C++11 lambda and template specialization

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忘了有多久
忘了有多久 2021-02-06 17:05

I would like to know what is the correct type definition for the lambda presented below, so that the following code will compile using a conformant c++11 compiler:



        
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  • 2021-02-06 17:45

    It's auto + decltype:

    auto l = [](int i) -> bool { printf("%d",i); return true; };
    foo<decltype(l)> fi(l);
    fi.fum();
    

    Every single lambda has a different, unique, unnamed type. You, as a coder, just can not name it.

    However, in your case, since the lambda doesn't capture anything (empty []), it is implicitly convertible to a pointer-to-function, so this would do:

    foo<bool(*)(int)> fi([](int i) -> bool { printf("%d",i); return true; });
    fi.fum();
    
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  • 2021-02-06 17:53

    It's std::function<bool(int)>. Or possibly just bool(*)(int) if you prefer, since the lambda doesn't capture.

    (The raw function pointer might be a bit more efficient, since std::function does (at least in some cases) require dynamic allocation for some type erasure magic.)

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