c# – 为什么是(双)0.6f>(双)(6 / 10f)？

您看到上述行为的原因在于编译器 – 如果您在反射器中查看编译的程序集,那么这里的内容有点清楚：

(更新我已经改变了代码,以便它不使用Console.WriteLine,因为我意识到编译器为你选择一个重载,这困惑的情况)

// As written in source
var j = (double)(float)0.6;
var k = (double)0.6f;
var l = (double)(6/10f);
var m = (double)(float)(6/10f);

// Code as seen by Reflector
double j = 0.60000002384185791;
double k = 0.60000002384185791;
double l = 0.6;
double m = 0.6;

为什么编译器选择以这种特定的方式编译是超出我的(fyi,这是所有的优化被关闭)

一些有趣的其他情况：

// Code
var a = 0.6;
var b = (double)0.6;
var c = 0.6f;
var d = (float)0.6;

var e = 6 / 10;
var f = 6 / (10f);
var g = (float)(6 / 10);
var h = 6 / 10f;
var i = (double)6 / 10;
// Prints out 0.60000002384185791

double n = (float)0.6;
double o = f;

// As seen by Reflector
double a = 0.6;
double b = 0.6;
float c = 0.6f;
float d = 0.6f;

int e = 0;
float f = 0.6f;
float g = 0f;
float h = 0.6f;
double i = 0.6;

double n = 0.60000002384185791;
double o = f;

编译器似乎在几个特殊情况下做了上述的伎俩,为什么只有当转换为双精度时才完全超出我的范围！

其余的时间似乎做了一些诡计,使浮点算术似乎工作,其实通常不会.