c++ - 该程序的输出是什么？

Question

#include<stdio.h>
void main()
{
    int i;
    float x=0.8;
    for(i=0; i<100; i++)
    x=x*(1-x)*5;
    printf("%f \n",x);
}

在我看来，答案0.80000000看起来很简单，但真正的答案是别的东西。其输出值“-inf”出现运行时错误。为什么？

Answer 1

这意味着负无穷大。

EEE 754 floating point numbers can represent positive or negative infinity, and NaN (not a number). These three values arise from calculations whose result is undefined or cannot be represented accurately. You can also deliberately set a floating-point variable to any of them, which is sometimes useful. Some examples of calculations that produce infinity or NaN:

这是您的代码的输出

0.800000 
0.800000 
0.800000 
0.800001 
0.799996 
0.800011 
0.799966 
0.800103 
0.799692 
0.800923 
0.797227 
0.808280 
0.774816 
0.872380 
0.556664 
1.233946 
-1.443383 
-17.633698 
-1642.905029 
-13503899.000000 
-911776526893056.000000 
-4156682286578109086379962007552.000000 
-inf 
after this -inf -inf

因为该值超出了 flat 的范围