简写赋值运算符，+ =，是真的吗? [英] Short hand assignment operator, +=, True Meaning?

问题描述

我了解到i+=2是i=i+2的简写.但是现在我对此表示怀疑. 对于以下代码，以上知识没有帮助:

I learnt that i+=2 is the short-hand of i=i+2. But now am doubting it. For the following code, the above knowledge holds no good:

byte b=0; b = b + 2;//错误:必填字节，找到int

byte b=0; b=b+2; //Error:Required byte, Found int

上面的代码是合理的，因为2是int类型，并且表达式返回int值.

The above code is justifiable, as 2 is int type and the expression returns int value.

但是，以下代码运行良好:

But, the following code runs fine:

byte b=0; b+=2; //b stores 2 after += operation

这迫使我怀疑+=速记运算符比我所知道的还要更多. 请赐教.

This is forcing me to doubt that the += short-hand operator is somewhat more than I know. Please enlighten me.

推荐答案

如有疑问，您可以随时检查Java语言规范.在这种情况下，相关部分是15.26.2，复合赋值运算符.

When in doubt, you can always check the Java Language Specification. In this case, the relevant section is 15.26.2, Compound Assignment Operators.

形式为E1 op = E2的复合赋值表达式是等效的 到E1 =(T)((E1)op(E2))，其中T是E1的类型，只是E1仅被评估一次.

A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T) ((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once.

因此，您几乎是正确的，除了还添加了强制类型转换.在您的情况下: b+=2;符合b=(byte)(b+2);

So you were almost correct, except that a cast is added as well. In your case: b+=2; qualifies to b=(byte)(b+2);

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