Can I stop c template argument from being deduced?-CodePudding

It's known that Template argument deduction takes place when you do not explicitly state the template argument, as in

template <typename T> void foo(T&& t) {};

Now we can invoke either as

foo(1);

foo('c')

foo<int>(1);

foo<char>('c')

In first set of invocations template argument deduction will happen. But will there be no template argument deduction in second case where template argument is explicitly specified?

CodePudding user response：

Yes you can. Here is one way:

template<typename T>
struct nodeduce {
    typedef T type;
};

Then you use typename nodeduce<T>::type instead of T in the parameter list

CodePudding user response：

You can use std::type_identify (which was added in C 20)

void foo(std::type_identity_t<T> && t) {}

But note that this changes the function signature since the argument is no longer a forwarding reference and it no longer accept lvalue arguments, unless you instantiate with an lvalue reference.

foo(1); // error. requires explicit instantiation

foo('c')  // error, requires explicit instantiation

int i;
foo<int>(i); // error, requires rvalue

foo<int&>(i); //fine
foo<int>(42); // fine