c++ - Template parameter deduction with function pointers and references -


possible duplicate:
why qualifiers of template arguments stripped when deducing type?

consider following c++ code:

void f(int&);  template <typename t> void tpl(void (*)(t), t);  void bar(int& x) {         tpl(&f, x); }

compilation using gcc 4.6.0 fails following error message:

fntpl.cpp: in function ‘void bar(int&)’: fntpl.cpp:7:11: error: no matching function call ‘tpl(void (*)(int&), int&)’ fntpl.cpp:7:11: note: candidate is: fntpl.cpp:3:46: note: template<class t> void tpl(void (*)(t), t)

if state template parameters explicitely (tpl<int&>(&f, x)), works. why doesn't template argument deduction work in case?

because these fundamentally different

void f(int&);

and

void (*)(t)

the compiler has deduced t int, looks for:

void f(int);

which nothing intention, change function pointer this:

template <typename t> void tpl(void (*)(t&), t);

and compiler happy...


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