c - 在C中循环后显示十进制值作为结果

Question

我正在尝试完成一个显示小数(作为百分比)的程序，其中从 1000 次中抽出一副牌中的方 block 、梅花、红心或黑桃。这是我完成的代码:

#include <stdio.h>

int main()
{
   unsigned int freq1 = 0;
    unsigned int freq2 = 0;
     unsigned int freq3 = 0;
      unsigned int freq4 = 0;

        for (unsigned int draw = 1; draw <= 1000; ++draw) {
            int face = 1 +rand() % 4;

            switch (face) {

                case 1:
                ++freq1;
                break;

                case 2:
                ++freq2;
                break;

                case 3:
                ++freq3;
                break;

                case 4:
                ++freq4;
                break;
            }
        }

        printf("Percent of diamonds is %u\n", freq1);
        printf("Percent of clubs is %u\n", freq2);
        printf("Percent of hearts is %u\n", freq3);
        printf("Percent of spades is %u\n", freq4);

}

当我运行代码时，它会从 1000 中获取正确的绘制数值。这是输入:

Percent of diamonds is 249                                                                                                                                                
Percent of clubs is 252                                                                                                                                                   
Percent of hearts is 258                                                                                                                                                  
Percent of spades is 241 

我试图让程序输入根据这些数字计算出的十进制值，如下所示:

Percent of diamonds is 0.25

到目前为止，我已尝试通过将结果声明为 float 并将每次绘制中计算的频率除以 1000 来纠正此问题，以便结果将是我尝试显示的十进制值。这是我尝试的代码:

 #include <stdio.h>

int main()
{
   unsigned int freq1 = 0;
    unsigned int freq2 = 0;
     unsigned int freq3 = 0;
      unsigned int freq4 = 0;
      float res1, res2, res3, res4;

        for (unsigned int draw = 1; draw <= 1000; ++draw) {
            int face = 1 +rand() % 4;

            switch (face) {

                case 1:
                ++freq1;
                break;

                case 2:
                ++freq2;
                break;

                case 3:
                ++freq3;
                break;

                case 4:
                ++freq4;
                break;
            }
        }

    res1= freq1/1000;
    res2= freq2/1000;
    res3= freq3/1000;
    res4= freq4/1000;

        printf("Percent of diamonds is %.2f\n", res1);
        printf("Percent of clubs is %.2f\n", res2);
        printf("Percent of hearts is %.2f\n", res3);
        printf("Percent of spades is %.2f\n", res4);
}

但是，当我尝试运行它时，我最终得到了以下结果输入:

Percent of diamonds is 0.00                                                                                                                                               
Percent of clubs is 0.00                                                                                                                                                  
Percent of hearts is 0.00                                                                                                                                                 
Percent of spades is 0.00   

对长篇文章表示歉意，但我希望能得到任何帮助来解决这个问题，因为这是我完成这个程序唯一需要解决的问题。谢谢!

Answer 1

问题是像 res1= freq1/1000 这样的东西首先被计算为整数，因为两个参数都是整数，然后转换为 float 。请改用 res1= freq1/1000.0 之类的内容，以便至少有一个参数是浮点型，第二个参数将自动转换为浮点型。