Question

Possible Duplicate:
Why are qualifiers of template arguments stripped when deducing the type?

Consider the 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 with the following error message:

fntpl.cpp: In function ‘void bar(int&)’:
fntpl.cpp:7:11: error: no matching function for call to ‘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 I state the template parameters explicitely (tpl<int&>(&f, x)), it works. Why doesn’t template argument deduction work in this case?

Answer 1

Because these are fundamentally different

void f(int&);

and

void (*)(T)

the compiler has only deduced that T is int, so it looks for:

void f(int);

which is nothing like your intention, change the function pointer to this:

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

And the compiler will be happy…