c++ - Why can't I deduce one of the class template arguments?

Question

I have some code here

template<typename T, std::size_t size, typename funcType>
struct foo
{
public:

    foo(const funcType& func) : m_func(func) {}
    ~foo() {}
    
    void m_call() { m_func(); }

private:
    const funcType& m_func;

    T x[size];
};

void printString() { std::cout << "some string
"; }

I can create an object

foo<int, 3, void(*)()> someObject(printString);

or

foo<int, 3, decltype(printString)> someObject(printString);

but when I try to do this:

foo<int, 3> someObject(printString);

I get this error on g++ 10.2

error: wrong number of template arguments (2, should be 3)
 foo<int, 3> someObject(printString);
           ^
note: provided for 'template<class T, long unsigned int size, class funcType> struct foo'
 struct foo
       

Why cant I do this? Doesn't the compiler know what type printString is?

And if I change the foo to

template<typename funcType>
struct foo
{
public:

    foo(const funcType& func) : m_func(func) {}
    ~foo() {}
    
    void m_call() { m_func(); }

private:
    const funcType& m_func;
};

I can create it normally

foo someObject(printString);

Am I missing something?

question from:https://stackoverflow.com/questions/65841372/why-cant-i-deduce-one-of-the-class-template-arguments

Answer 1

Use template function to create the object and to deduct missing template arguments from function call. Something like this.

template<typename T, std::size_t Size, typename FunctionT>
foo<T, Size, FunctionT> create_foo(const FunctionT &func) {
    return foo<T, Size, FunctionT>(func);
} 

auto foo_obj = create_foo<int, 3>(printString);