Can an lvalue reference non-type template parameter be inferred?

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深忆病人
深忆病人 2021-01-02 03:07

I have the following code, which I cannot get to work:

struct foo {};
foo foo1 = {};

template 
class FooClass {};

template 

        
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  • 2021-01-02 03:50

    This is precisely covered by CWG 2091:

    According to 14.8.2.5 [temp.deduct.type] paragraph 17,

    If P has a form that contains <i>, and if the type of the corresponding value of A differs from the type of i, deduction fails.

    This gives the wrong result for an example like:

    template<int &> struct X;
    template<int &N> void f(X<N>&);
    int n;
    void g(X<n> &x) { f(x); }
    

    Here, P is X<N>, which contains <i>. The type of i is int&. The corresponding value from A is n, which is a glvalue of type int. Presumably this should be valid.

    I think this rule means to say something like,

    If P has a form that contains <i>, and the type of i differs from the type of the corresponding template parameter of the template named by the enclosing simple-template-id, deduction fails.

    As noted by @dyp, [temp.deduct.type]/17 should be more permissive. In your example, the argument in FooClass<F> (F) does not have reference type - it's an lvalue of type foo. The template parameter of FooClass is a reference. The DR was resolved last year.

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