Boost::Function: Getting a pointer to a template function

Hi! I want to use a boost::function pointer to a template function. The template type should be determinated at the time the boost::function object is initialized. Example: template<class T> inline string addNumber(T inputNumber) { return lexical_cast<string>(inputNumber + 2); } template<class numType> inline boost::function<string(numType)> getFunction(numType number) { boost::function<string(numType)> f; f = //Which code can I use here? If I call f() I want to call addNumber<numType>(number) } ---------------------- I want to use it like that: void callFunction() { boost::function<string(double)> f; f = getFunction<unsigned long>(); for(int i = 0; i < 10000;i++) { std::cout << f(rand()) << endl; } } Thanks in advance, Hans

AMDG Hans Mull wrote:
Hi! I want to use a boost::function pointer to a template function. The template type should be determinated at the time the boost::function object is initialized. Example:
template<class T> inline string addNumber(T inputNumber) { return lexical_cast<string>(inputNumber + 2); }
template<class numType> inline boost::function<string(numType)> getFunction(numType number) { boost::function<string(numType)> f; f = //Which code can I use here? If I call f() I want to call addNumber<numType>(number) }
f = static_cast<string(*)(numType)>(&addNumber<numType>); In Christ, Steven Watanabe

Steven Watanabe schrieb:
AMDG
Hans Mull wrote:
Hi! I want to use a boost::function pointer to a template function. The template type should be determinated at the time the boost::function object is initialized. Example:
template<class T> inline string addNumber(T inputNumber) { return lexical_cast<string>(inputNumber + 2); }
template<class numType> inline boost::function<string(numType)> getFunction(numType number) { boost::function<string(numType)> f; f = //Which code can I use here? If I call f() I want to call addNumber<numType>(number) }
f = static_cast<string(*)(numType)>(&addNumber<numType>);
In Christ, Steven Watanabe
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Thank you, that helped very well! Best regards, Hans
participants (2)
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Hans Mull
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Steven Watanabe