推导类型

Question

I am trying to understand type deduction while walking through Scott Meyer's Effective Modern C++.

Consider the code snippet below:

template<typename T>
void f(const T& param); // param is now a ref-to-const; paramType is const T&

int x = 27; // as before
const int cx = x; // as before
const int& rx = x; // as before

f(x); // T is int, param's type is const int&
f(cx); // T is int, param's type is const int&
f(rx); // T is int, param's type is const int&

He says that since the paramType is a reference, we can follow a two step procedure to deduce the type of T :

1. Ignore references (if any) in expr (ie, x , cx and rx )
2. Pattern match the type of expr and paramType

Now when cx is a const int :

cx -> const int

paramType -> reference to const int

So, according to the logic mentioned, shouldn't T be a const int due to pattern matching (and not just int )? I understand that the const ness of cx has been passed over to paramType , but is what he says, wrong? Is this 2 step procedure that he has mentioned not to be followed as a rule of thumb? How do you do it?

Thanks!

Answer 1

In his book Scott uses this "notation":

template<typename T>
void f(ParamType param); // where `ParamType` depends on T

So let's do pattern matching for ParamType when param is const int . We have:

const T & <----> const int // <----> is symbolic notation for pattern matching

So T is deduced as int , hence ParamType is const int& .

Answer 2

当cx为const int ，则T推导为int因此const T& param为const int& param ，即param为const int&类型。