static function calling nonstatic method ? (or something like that)

[Lucy Ludmiller]

I have code like this :

void foo(int i, void* p) { .. }
class A
{
public:
A();
~A();

private:
void foobar(int i, char* c);
};

Is there a way for foo to call foobar() ?

[Thomas Tutone]

Lucy Ludmiller wrote:

I have code like this :

void foo(int i, void* p) { .. }

class A
{
public:
A();
~A();
private:
void foobar(int i, char* c);
};

Is there a way for foo to call foobar() ?

Sure:

void foo(int i, void* p)
{
A a;
a.foobar(i, static_cast<char*>(p));
}

But if your question is, can foo call foobar without an instance of
class A to call it with, the answer is no, unless you change foobar to
be a static member of A.

Best regards,

Tom

[Radu]

Thomas Tutone wrote:

Lucy Ludmiller wrote:

I have code like this :

void foo(int i, void* p) { .. }

class A
{
public:
A();
~A();
private:
void foobar(int i, char* c);
};

Is there a way for foo to call foobar() ?

Sure:

void foo(int i, void* p)
{
A a;
a.foobar(i, static_cast<char*>(p));
}

.... if class A declares foo as a friend:
class A
{
friend void foo(int i, void* p) ;
//.....
};

HTH
Radu

>
But if your question is, can foo call foobar without an instance of
class A to call it with, the answer is no, unless you change foobar to
be a static member of A.

you can declaring

Best regards,

Tom

[Thomas Tutone]

Radu wrote:

Thomas Tutone wrote:

Lucy Ludmiller wrote:

I have code like this :
>
void foo(int i, void* p) { .. }
>
class A
{
public:
A();
~A();
private:
void foobar(int i, char* c);
};
>
Is there a way for foo to call foobar() ?

Sure:

void foo(int i, void* p)
{
A a;
a.foobar(i, static_cast<char*>(p));
}

... if class A declares foo as a friend:
class A
{
friend void foo(int i, void* p) ;
//.....
};

Good point - I missed that foo was private. Thanks for the
clarification.

Best regards,

Tom