Template type deduction for char array

Question

template<typename T>
class X;

int main() {
    X<decltype("")> x;
}

Why does g++ deduce T as const char (&)[1] and not simply const char[1]?

NathanOliver · Accepted Answer · 2019-01-10T20:34:28.883

7

Unlike every other literal that is an rvalue, string literals are lvalues. decltype applied to an lvalue expression gives you a reference so const char (&)[1] is the correct behavior.

edited Jan 10 '19 at 20:34

answered Jan 10 '19 at 20:28

NathanOliver

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I find it odd that string literals are lvalues. Do you know of any particular reason why the language does this? Edit : I guess you don't want it to bind to a `char(&&)[N]`. – François Andrieux Jan 10 '19 at 20:31
1

@FrançoisAndrieux see: https://stackoverflow.com/questions/10004511/why-are-string-literals-l-value-while-all-other-literals-are-r-value – NathanOliver Jan 10 '19 at 20:33
1

If string literals were not lvalues, the pointers they decay to would immediately become dangling. – Brian Bi Jan 10 '19 at 20:36
@Brian Yeah, it seems obvious after reading through the linked question. – François Andrieux Jan 10 '19 at 20:37

Template type deduction for char array

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