using out of scope variables in C++11 lambda expressions

Question

I'm playing with C++11 for fun. I'm wondering why this happens:

//...
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();});

Compilation terminates with this error:

error: 'requestPacket' has not been declared

Which is declared earlier in code. I tried ::requestPacke and it doesn't worked too.

How can I use an external scope variable inside a lambda function?

They are inside a method of a class. That's not a good title, maybe I should change it to out of current scope... — Soroush Rabiei
– Soroush Rabiei, Commented Jun 4, 2013 at 12:55

Community · Accepted Answer · 2017-05-23 11:55:05Z

40

You need to capture the variable, either by value (using the [=] syntax)

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

or by reference (using the [&] syntax)

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

Note that as @aschepler points out, global variables with static storage duration are not captured, only function-level variables:

#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";
}

edited May 23, 2017 at 11:55

CommunityBot

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answered Jun 4, 2013 at 12:50

TemplateRex

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1 Comment

aschepler Over a year ago

Lambdas never capture globals, only function-local variables.

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