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()$ c的函数指针$ c>?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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