c++ - C++0x lambda, how can I pass as a parameter? -

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please @ following c++0x lambda related code:

typedef uint64_t (*weight_func)(void* param); typedef std::map<std::string, weight_func> callbacktable;  callbacktable table; table["rand_weight"] = [](void* param) -> uint64_t {   return (rand() % 100 + 1); };

i got error (in visual studio 2010) lambda couldn't converted type of weight_func. know answer: using std::function object:

typedef std::function<uint64_t (void*)>  weight_func;

however, want know how can receive type of lambda without using std::function. type should be?

the conversion function pointer relatively new: introduced n3043 on february 15, 2010.

while e.g. gcc 4.5 implements it, visual studio 10 released on april 12, 2010 , didn't implement in time. james pointed out, will fixed in future releases.

for moment have use 1 of alternative solutions provided here.

technically following workaround work, without variadic templates no fun generalize (boost.pp rescue...) , there no safety net against passing capturing lambdas in:

typedef uint64_t (*weightfunc)(void* param);  template<class func> weightfunc make_function_pointer(func& f) {     return lambda_wrapper<func>::get_function_pointer(f); }  template<class f> class lambda_wrapper {     static f* func_;     static uint64_t func(void* p) { return (*func_)(p); }         friend weightfunc make_function_pointer<>(f& f);         static weightfunc get_function_pointer(f& f) {         if (!func_) func_ = new f(f);         return func;     } };  template<class f> f* lambda_wrapper<f>::func_ = 0;  // ... weightfunc fp = make_function_pointer([](void* param) -> uint64_t { return 0; });

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