浮点常数

问题描述

限时删除！！

对于以下代码:

#include<stdio.h>

int main()
{
    float a = 0.7;
    printf("%.10f %.10f\n", 0.7f, a);
    return 0;
}

我得到的输出是:

请解释为什么将a打印为0.6999999881而将文字常量打印为0.7000000000吗?

Please explain why a is printed as 0.6999999881 while the literal constant is printed as 0.7000000000 ?

在这种情况下，是否使用浮点常量取决于编译器?

Is the use of a floating point constant in this case compiler-dependent?

推荐答案

如果编译器将FLT_EVAL_METHOD定义为1或2，则为printf("%.10f",0.7f)打印的字符串获得"0.7000000000"是正常现象.

Obtaining "0.7000000000" as the string printed for printf("%.10f",0.7f) is normal behavior if the compiler defines FLT_EVAL_METHOD as 1 or 2.

实际上，在那种模式下，浮点常量可以精度为

Indeed, in that mode, floating-point constants can be represented at a precision beyond that of their type (C11 5.2.4.2.2:9):

换句话说，对于下面的修改程序，打印0.7000000000 0.6999999881 FLT_EVAL_METHOD=2 是一种可能的行为.

In other words, printing 0.7000000000 0.6999999881 FLT_EVAL_METHOD=2 is one possible behavior for the modified program below.

#include<stdio.h>
#include <float.h>

int main()
{
    float a=0.7;
    printf("%.10f %.10f FLT_EVAL_METHOD=%d\n",0.7f, a, (int)FLT_EVAL_METHOD);
    return 0;
}

因为编译器将FLT_EVAL_METHOD定义为2，所以将printf("%.10f %.10f", 0.7f, 0.7)视为printf("%.10f %.10f", (double)0.7L, (double)0.7L).同样，printf("%.60Lf %.60Lf\n", (long double)0.7f, (long double)0.7)等同于printf("%.60Lf %.60Lf\n", 0.7L, 0.7L).

Because the compiler defines FLT_EVAL_METHOD to 2, printf("%.10f %.10f", 0.7f, 0.7) is treated as if it was printf("%.10f %.10f", (double)0.7L, (double)0.7L). Similarly, printf("%.60Lf %.60Lf\n", (long double)0.7f, (long double)0.7) is equivalent to printf("%.60Lf %.60Lf\n", 0.7L, 0.7L).

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