C++ initialize variable with lambda

Question

#include <iostream>

using namespace std;

int main()
{

    static bool temp([]{ 
        cout <<"Hi ";
        return false;});


   cout <<"temp "<< temp;

   return 0;
}

It does not execute lambda. But if we declare lambda separately like:

#include <iostream>

using namespace std;

int main()
{
    auto lambda = []{ 
        cout <<"Hi ";
        return false;};

    static bool temp(lambda());


   cout <<"temp "<< temp;

   return 0;
}

It will execute it. What am I missing here?

You're not calling it in the first example.

David G
– David G

2019-12-18 01:16:16 +00:00
Commented Dec 18, 2019 at 1:16 — David G
– David G, Commented Dec 18, 2019 at 1:16
Try changing [] to [=] and see what happens.

Kostas
– Kostas

2019-12-18 01:37:31 +00:00
Commented Dec 18, 2019 at 1:37 — Kostas
– Kostas, Commented Dec 18, 2019 at 1:37

songyuanyao · Accepted Answer · 2019-12-18 01:28:29Z

12

You need to invoke the lambda, just as the 2nd code snippet does.

static bool temp([]{ 
    cout <<"Hi ";
    return false;}());
//                ^^

LIVE

PS: In the 1st code snippet temp will always be initialized as true, because lambda without capture list could convert to function pointer implicitly; which is a non-null pointer and then could convert to bool with value true.

edited Dec 18, 2019 at 1:28

answered Dec 18, 2019 at 1:16

songyuanyao

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