-1.#IND和-1.#INF等“无效”浮点数说明及其判断

常用的浮点数INF&NaN判断方法:

 Windows中，使用_isnan判断NaN情况；使用!_isnan && !_finite判断INF情况；使用!_finite判断NaN和INF两种情况

   //0/0的情况下或得到NAN，非零除以零会得到INF
    double spread=0.0;
    spread/=0.0;
    if (_isnan(spread))
    {
        cout<<"not a number"<<endl;
    }
    if(!_isnan(spread) && !_finite(spread))
    {
        cout<<"finite"<<endl;
    }
    if(!_finite(spread))
    {
        cout<<"nan and inf"<<endl;
    }

依据以下特性也可自己实现判断:

1.   NaN是唯一与自身不等的浮点数类型，所以IsNan可以用 x != x来判断。Anything compared with NAN is false, so NAN == NAN is false.

2.   INF阶码全1，尾数全0

int isNaN(double x) {return x != x; }
#define _INF_DEFINE 1.0/0.0
int isInf(double n)
{
    static double pinf = 0.0;
    static double ninf = 0.0;
    if (pinf == 0.0) {
            pinf = _INF_DEFINE;
            ninf = -pinf;
    }
    return memcmp(&n, &pinf, sizeof(n)) == 0
            || memcmp(&n, &ninf, sizeof(n)) == 0;
}