const keyword position in function declaration [duplicate]

Question

Possible Duplicate:
Difference between const declarations in C++

#include <iostream>

class Bar{};

void foo(const Bar x){}  //l5
void foo(Bar x){}        //l6
void foo(Bar const x){}  //l7

////pointer functions

void foo(const Bar* x){} //l11
void foo(Bar* x){}       //l12
void foo(Bar* const x){} //l13

Compiler output: (long story short l5,l6,l7 conflict; but only l12,l13 conflict)

untitled.cpp:6:6: error: redefinition of ‘void foo(Bar)’
untitled.cpp:5:6: error: ‘void foo(Bar)’ previously defined here
untitled.cpp:7:6: error: redefinition of ‘void foo(Bar)’
untitled.cpp:5:6: error: ‘void foo(Bar)’ previously defined here
untitled.cpp:13:6: error: redefinition of ‘void foo(Bar*)’
untitled.cpp:12:6: error: ‘void foo(Bar*)’ previously defined here

What going on?

1. What is the meaning of each of the declarations
2. Why does all 3 declarations conflict with object functions but only 2 with pointer functions?
3. Please elaborate that conflict is between l12 and l13, even though l12 does not contain const keyword
4. Really sorry if trivial question

Answer 1

The "problem" is that constness of a parameter's value doesn't participate in overloading!

First, Bar const and const Bar are already identical meaning, so they would automatically have a problem. But as a function parameter the const doesn't apply to overloading so the Bar version of the function also looks the same too. The const in the paremeter only tells the compiler that you don't intend to modify it in the function body.

For the same reason, Bar* and Bar* const are treated the same: The const applies to the value of the parameter (not what's pointed to) and does not participate in overloading, so you've defined the same function.

On the other hand const Bar* means something totally different: A non-const pointer to a const object (of type Bar). Since the type is different it does participate in overloading and allows that function to be unique.