传递const类型的地址时模板参数推断失败

Question

template <typename T>
T go(T a, T *b){ T *t; return *t;}

int main() {
    const int x = 10;
    go(x, &x);
    return 0;
}

Gives compiler error:

error: no matching function for call to 'go(const int&, const int*)'

Why is the first argument a reference type const int& instead of just const int ?

To fix this compilation error, I overrode the compiler deduction process by specifying the type of arguments go<const int>(x, &x); , but again why do I need to do that?

Answer 1

It's a conflict of type deduction. T is deduced as int from the first argument, and as const int from the second argument. Hence type deduction fails (and the compiler presents a message which may or may not make the underlying cause clear).

If you want to make this function work without having to specify the template argument explicitly, you could make it so that only the second function argument drives the deduction:

template <class T>
struct NonDeduced { using type = T; }

template <class T>
T go(typename NonDeduced<T>::type a, T *b) { T *t; return *t; }

That way, T will only be deducible from the second argument, and the first parameter will use the deduced T without modification.

Answer 2

Because a is declared to be passed by value, then in template argument deduction :

c) otherwise, if A is a cv-qualified type, the top-level cv-qualifiers are ignored for deduction:

That means, for go(x, &x); , for the 1st argument x , the template parameter T will be deduced as int , not const int . For the 2nd argument T will be deduced as const int , because b is declared to be passed by pointer (and the cv-qualifiers on the object being pointed are reserved; the same thing happens for pass-by-reference). Then deduction fails.

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BTW: clang gives quite clear message for it:

prog.cc:4:3: note: candidate template ignored: deduced conflicting types for parameter 'T' ('int' vs. 'const int')