是否可以安全地假设相同的lambda表达式具有不同的类型? [英] Is it safe to assume that identical lambda expressions have different types?

问题描述

我正在尝试使用lambda和不同的lambda表达式具有不同类型的事实，即使它们是相同的.考虑一下这段代码

I am experimenting with lambdas and the fact that different lambda expressions have different types, even though they are the same. Consider this code

#include <iostream>

template <typename T> void once(T t){
    static bool first_call = true;
    if (first_call) t();
    first_call = false;
}

int main() {    
    int counter = 0;
    auto a = [&counter](){counter++;};
    once(a);
    once(a);
    std::cout << counter;              // 1

    auto b = a;                        // same type
    once(b);
    std::cout << counter;              // 1

    auto c = [&counter](){counter++;}; // different type
    once(c);
    once(c);               
    std::cout << counter;              // 2
}

此打印品112，即a和b当然是相同的类型，而c具有不同的类型.

This prints 112, ie a and b are of course of same type and c has a different type.

是否允许编译器使c与a具有相同的类型?

Is the compiler allowed to let c be of the same type than a?

我的意思是表达式是相同的，这将是显而易见的优化.

I mean the expressions are identical and it would be an obvious optimization.

PS:如果捕获阻止了这种优化，那么没有捕获的lambda呢?

PS: In case the capture prevents such an optimization, then what about lambdas without capture?

相关:什么是类型签名一个c ++ 11/1y lambda函数?和

推荐答案

是否允许编译器使c与a具有相同的类型?

Is the compiler allowed to let c be of the same type than a?

不. [&counter](){counter++;}是lambda表达式，并且每个 [expr. prim.lambda.closure]/1 :

No. [&counter](){counter++;} is a lambda expression and per [expr.prim.lambda.closure]/1:

lambda表达式的类型(也是闭包对象的类型)是一种唯一的，未命名的非工会类类型，称为闭包类型，其属性如下所述.

The type of a lambda-expression (which is also the type of the closure object) is a unique, unnamed non-union class type, called the closure type, whose properties are described below.

因此，对于每个lambda表达式，即使它与先前的表达式相同，您也将获得唯一的类型.

So, for each lambda expression, even if it is identical to a previous one, you will get a unique type.

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