Lambda表达式中的变量赋值 [英] Variable assignment in lambda expression

问题描述

我有以下代码片段:

    SomeClass someClass;
    switch (type) {
        case FIRST:
            someClass = new SomeClass();
            break;
        case SECOND:
            OptionalLong optional = findSomeOptional();
            optional.ifPresent(value -> someClass = new SomeClass(value));
    }

并且我试图在lambda表达式中将新对象分配给 someClass 引用，但是随后出现错误消息:在lambda中使用的变量应该有效地是最终的" .

And I'm trying to assign new object to someClass reference in lambda expresion but then I've got error message: "variable used in lambda should be effectively final".

当我在someClass的声明中添加 final 时，出现了另一个错误无法为最终变量赋值"

When I add final to declaration of someClass I got another error "cannot assign value to final variable"

那么我该如何巧妙地处理lamda中的此类征兆?

So how can I smartly deal with such assigment in lamdas?

推荐答案

简单的答案是您不能在lambda表达式中分配上层的局部变量.

您可以将变量变成实例成员，或者使用简单的if语句:

The simple answer is you cannot assign local variables from upper levels in lambda expressions.

Either, you turn your variable into an instance member, or use an simple if statement:

SomeClass someClass;
switch (type) {
  case FIRST:
    someClass = new SomeClass();
  break;
  case SECOND:
    OptionalLong optional = findSomeOptional();
    if(optional.isPresent()) {
      someClass = new SomeClass(optional.getAsLong());
    }
}

最后一个选择是使用AtomicReference.

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