为什么隐式“lambda到函数指针转换”禁止“通过引用”捕获静态成员？ [英] Why does the implicit "lambda to function pointer conversion" forbid the "by reference" capture of static members?

问题描述

C ++ 11标准（或至少是我的版本，不是最后一个）：

The C++11 standard says (or at least, the version I have - not the final one) :

类型为具有无lambda捕获的lambda表达式具有
公共非虚拟非显式const转换函数指针
到具有相同参数和返回类型的函数关闭
类型的函数调用操作符。

The closure type for a lambda-expression with no lambda-capture has a public non-virtual non-explicit const conversion function to pointer to function having the same parameter and return types as the closure type’s function call operator.

我理解为什么不能从有状态的lambda因为函数指针本身不能保存任何数据。

I understand why it is not possible to get a function pointer from a stateful lambda since a function pointer cannot hold any data by itself.

但是当捕获的对象只是静态成员/静态变量时，没有这样的限制，因为引用捕获的对象可以在函数本身中硬连线。

But when the captured objects are just a static members /static variable, there is no such limitation since the references to the captured objects can be hardwired in the function itself.

struct A {
    static int count = 0;
    void foo() {
         static int bar = 0;
         auto fun = [&]()->void {
             count++;
             bar++;
         };
         void(*ptrFun)();
         ptrFun = fun; // forbidden by the quoted wording
    }
};

为什么总是不可能将lambda转换为函数指针无国籍？

Why isn't it always possible to convert a lambda to a function pointer as soon as the former is stateless ? Am I missing something or does the committee forgot this specific point ?

推荐答案

A :: count

A::count does not need to be captured at all. Only this and local variables need to be captured. Variables with static storage duration (e.g., static data members, namespace-scope variables, or function-local static variables) do not need to be captured because they are "unique." There is exactly one instance of each such variable, so a reference to the object does not need to be captured.

如果从lambda中删除默认捕获（即，更改 [&] 到 [] ）并定义 count ，它应该编译没有错误。 （我已经验证Visual C ++ 2012 RC和g ++ 4.5.1接受代码;唯一的变化，我要做的是移动内联初始化 count ，因为这些编译器都不支持C ++ 11的功能。）

If you remove the default capture from your lambda (i.e., change [&] to []) and define count, it should compile without error. (I've verified that both Visual C++ 2012 RC and g++ 4.5.1 accept the code; the only change I had to make was to move the inline initialization of count, since neither of those compilers supports that C++11 feature yet.)

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