Java的一元加运算符的目的是什么？

发布于 2024-08-28 18:22:53 字数 485 浏览 7 评论 0 原文

Java 的一元加运算符似乎是通过 C++ 从 C 而来的。

int result = +1;

它似乎具有以下效果：

如果它是包装对象，则将其操作数拆箱
如果它还不是 int 或更宽的
操作数，则将其提升为 int 使解析稍微复杂化包含大量连续加号的邪恶表达式

在我看来，有更好/更清晰的方法来完成所有这些事情。

在这个SO问题中，关于C#中的对应运算符，有人说“如果你觉得有需要，它就可以超载。”

然而，在 Java 中，不能重载任何运算符。那么这个一元加运算符在 Java 中存在只是因为它在 C++ 中存在吗？

原文

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

int result = +1;

It appears to have the following effects:

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

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

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

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

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つ低調成傷 2024-09-04 18:22:53

当操作数的类型为 byte、char 或 short 时，一元加运算符会自动转换为 int >。这称为一元数字提升，它使您能够执行以下操作：

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

当然，它的用途有限。但它确实有一个目的。请参阅规范，特别是 §15.15.3 和 §5.6.1。

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;

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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无戏配角 2024-09-04 18:22:53

下面是一元加号对 Character 变量执行的操作的简短演示：

private static void method(int i){
    System.out.println("int: " + i);
}

private static void method(char c){
    System.out.println("char: " + c);
}

public static void main(String[] args) {
    Character ch = 'X';
    method(ch);
    method(+ch);        
}

运行该程序的输出是：

char: X
int: 88

工作原理：一元 + 或 - 取消装箱其操作数，如果它是包装对象，则将它们的操作数提升为 int（如果还不是 int 或更宽的值）。因此，正如我们所看到的，虽然第一次调用 method 将选择 char 重载（仅拆箱），但第二次调用将选择 int > 方法的版本。由于应用了一元加法，Character 类型的变量 ch 将作为 int 参数传递到 method 中。

Here's a short demonstration of what the unary plus will do to a Character variable:

private static void method(int i){
    System.out.println("int: " + i);
}

private static void method(char c){
    System.out.println("char: " + c);
}

public static void main(String[] args) {
    Character ch = 'X';
    method(ch);
    method(+ch);        
}

The output of running this programme is:

char: X
int: 88

How it works: Unary + or - unbox their operand, if it's a wrapper object, then promote their operand to int, if not already an int or wider. So, as we can see, while the first call to method will choose the char overload (unboxing only), the second call will choose the int version of method. Variable ch of type Character will be passed into method as int argument because of the applied unary plus.