Java的一元加运算符的目的是什么? [英] What is the purpose of Java's unary plus operator?

问题描述

Java 的一元加号运算符似乎是通过 C++ 从 C 过来的.

Java's unary plus operator appears to have come over from C, via C++.

int result = +1;   

它似乎有以下效果:

• 拆箱它的操作数，如果它是一个包装对象
• 将其操作数提升为 int，如果它不是 int 或更宽的
• 使包含大量连续加号的邪恶表达式的解析稍微复杂化

• Unboxes its operand, if it's a wrapper object
• Promotes its operand to int, if it's not already an int or wider
• Complicates slightly the parsing of evil expressions containing large numbers of consecutive plus signs

在我看来，有更好/更清晰的方法来完成所有这些事情.

It seems to me that there are better/clearer ways to do all of these things.

在 this SO question中，关于 C# 中的对应运算符，有人说如果你觉得需要，它就会超载."

In this SO question, concerning the counterpart operator in C#, someone said that "It's there to be overloaded if you feel the need."

然而，在Java中，不能重载任何运算符.那么这个一元加号运算符是否仅仅因为存在于 C++ 中就存在于 Java 中?

However, in Java, one cannot overload any operator. So does this unary plus operator exist in Java only because it existed in C++?

推荐答案

当其操作数的类型为 byte 时，一元加运算符执行自动转换为 int，char 或 short.这称为一元数字提升，它使您能够执行以下操作:

The unary plus operator performs an automatic conversion to int when the type of its operand is byte, char, or short. This is called unary numeric promotion, and it enables you to do things like the following:

char c = 'c';
int i = +c;

当然，它的用途有限.但它确实是有目的的.请参阅规范，特别是部分 §15.15.3 和 §5.6.1.

Granted, it's of limited use. But it does have a purpose. See the specification, specifically sections §15.15.3 and §5.6.1.

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