argc!=2)
    {
        printf("Usage: %s <matrix_width>\n", argv[0]);
        return 0;
    }

    int w;
    w = atoi( argv[1] );

    getrand(w);
        return 0;
}

仍会打印：

$ ./a.out 5

-2.2799999714 3.5999999046 15.5200004578 -0.9900000095 -1.3099999428
  -1.8799999952 1.4700000286 2.0499999523 0.4900000095 2.0499999523
  0.9599999785 -0.4199999869 -0.4499999881 4.3200001717 -0.7699999809
  -2.9300000668 2.5099999905 -1.5900000334 -0.4199999869 -0.6800000072
  -0.2300000042 -1.7599999905 3.1800000668 0.6000000238 0.6000000238

Answer 1

您可以尝试使用trunc功能。

可能你看起来像这样。举个例子。

float a = 34.123456000000000000000;
a = trunc(1000000*a)/1000000; 
printf("%f\n", a);// output = 34.123455

注意 - 请在程序中加入math.h标题

How to truncate a floating point number after a certain number of decimal places (no rounding)?

可能重复

如何将随机生成的float截断为6位小数？

1 个答案: