Function template overload resolution, dependent and non-dependent parameters

This is bog-standard partial ordering. We substitute unique types into one of the function templates and try to deduce the other against it. Do it both ways and if deduction only succeeds in one direction, we have an order. If you want to read the arcane rules, see [temp.func.order] and [temp.deduct.partial].

So here,

• Substituting T1=U1, T2=U2 into the first overload's function type produces int f(U1, U2); Can we deduce T1 and T2 in the second overload from this? No; both are in non-deduced contexts. Ergo, deduction fails.
• Substituting T1=U1, T2=U2 into the second overload produces int f(id<U1>::type, id<U2>::type) (this is conducted in the definition context so we can't substitute further into id - there may be a specialization somewhere). Can we deduce the T1 and T2 in the first overload from this? Yes, by deducing T1 = id<U1>::type and T2 = id<U2>::type. Deduction succeeds.

Since deduction succeeds only in one direction - deducing the first from transformed second - the second is more specialized than the first and is preferentially picked by overload resolution.

The alias template case changes nothing.

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These templates are neither equivalent nor functionally equivalent.

The following func overload

// Denote as overload F.
template<class T1, class T2>
int func(typename id<T1>::type, typename id<T2>::type) { return 1; }

is more specialized than the following func overload

// Denote as overload G.
template<class T1, class T2>
int func(T1, T2) { return 0; }

thus, the former is chosen by overload resolution.

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_{(All ISO Standard references below refer to N4659: March 2017 post-Kona working draft/C++17 DIS)}

The partial ordering of the G and F overloads of func is governed by:

• [temp.func.order]/2, [temp.func.order]/3 and [temp.func.order]/4, and
• [temp.deduct.partial]/2 and [temp.deduct.partial]/10.