How does template argument deduction work when an overloaded function is involved as an argument?

Question

This is the more sophisticated question mentioned in How does overload resolution work when an argument is an overloaded function?

Below code compiles witho

Answer 1

As noted in the answer to your linked question:

[temp.deduct.call]/6:When P is a function type, pointer to function type, or pointer to member function type:

— If the argument is an overload set containing one or more function templates, the parameter is treated as a non-deduced context.

Since the overload set contains a function template, the parameter is treated as a non-deduced context. This causes template argument deduction to fail:

[temp.deduct.type]/4: [...]If a template parameter is used only in non-deduced contexts and is not explicitly specified, template argument deduction fails.

And this failed deduction gives you your error. Note that if you explicitly specify the arguments, the code compiles successfully:

bar<int,double>(foo);

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