为什么for_each + lambda触发器-Waggregate-return警告？ [英] Why does for_each + lambda trigger -Waggregate-return warning?

问题描述

当使用gcc和 -Waggregate-return 标志尝试以下示例时，警告：函数调用具有聚合值 triggers：

When trying the following example with gcc and the -Waggregate-return flag the warning: function call has aggregate value triggers:

struct Element {
// ... stuff ...
}

Container<Element> elements(10);
for_each(begin(elements),end(elements),[](Element& e){

// ... modify elements ...

});

据我所知， -Waggregate-return flag如果任何返回结构或联合的函数被定义或调用，警告，因为，如果我理解正确，你可能通过返回一个足够大的对象来溢出堆栈。

As far as I could find out, the -Waggregate-return flag "Warns if any functions that return structures or unions are defined or called", because, if I understood correctly, you could potentially overflow the stack by returning a large enough object.

但是， for_each 返回 lambda类型，类型为 void 。为什么会触发警告？我错过了什么？

However, for_each returns the type of the lambda, whose type is void. Why does it trigger the warning? What have I missed? How can I improve my code?

推荐答案

为什么会触发警告？

Why does it trigger the warning?

大概，通过返回类或联合类型的任何东西来触发警告。 Lambdas有类类型， for_each 返回其函数参数，因此将触发警告。

Presumably, the warning is triggered by returning anything of class or union type. Lambdas have class type, and for_each returns its function argument, so that will trigger the warning.

由 begin（elements）和 end（elements）返回的迭代器类型可能会触发警告， 容器类型实现迭代器。

It's also possible that the iterator type returned by begin(elements) and end(elements) might trigger the warning, depending on how the Container type implements iterators.

如何改进我的代码？ >

How can I improve my code?

我会禁用该警告;它不是真正地兼容惯用的C ++，因为它是很常见的从函数返回小类对象。它也会被例如 std :: map :: insert（）触发，它返回一个对和许多其他标准库函数。

I'd disable that warning; it's not really compatible with idiomatic C++, since it's very common to return small class objects from a function. It would also be triggered by, for example, std::map::insert(), which returns a pair, and many other standard library functions.

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