Template type deduction failure when using reference types
in [C++]
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From: Marcus on 2 Aug 2010 22:18 Hello all, Can anyone help me with a particularly obscure error? The following code fails to compile on g++ template<typename retT, typename argT> void func(retT (*funcPtr)(argT), argT argument) {} void func1(int a) {} void func2(int& a) {} int main() { int val = 1; int& valref = val; func(&func1, valref);//Succeeds func(&func2, valref);//Fails } The compiler appears to lose the ampersand from argT when it creates the function. I'm completely stuck on this one so any help would be greatly appreciated. Thanks, Marcus -- [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ]
From: Johannes Schaub (litb) on 3 Aug 2010 07:23 Marcus wrote: > Hello all, > Can anyone help me with a particularly obscure error? The following > code fails to compile on g++ > > template<typename retT, typename argT> > void func(retT (*funcPtr)(argT), argT argument) > {} > > void func1(int a) {} > void func2(int& a) {} > > int main() > { > int val = 1; > int& valref = val; > > func(&func1, valref);//Succeeds > func(&func2, valref);//Fails > } > > The compiler appears to lose the ampersand from argT when it creates > the function. I'm completely stuck on this one so any help would be > greatly appreciated. "argT" is deduced to different types in the second function call. It's "int" for the right function parameter, but it is "int&" for the left function parameter. Thus you have a contradiction, and type deduction fails. You are going to need to make the left parameter a non-deduced context so that it won't participate template<typename T> struct identity { typedef T type; }; template<typename retT, typename argT> void func(retT (*funcPtr)(argT), typename identity<argT>::type argument) {} -- [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ]
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