const pointer to class problem [C++]

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From: David Pol on 18 Apr 2008 08:15

On 18 abr, 03:33, t.wa...(a)nci-sw.com wrote:
> I'm trying to slowly convert some C code to C++ and have encounter
> what's probably a trivial problem.
>
> The easiest way to describe the problem is with an example:
>
> class FOO
> {
> public:
> int Version(void){return m_ver;};
> FOO(){m_ver = 1;};
> private:
> int m_ver;
> };
>
> void PrintVer(const FOO * fp)
> {
> int v = fp->Version(); /* error here */
>
> cout << "The version is " << v << endl;
> }
>
> When I try to compile, I get the error
> 'FOO::Version' cannot convert 'this' pointer from 'const FOO' to 'FOO
> &'
>
> on the line containing fp->Version().
>
> If I remove the const from the routine declaration, it works fine, but
> I'd rather not do that. I guess I don't understand what the compiler
> is trying to tell me, and have no idea of how to work around this
> without removing the const.

Hello,

FOO::Version() is not a const member function, so it cannot be invoked
by a const FOO object, or via a pointer (or a reference) to a const
FOO object. It makes sense to mark FOO::Version() as a const member
function, because it does not mutate its object (it merely inspects
the m_ver member variable of its object):

#include <iostream>

class FOO
{
public:
int Version() const { return m_ver; }
FOO() : m_ver(1) {}

private:
int m_ver;
};

void PrintVer(const FOO& fp)
{
int v = fp.Version();

std::cout << "The version is " << v << std::endl;
}

int main()
{
FOO f;
PrintVer(f);

return 0;
}

Note the use of an initialization list in the default constructor of
FOO [1] and the change of the parameter type of PrintVer() to a
reference to a const FOO (it is generally preferable to pass an
argument via a reference instead of passing it via a pointer unless
the argument can be legitimately null).

[1] See: http://www.parashift.com/c++-faq-lite/ctors.html#faq-10.6

Regards,
David

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From: rwf_20 on 18 Apr 2008 08:17

On Apr 17, 9:33 pm, t.wa...(a)nci-sw.com wrote:
> When I try to compile, I get the error
> 'FOO::Version' cannot convert 'this' pointer from 'const FOO' to 'FOO
> &'
>
> on the line containing fp->Version().

fp is a pointer to const FOO, and you are calling a non-const method
Version(). Declare Version as a const member function:

int Version(void) const { return m_ver; }

Ryan

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From: Mathias Gaunard on 18 Apr 2008 08:18

On 18 avr, 03:33, t.wa...(a)nci-sw.com wrote:

> The easiest way to describe the problem is with an example:
>
> class FOO
> {
> public:
> int Version(void){return m_ver;};
> FOO(){m_ver = 1;};
> private:
> int m_ver;
> };
>
> void PrintVer(const FOO * fp)
> {
> int v = fp->Version(); /* error here */
>
> cout << "The version is " << v << endl;
> }

class Foo
{
public:

Foo() : m_ver(1)
{
}

int Version() const
{
return m_ver;
}

private:
int m_ver;
};

void PrintVer(const Foo& f)
{
std::cout << "The version is " << f.Version() << std::endl;
}

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From: Todd Mars on 18 Apr 2008 08:18

From: SeanW on 18 Apr 2008 08:17

> class FOO
> {
> public:
> int Version(void){return m_ver;};
> FOO(){m_ver = 1;};
> private:
> int m_ver;
> };
>
> void PrintVer(const FOO * fp)
> {
> int v = fp->Version(); /* error here */
>
> cout << "The version is " << v << endl;
> }
>
> When I try to compile, I get the error
> 'FOO::Version' cannot convert 'this' pointer from 'const FOO' to 'FOO
> &'
>
> on the line containing fp->Version().
>
> If I remove the const from the routine declaration, it works fine, but
> I'd rather not do that. I guess I don't understand what the compiler
> is trying to tell me, and have no idea of how to work around this
> without removing the const.

It *is* a bad error message, but it's trying
to tell you that you're not respecting the
promise you made with by qualifying the pointer
with "const" in the parameter list: not to
change the object either by direct access to
its data or by calling a method on it that hasn't
made that promise, such as Version().

The way to tell the compiler that Version() promises
not to change the object it's called on is
with this unfortunate syntax:

int Version(void) const
{return m_ver;}

That "const" after the parameters means that
the implicit "this" pointer will be treated as
if declared "const FOO*" rather than just "FOO*",
which is the mismatch that the compiler is ineptly
trying to tell you about. It just throws in
the ampersand to throw you off.

Sean

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