它对于一个lambda是有效的，本质上，关闭它自己吗？ [英] Is it valid for a lambda to, essentially, close over itself?

问题描述

这个lambda递归是否有效？

Is this lambda recursion valid?

#include <functional>
#include <iostream>

int main() {
   std::function<int(int)> g = [&g](int k) {
       return (k ? k * g(k-1) : 1);
   };

   std::cout << g(10); // 3628800
}

它似乎编译并运行确定，但我很紧张关闭 g 在我初始化它的语句。严格有效性为1-10 ...？

It appears to compile and run ok, but I'm nervous about closing over g in the same statement that I initialise it. Strict validity on a scale of 1-10...?

推荐答案

在捕获 g 通过引用已声明，因此名称可供使用：

At the point at which you capture g by reference, it has been declared, so the name is available for use:

/ 1一个名字的声明点紧接在其完整的声明器（第8条）之后和它的初始化器之前

3.3.2/1 The point of declaration for a name is immediately after its complete declarator (Clause 8) and before its initializer

在初始化之前以有限的方式使用对象 - 基本上，不依赖于值的任何东西都可以：

You are allowed to use objects in limited ways before they are initialised - basically, anything that doesn't depend on the value is OK:

3.8 / 6 before对象的生命周期已经开始，但在对象将占据的存储之后
已经被分配[...]可以使用引用原始对象的任何glvalue，但只能在有限的
方法中使用。 [...]使用不依赖于其值的glvalue的属性是明确定义的。

3.8/6 before the lifetime of an object has started but after the storage which the object will occupy has been allocated [...] any glvalue that refers to the original object may be used but only in limited ways. [...] using the properties of the glvalue that do not depend on its value is well-defined.

，你正在做的是定义明确的。

So by my understanding, what you are doing is well-defined.

（虽然，超级的，我不认为它是指定的自动对象的存储分配，8.3 .2 / 5说引用应该被初始化为引用一个有效的对象，而不定义有效，所以有范围可以认为它没有被很好地定义）。

(Although, being ultrapedantic, I don't think it's specified when the storage for an automatic object is allocated, and 8.3.2/5 says that "a reference shall be initialized to refer to a valid object" without defining "valid", so there's scope to argue that it's not well-defined).

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