c++ - 在 C++11 lambda 表达式中使用超出范围的变量

Question

我玩 C++11 是为了好玩。我想知道为什么会这样:

//...
std::vector<P_EndPoint> agents;
P_CommunicationProtocol requestPacket;
//...
bool repeated = std::any_of(agents.begin(), agents.end(),
                    [](P_EndPoint i)->bool 
                    {return requestPacket.identity().id()==i.id();});

编译因以下错误而终止:

error: 'requestPacket' has not been declared

在前面的代码中已声明。我试过 ::requestPacke 也没有用。

如何在 lambda 函数中使用外部范围变量？

Answer 1

您需要 capture the variable , 按值(使用 [=] 语法)

bool repeated = std::any_of(agents.begin(), agents.end(),
                    [=](P_EndPoint i)->bool                          
                    {return requestPacket.identity().id()==i.id();});

或通过引用(使用 [&] 语法)

bool repeated = std::any_of(agents.begin(), agents.end(),
                    [&](P_EndPoint i)->bool 
                    {return requestPacket.identity().id()==i.id();});

请注意，正如@aschepler 指出的那样，global variables with static storage duration are not captured , 只有函数级变量:

#include <iostream>

auto const global = 0;

int main()
{
    auto const local = 0;

    auto lam1 = [](){ return global; }; // global is always seen
    auto lam2 = [&](){ return local; }; // need to capture local

    std::cout << lam1() << "\n";
    std::cout << lam2() << "\n";
}