Re: function template explicit specialization not chosen

=?UTF-8?B?RGFuaWVsIEtyw7xnbGVy?= <>
Thu, 19 Apr 2012 22:46:17 -0700 (PDT)
Am 19.04.2012 20:48, schrieb Johannes Sixt:

On 19 Apr., 10:43, Daniel Kr??gler<>

template<typename T> int hash (T);
template<typename T> int hash (T t) { return t; } // (1)
template<> int hash (dummy const&) { return 0; } // (2)

int main ()
    dummy a;
    dummy const&b = a;
    hash(b); // expect (2) called


(1) What does the standard (c++11) say about the call to hash with
b, should T be deduced to be dummy const&, or should it be dummy

It won't be deduced to be dummy const&, that is for sure. The rules
for selection of function templates in this context is uniformly
described by []. The function parameter
type P here is not a reference type, therefore we need to consider

"If P is not a reference type: [..] If A is a cv-qualified type,
the top level cv-qualifiers of A???s type are ignored for type

So even though the argument type A is a cv-qualified type (The type
is 'dummy const'), this is irrelevant here. As a rule of thumb one
should remember that a function template that looks like using
"by-value" arguments, will have this way, unless you provide
explicit template parameters that would change that.

Can you please explain why the type is 'dummy const', not 'dummy
const&'? The function argument b clearly is a reference. Is there
some rvalue- to- lvalue conversion involved, and if so, why does it
happen before the function type is completely deduced?

Yes, the *declared* type of b is a reference type, but this point is
irrelevant here. The compiler interprets the expression 'b' within the
context of a function call expression 'hash(b)'. The meaning of 'b' is
specified in Clause 5 [expr] p5:

"If an expression initially has the type ???reference to T??? (8.3.2,
8.5.3), the type is adjusted to T prior to any further analysis."

Therefore in regard to expression analysis we can say the following
about the sub-expression 'b':

a) It is an lvalue.

b) It has type dummy const.

The const-qualifier is conserved here, because dummy is a class type
(According to [basic.lval] p4 this does not apply to non-class types,
even though it does apply for array types as well).

HTH & Greetings from Bremen,

Daniel Kr??gler

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