const参数的模板参数推导

Question

Suppose we have

template <typename T>
void foo(T a)
{
    a = 10;
}

int main()
{
    const int a = 5;
    foo(a);
}

Why T is deduced as int and not const int, why should I can modify a in function? How deduction works in this case?

Here's a working sample.

Answer 1

Why should the fact that the object outside the function is const matter to the internals of the function? It is getting it's own copy of the object, so it can choose for itself whether to treat that copy as const or not. Modifying a inside the function will not affect the object outside. This is the reason that top-level cv-qualifiers are ignored for type deduction.

If you want the parameter to be const , you need to ask for it.

Answer 2

Why T is deduced as int and not const int, why should I can modify a in function?

Because you are passing by value and receiving by value (not reference). According to language grammar such deduction would always result in priority to non-const over const .
For better understanding try below C++11 code and print the typeid :

const int a = 5;  // `a` typeid is `const int`
auto b = a;    // `b` typeid is `int`

How deduction works in this case?

If you want a to be non modifiable and respect the const ness of the object passed then receive by reference in the function. ie

template<typename T>
void foo (T& a) 
{
  a = 10; // this will result in error if a `const int`
}

[Note: Passing a literal wouldn't never work with above signature.]