将double转换为float时精度损失为1

Question

对于下面的程序,我得到的精度损失为1,我无法理解.需要帮忙.

void main()
{
    typedef std::numeric_limits< double > dbl;
    cout.precision(dbl::digits10);

    double x = -53686781.0;
    float xFloat = (float) x;

    cout << "x :: " << x << "\n";
    cout << "xFloat :: " << xFloat << "\n";
}


Outpput:
x :: -53686781
xFloat :: -53686780

Answer 1

53686781在二进制文件中看起来像这样:11001100110011000111111101.那是26位.

您的浮点数在其尾数部分最多只能存储24位,因此,最终存储了110011001100110001111111.最后两个二进制数字01被截断.

11001100110011000111111100是53686780.

就如此容易.