Java:空间在编译中有所作为? [英] Java : space makes a difference in compilation?

问题描述

我正在编写一个程序(有点像Piglatin ...)，在其中我无意中错过了语句中的变量:

I was making a program (A Piglatin sort of...), in which I unintentionally missed a variable in the statement:

String a = "R"++'a';

它实际上应该是 String a ="R" + text +'a'; .编译器产生一个错误.但是，当我做到这一点时:

It should actually have been String a = "R"+text+'a';. The compiler produced an error. But, when I made it:

String a = "R"+ +'a';

程序已编译.

我想知道，为什么Java不在乎是否在某些语句中放置空格，例如: String a ="ABCD"; 与字符串a ="ABCD";

I am wondering why putting a space made the difference even though Java does not care whether you put a space or not in certain statements, like : String a="ABCD"; is the same as String a = "ABCD";

有人可以解释这种行为吗?

Can someone please explain this behavior?

推荐答案

++ 本身就是 operator (先递增或后递增).

++ is an operator in its own right (pre or post increment).

将其放在字符串和char文字之间在语法上无效.

Putting it between a string and a char literal is not syntactically valid.

但是使用"R" + +'a'，第二个 + 将绑定到字符常量 a 并将其用作一元加号运算符(此运算符具有很高的优先级).这不是不是的禁止操作:在Java中，它具有将 a 的 type 提升为 int .这种类型的提升表示输出将是 R97 而不是 Ra (97是 a 的ASCII码).其余的 + 充当字符串连接器.

But with "R"+ +'a', the second + will bind to the char literal a and will act as the unary plus operator (this operator has a very high precedence). This is not a no-op: in Java it has the effect of promoting the type of a to an int. This type promotion means that the output will be R97 rather than Ra (97 is the ASCII number for a). The remaining + acts as the string concatenator.

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