强制Java Lambda表达式捕获Java中的非最终变量

问题描述

是否可以让lambda表达式捕获不是有效的final变量?

Is it possible to have a lambda expression capture a variable that is not effectively final?

我知道我的代码可以捕获它们，因此可以很好地工作，目前，我必须创建一个新变量，该变量将要传递给lambda表达式的非最终变量复制.

I know that my code would work perfectly if it could capture them, currently I have to create a new variable that copies the non-final variable I want to pass into a lambda expression.

final SomeClass otherObj;

for(int i = 0; i < 10; i++) {
    int a = i;
    someObject.method(() -> otherObj.process(a, a+1))
}

我想传递"i"，而不用创建一个新的临时变量.

I'd like to pass in 'i' instead of having to create a new temporary variable.

推荐答案

您可以使用java8进行尝试，

You can try this with java8,

IntStream.range(0, 10).forEach(i ->{
   someObject.method(() -> otherObj.process(i, i+1))
});

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