为什么 printf 舍入浮点数? [英] Why printf round floating point numbers?

问题描述

我正在尝试使用 printf 打印一些浮点数.例如:

I am trying to print some floating point numbers using printf. For example:

int main()
{
    printf("%.1f",76.75); 
    return 0;
}

输出:76.8

我对结果有一些疑问.

首先，为什么不打印 76.7?

First of all, why didn't it print 76.7?

第二，如何四舍五入?

推荐答案

C99 §7.19.6.1 fprintf 函数

f,F

表示浮点数的 double 参数以 [−]ddd.ddd 样式转换为十进制表示法，其中小数点后的位数-点字符等于精度规范.如果精度缺失，则取6；如果精度为零且未指定 # 标志，则不会出现小数点字符.如果出现小数点字符，则在其前面至少出现一位数字.该值四舍五入到适当的位数.

A double argument representing a floating-point number is converted to decimal notation in the style [−]ddd.ddd, where the number of digits after the decimal-point character is equal to the precision specification. If the precision is missing, it is taken as 6; if the precision is zero and the # flag is not specified, no decimal-point character appears. If a decimal-point character appears, at least one digit appears before it. The value is rounded to the appropriate number of digits.

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