为什么通过引用捕获变量的lambda不能转换为函数指针？ [英] Why isn't a lambda that captures variables by reference convertible to a function pointer?

问题描述

如果我有一个lambda可以通过引用捕获所有自动变量（ [&] {} ），为什么不能将其转换为函数指针？常规函数可以像通过引用捕获所有内容的lambda一样修改变量，所以为什么不一样呢？

If I have a lambda which captures all automatic variables by reference ([&] {}), why can't it be converted to a function pointer? A regular function can modify variables just like a lambda that captures everything by reference can, so why is it not the same?

我想换句话说，函数是什么

I guess in other words, what is the functional difference between a lambda with a & capture list and a regular function such that the lambda is not convertible to a function pointer?

推荐答案

让我们以一个小小的lambda为例：

So let's take the example of a trivial lambda:

Object o;
auto foo = [&]{ return o; };

foo 的类型是什么样的？可能看起来像这样：

What does the type of foo look like? It might look something like this:

struct __unique_unspecified_blah
{
    operator()() const {
        return o;
    }

    Object& o;
};

是否可以创建指向该 operator（）？不，你不能。该功能需要来自其对象的一些额外信息。这是您无法将典型的类方法转换为原始函数指针的原因（没有 this 所在的第一个自变量）。假设您确实创建了此指针-它如何知道从哪里获取 o ？

Can you create a function pointer to that operator()? No, you can't. That function needs some extra information that comes from its object. This is the same reason that you can't convert a typical class method to a raw function pointer (without the extra first argument where this goes). Supposing you did create some this pointer - how would it know where to get o from?

引用问题的一部分不相关-如果您的lambda捕获了任何东西，则其 operator（）将需要引用对象中的某种存储方式。如果需要存储，则无法转换为原始函数指针。

The "reference" part of the question is not relevant - if your lambda captures anything, then its operator() will need to reference some sort of storage in the object. If it needs storage, it can't convert to a raw function pointer.

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