Re: why arguments are not incompatible

From:

"Eric Pruneau" <eric.pruneau@cgocable.ca>

Newsgroups:

comp.lang.c++

Date:

Fri, 6 Jun 2008 10:29:23 -0400

Message-ID:

<5Rb2k.1691$Mc.1351@read1.cgocable.net>

"sumsin" <sumsin123@gmail.com> a ?crit dans le message de news:
cdb01b08-0268-41c4-a160-6e403c67cd69@p25g2000pri.googlegroups.com...

Why the below two code snippet behave differently

Case 1:
#include <iostream>

void foo(const int *i)
{
int *local = i;
}

int main(void)
{
return 0;
}

and Case 2:
#include <iostream>

void foo(const int i)
{
int local = i;
}

int main(void)
{
return 0;
}

compilation command
g++ -Wall test03.cpp -o test03

In case 1 I got the below error.
error: invalid conversion from `const int*' to `int*'

In my understanding for an assignment:
- first the lvalue are rvalue must be of compatible type without
considering the qualifiers.
- then the lvalue must have all or more qualifiers than rvalue.

So its ok that I getting an error for case 1 but why its not happening
in case 2.

please clarify...

in case 2, local is a copy of i, so local has nothing to do with i except
that it is initialized with the same value.
So changing the value of local doesn't affect i.

But if you do

void foo(const int i)
{
int& local = i;
}

it wont work cause now local is a reference to i and therefore changing
local cause i to change. It was not the case with your original example.