c# - 为什么 (int)==(float) 总是编译为 (float)==(float)

Question

我正在研究 C# 编译器并试图理解数学运算规则。

我发现在两种不同的原始类型之间使用 == 运算符时会出现难以理解的行为。

int a = 1;
float b = 1.0f;        
Console.WriteLine(a == b);

这实际上编译为

.locals init (
    [0] int32,
    [1] float32
)

IL_0000: nop
IL_0001: ldc.i4.1
IL_0002: stloc.0
IL_0003: ldc.r4 1
IL_0008: stloc.1
IL_0009: ldloc.0
IL_000a: conv.r4
IL_000b: ldloc.1
IL_000c: ceq

这意味着

(float)a == (float)b

我的期望是(int)a == (int)b，因为左值是一个整数。

这个结果有什么原因吗？

---

• 这是我的猜测:int->float 比 float->int 快

Answer 1

这真的与速度无关(正如您所建议的)，更多的是与隐式转换有关，您可以在<主题下找到相关信息定义C# Specifications 中的 em>数字促销

12.4.7 Numeric promotions

Numeric promotion consists of automatically performing certain implicit conversions of the operands of the predefined unary and binary numeric operators. Numeric promotion is not a distinct mechanism, but rather an effect of applying overload resolution to the predefined operators. Numeric promotion specifically does not affect evaluation of user-defined operators, although user-defined operators can be implemented to exhibit similar effects.

As an example of numeric promotion, consider the predefined implementations of the binary * operator:

int operator *(int x, int y);
uint operator *(uint x, uint y);
long operator *(long x, long y);
ulong operator *(ulong x, ulong y);
float operator *(float x, float y);
double operator *(double x, double y);
decimal operator *(decimal x, decimal y);

When overload resolution rules (§12.6.4) are applied to this set of operators, the effect is to select the first of the operators for which implicit conversions exist from the operand types.

更进一步

Binary numeric promotion occurs for the operands of the predefined +, –, *, /, %, &, |, ^, ==, !=, >, <, >=, and <= binary operators. Binary numeric promotion implicitly converts both operands to a common type which, in case of the non-relational operators, also becomes the result type of the operation. Binary numeric promotion consists of applying the following rules, in the order they appear here:

• If either operand is of type decimal, the other operand is converted to type decimal, or a bindingtime error occurs if the other operand is of type float or double.
• Otherwise, if either operand is of type double, the other operand is converted to type double.
• Otherwise, if either operand is of type float, the other operand is converted to type float.
• Otherwise, if either operand is of type ulong, the other operand is converted to type ulong, or a binding-time error occurs if the other operand is of type sbyte, short, int, or long.
• Otherwise, if either operand is of type long, the other operand is converted to type long.
• Otherwise, if either operand is of type uint and the other operand is of type sbyte, short, or int, both operands are converted to type long.
• Otherwise, if either operand is of type uint, the other operand is converted to type uint.
• Otherwise, both operands are converted to type int.

你可以通过他们展示的例子来感受这一点

byte b = 1;
short a = 2;
WriteLine((int)b==(int)s); // promotes both to int

int i = 1;
double d = 2;
WriteLine((double)i==d); // promotes i to double

或者你的例子

int a = 1;
float b = 1.0f; 
WriteLine((float)a==b); // promotes a to float