Lambda捕获：使用还是不使用初始化程序？ [英] Lambda capture: to use the initializer or not to use it?

问题描述

请考虑以下最小示例：

int main() {
    int x = 10;    
    auto f1 = [x](){ };
    auto f2 = [x = x](){};
}

我不止一次看到这样的初始化器 [x = x] ，但我不能完全理解它，为什么我应该使用它代替 [x] 。 br>
我可以得到 [& x = x] 或 [x = x + 1]之类的含义（如文档所示，以及为什么它们与 [x] 当然，但是仍然无法弄清示例中的lambda之间的差异。

I've seen more than once such an use of the initializer [x = x], but I can't fully understand it and why I should use it instead of [x].
I can get the meaning of something like [&x = x] or [x = x + 1] (as shown in the documentation and why they differ from [x], of course, but still I can't figure out the differences between the lambdas in the example.

它们是完全可互换的还是我看不到有什么区别？

Are they fully interchangeable or is there any difference I can't see?

推荐答案

有各种各样的极端情况，几乎都是沸腾的降至 [x = x] 衰减； [x] 不会衰减。

There are various corner cases that pretty much boils down to "[x = x] decays; [x] doesn't".

• 捕获对函数的引用：

• capturing a reference to function:

void (&f)() = /* ...*/;
[f]{};     // the lambda stores a reference to function.
[f = f]{}; // the lambda stores a function pointer

捕获数组：

capturing an array:

int a[2]={};
[a]{}     // the lambda stores an array of two ints, copied from 'a'
[a = a]{} // the lambda stores an int*

捕获具有简历资格的内容：

capturing a cv-qualified thing:

const int i = 0; 
[i]() mutable { i = 1; } // error; the data member is of type const int
[i = i]() mutable { i = 1; } // OK; the data member's type is int

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