pointer to function

[Gernot Frisch]

static short gfunc(short a, short b)
{return 0;}

class B{
B();
short(m_func)(short, short);
}

B::B()
{
m_func = gfunc;
}


error C2659: '=' : overloaded function as left operand

I'm not very familiar with pointer to functions. Can you help?

Thank you,
Gernot

--
-Gernot
int main(int argc, char** argv) {printf
("%silto%c%cf%cgl%ssic%ccom%c", "ma", 58, 'g', 64, "ba", 46, 10);}

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[Senapathy]

static short gfunc(short a, short b)
{return 0;}

// 1. Create a type for the function pointer...

typedef short(*pfn)(short, short);

class B{
B();
short(m_func)(short, short);

// 2. Have a member variable of that type

pfn m_func;

}

B::B()
{
m_func = gfunc;
}

Rest should be fine:

// invoke the function
B bObj;
bObj.m_func(1, 2);

Regards,
Senapathy

 

[Senapathy]

// 1. Create a type for the function pointer...

typedef short(*pfn)(short, short);

Ok, the type creation is not really required.... I just used it the way I am
always used to creating fn ptrs.

Just change this to short( * m_func)(short, short);
^^^^
After all you want to declare a function _pointer_ !

Regards,
Senapathy

 

[Gernot Frisch]

Just change this to short( * m_func)(short, short);
^^^^
After all you want to declare a function _pointer_ !

Ah, all right. What good would "void (a)()" be for? Would it mean the
same as "void a();" ?
Anyway, thank you.
Gernot

 

[JKop]

Gernot Frisch posted:

Ah, all right. What good would "void (a)()" be for? Would it mean the
same as "void a();" ?
Anyway, thank you.
Gernot

Yes, it would.

int (main)(void)
{

}


-JKOp