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I am playing around with the c++11 functional features. One thing I find odd is that the type of a lambda function is actually NOT a function<> type. What's more, lambda's do not seem to play really well with the type-inferencing mechanism. Attached is a small example in which I tested flipping the two arguments of a function for adding two integers. (The compiler I used was gcc 4.6.2 under MinGW.) In the example, the type for addInt_f has been explicitly defined using function<> while addInt_l is a lambda whose type is type-inferenced with auto . When I compiled the code, the flip function can accept the explicitly type-defined version of addInt but not the lambda version, giving an error saying that, testCppBind.cpp:15:27: error: no matching function for call to 'flip(<lambda(int, int)>&)' The next few lines show that the lambda version (as well as a 'raw' version) can be accepted if it's explicitly cast to the appropriate function<> type. So my questions are: Why is it that a lambda function does not have a function<> type in the first place? In the small example, why does not addInt_l have function<int (int,int)> as the type instead of having a different, lambda type? From the perspective of functional programming, what's the difference between a function/functional object and a lambda? If there is a fundamental reason that these two have to be different. I heard that lambda's can be converted to function<> but they are different. Is this a design issue/defect of C++11, an implementation issue or is there a benefit in distinguishing the two as the way it is? It seems that the type-signature of addInt_l alone has provided enough information about the parameter and return types of the function. Is there a

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