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Possible Duplicate: Disambiguating calls to functions taking std::functions Isn't the template argument (the signature) of std::function part of its type? I want to overload a function so that it can be called with a variety of different lambdas (generally with more or fewer arguments) naturally. The obvious thing I tried was: #include <functional> #include <iostream> extern void fn(std::function<void(int)>); extern void fn(std::function<void(int, int)>); void test() { fn([](int a) { std::cout << "lambda with 1 arg " << a << std::endl; }); } However, this fails with g++ (tried v4.6.2 and v4.7.1) with the error: test.cc: In function ‘void test()’: test.cc:9:74: error: call of overloaded ‘fn(test()::<lambda(int)>)’ is ambiguous test.cc:9:74: note: candidates are: test.cc:4:13: note: void fn(std::function<void(int)>) test.cc:5:13: note: void fn(std::function<void(int, int)>) Now I found an alternate (and much more complex) approaches here and here , but my question is, why does the above code fail? Is there something in the standard that says it can't work, or is this merely a bug/limitation of g++?

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