Re: Deducing reference template argument
On 2??10??, ????6??50??, Piotr Jachowicz <pjach...@gmail.com> wrote:
Hello,
Following code produces "string" on the console (gcc v4.3 and Comeau
4.3.10.1 beta):
# include <string>
# include <iostream>
using namespace std;
template<class T>
struct type {
static string name() {
return "unknown";
}
};
template<>
struct type<string> {
static string name() {
return "string";
}
};
template<>
struct type<string&> {
static string name() {
return "string&";
}
};
template<class T>
void deduce_type(T& t) {
cout << type<T>::name() << endl;
}
int main() {
string s1;
string& s2 = s1;
deduce_type(s2);
}
Why? According to Standard 14.8.2.1 [temp.deduct.call] template
deduction follows:
1. P = T&, A = string&
2. P is reference type, so type referred to by P is used for
deduction: P = T
3. Deducion can be performed: P = T, A = string&, so T = string& makes
P = A
Apparently compiler's deduction differs. It substitutes T = string.
Where made I an error?
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I think T&& is deduced to T& by compiler
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