字符串隐式转换列表，例如+ = [英] List of String implicit conversions like +=

问题描述

请考虑以下代码段. + =不是java.lang.String的成员，所以我猜正在进行某种隐式转换.如何找到作用在String上的此类预定义隐式转换的列表?

Consider the following snippet. += is not a member of java.lang.String, so I guess there is some sort of Implicit conversion going on. How do I find a list of such predefined implicit conversions acting on String?

scala> var x = "asdf"
x: java.lang.String = asdf

scala> x += x

scala> x
res2: java.lang.String = asdfasdf

推荐答案

您选择了一个特别糟糕的示例.从某种意义上说，+=是String的一部分.请参阅Javadoc上的java.lang.String注释:

You picked a particularly bad example. += is, in a sense, part of String. See this comment on the Javadoc for java.lang.String:

Java语言为字符串提供了特殊的支持 并置运算符(+)，并将其他对象转换为 字符串.

The Java language provides special support for the string concatenation operator ( + ), and for conversion of other objects to strings.

您必须查找 Java语言规范，以查找有关该规范的更多信息(15.18.1 ).但是，同样，Scala不是Java，因此+也是

You'll have to look up Java language specification to find more information about it (15.18.1). But, then again, Scala is not Java, so + is also part of the Scala language specification (12.3.1).

到目前为止，我所说的是+，而不是+=.但是，Scala具有特殊的语法糖来分配.如6.12.4节所述，除了<=，>=，!=和以=开头的运算符外，任何以等号结尾的运算符(请参见第1章中的运算符")如果它不存在，则重新解释.具体来说，

So far I have spoken of +, not +=. However, Scala has a special syntactic sugar for assignment. As described in section 6.12.4, except for <=, >=, != and operators starting with an =, any operator symbol (see "operator characters" in chapter 1) that ends in an equal sign will be reinterpreted if it does not exist as a method. Specifically,

x += 1

将重新解释为

x = x + 1

无论x是否为var，都将发生这种情况，因此有时可能会看到错误消息"reassignment to val".

That will happen regardless of whether x is a var, so one might occasionally see an error message "reassignment to val".

因此，如您所见，通过在Scala规范中复制的Java规范中的异常以及一些语法糖，+=确实是String的一部分.

So, as you can see, += is really part of String, through an exception in Java specification that was replicated in Scala specification, plus a bit of syntactic sugar.

这并不意味着java.lang.String中没有没有可以通过隐式转换与之配合使用的方法.但是，我将其留给其他答案.如果我是你，我将更改问题中的方法以使其正确.此外，+=在堆栈溢出中不可搜索.

Which doesn't mean there aren't methods not in java.lang.String that can be used with it through implicit conversions. I'll leave that to the other answers, however. If I were you, I'd change the method in the question, to make it correct. Besides, += is unsearchable in Stack Overflow.

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