C语言中的浮点数和双精度 [英] Float and double precision in C

问题描述

在C语言中，双精度数比浮点数更精确，根据"C Primerplus第六版"书(第80页)，浮点数可以表示至少6个有效数字，而双精度数可以表示至少13个有效数字.因此，我尝试通过以下简单示例进行验证:

In C, double has more precision than float, and according to "C primerplus sixth edition" book (page 80), a float can represent at least 6 significant figures and a double can represent at least 13 significant figures. So I tried to verify that with this simple example:

#include<stdio.h>

int main(void){
    float a = 3.3333333; // 7 significant digits
    double b = 3.33333333333333;// 14 significant digits

    printf("\nFloat:  %f\n", a);
    printf("Double: %f\n", b);

    return 0;
} 

这是该程序的输出:

Float : 3.333333
Double: 3.333333

为什么double值与float值具有相同的精度，而不显示更多的有效数字?

Why does the double value have the same precision as the float value, instead of displaying more significant digits?

推荐答案

可以通过参考C标准来回答大多数此类问题:

Most questions like this can be answered by consulting the C standard:

每个转换规范均以'％'字符...开头，随后依次出现以下内容:

Each conversion specification is introduced by the '%' character ... after which the following appear in sequence:

...

• 一种可选的精度，它给出...在a，A，e，E，f和F转换说明符的基数字符后出现的位数.

描述f说明符:

双精度参数应转换为"[-] ddd.ddd"格式的十进制表示法，其中基数字符后的位数等于精度规范. 如果精度丢失，则应取为6 .

因此，只需使用%f，即指示printf在.之后打印六位数字.如果要查看更多数字，则需要指定精度:例如%.15f.

So, by simply using %f, you are instructing printf to print six digits after the .. If you want to see more digits, you need to specify the precision: %.15f, for example.

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