Can C++ deduce argument type from default value?

user_17507361 0
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I tried to write this function with a default template argument:

template<typename A, typename B>
void func(int i1, int i2, A a, B b = 123){
    ...
}

In my mind I can call it like this: func(1, 2, 3) and compiler should deduce type B as int from default value, but I get no instance of overloaded function.
Is it incorrect C++ construction and compiler can't deduce type in this case?

template-argument-deduction default-arguments default templates c++
2 Answers

The type of a template parameter in a function can't be deduced from a default argument. As shown in the example on cppreference.com:

Type template parameter cannot be deduced from the type of a function default argument:

template<typename T> void f(T = 5, T = 7); 

void g()
{
    f(1);     // OK: calls f<int>(1, 7)
    f();      // error: cannot deduce T
    f<int>(); // OK: calls f<int>(5, 7)
}

However, you can specify a default argument for the template parameter:

template<typename A, typename B = int>
void func(int i1, int i2, A a, B b = 123){
    ...
}
user_33097900

As often when default arguments don't work you can use overloads:

template<typename A, typename B>
void func(int i1, int i2, A a, B b){
    ...
}
template<typename A>
void func(int i1, int i2, A a){
    func(i1,i2,a,123);
}
user_41177280
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