C＃字节加法运算 [英] C# byte addition operation

问题描述

可能有人点我，为什么在这里：

Could someone point me, why here:

Byte b = 100; 
b = (Byte)(b+200);

我必须使用显式类型转换。但在这里

I have to use explicit type conversion. But here

Byte b = 100;
b += 200;

我不需要做到这一点？

I don't need to do this?

难道编译器生成不同的IL $ C $下这两个案？而这种情况下，是更好吗？

Does compiler generate different IL code for this two cases? And which case is better?

推荐答案

由于在标准允许的话（见第二下面的情况下）：

Because the standard permits it (see the second case below):

的形式的操作X 的运算的 = 是通过应用二元运算符重载解析处理（§14.2.4），好像操作是写 X 的运算的是。然后，

14.14.2 Compound assignment

An operation of the form x op= y is processed by applying binary operator overload resolution (§14.2.4) as if the operation was written x op y. Then,

  

• 如果选定的运算符的返回类型可隐式转换到 X型，该运算按 X = X 的运算的是，但 X 只计算一次

• If the return type of the selected operator is implicitly convertible to the type of x, the operation is evaluated as x = x op y, except that x is evaluated only once.

否则，如果选定的运营商为p $ pdefined运营商，如果所选运算符的返回类型是显式转换到 X A型$，如果是隐式转换为对 X型或运算符是移位运算符，则操作进行评估为 X =（T）（X 的运算的 Y），其中 T 是 X ，但 X 只计算一次。

Otherwise, if the selected operator is a predefined operator, if the return type of the selected operator is explicitly convertible to the type of x, and if y is implicitly convertible to the type of x or the operator is a shift operator, then the operation is evaluated as x = (T)(x op y), where T is the type of x, except that x is evaluated only once.

在IL code应该是在这种情况下基本相同。当然，如果评估 B 有副作用，它会在两次评估B =（字节）B + 200 情况下，只有一次使用复合赋值的时候。

The IL code should be essentially identical in this case. Of course, if evaluating b has side effects, it will be evaluated twice in the b = (byte)b + 200 case and only once when using compound assignment.

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