Const keyword with function signature - c++

Question

Here is my simple code:
#include <iostream>
using namespace std;
class alloc { };
template <typename T, typename Alloc = alloc>
class vector
{
public:
void swap(vector<T,Alloc> &v) { cout << "swap()" << endl; }
};
template <typename T, typename Alloc>
void swap(const vector<T,Alloc> &v1,const vector<T,Alloc> &v2)
{
v1.swap(v2);
}
int main()
{
vector<int> x;
vector<int> y;
swap(x,y);
return 0;
}
The code snippet run without problem.But i can't get any outputs
Then I delete const keyword.
void swap(vector<T,Alloc> &v1,vector<T,Alloc> &v2)
I get output swap()
I have read "The reason is that const for the parameter only applies locally within the function, since it is working on a copy of the data. This means the function signature is really the same anyways."
So i thought there is no difference between write or not write const.if i insist to write const here,how do i modify the code to get the output swap()

Answer 1

This is an excellent illustration of why using std should be avoided.
In order to debug this problem, remove using std, and add std:: in places where you want the behavior from the standard library. Fortunately, there is only one such place, i.e. the swap function in the template class:
void swap(vector<T,Alloc> &v) { std::cout << "swap()" << std::endl; }
Now try compiling again to see the error preventing your swap with const from being used:
prog.cpp:19:5: error: passing const vector<int> as this argument discards qualifiers
When your program was using std C++ had an alternative of picking std::swap over your own swap function when your function was not applicable. That is exactly what it did, without any warning, because it assumed that it is what you want it to do.
The error also tells you what to do in order to make const-qualified vectors be accepted: add const to the parameters of vector::swap, like this:
void swap(const vector<T,Alloc> &v) const { std::cout << "swap()" << std::endl; }
Now your program compiles and runs again (demo).