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pointer to a static function

Posted on 2003-11-12
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Last Modified: 2010-04-01
To C++ experts,

If I have a class, and a function like :
class X{} ;
static X* fun(int*) ;

How do I declare a pointer to a function which can point to fun ?
thanks.

meow.
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Question by:meow00
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7 Comments
 
LVL 17

Accepted Solution

by:
rstaveley earned 20 total points
ID: 9737786
Use of the word static for a function to prevent it from having public linkage is now deprecated (though I must confess I do it all the time). You should really use an unnamed namespace instead, if you want to prevent the function from having public linkage.

--------8<--------
#include <iostream>

class X {};

namespace {
        class X x;
        X* fun(int*)
        {
                return &x;
        }
}

int main()
{
        // Get a pointer to the function
        X* (*ptr)(int*) = fun;

        // Use it
        int myint = 123;
        X* xp = (*ptr)(&myint);
}
--------8<--------
0
 
LVL 3

Expert Comment

by:freewell
ID: 9737793
int nData = 10;
X* (*pMyFunction)(int*);

pMyFunction = fun;
pMyFunction(&nData);
0
 
LVL 5

Expert Comment

by:migoEX
ID: 9737799
Exactly if there was no 'static' keyword - this only means the function is accessible only from this file. Even when you declare static function inside the class, the syntax would be the same.

X* (* tFunnc)(int*);

tFunc myFunc = fun;
0
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LVL 1

Author Comment

by:meow00
ID: 9741140
Hi,

    Could anyone please tell me that why not use
 "(static X*)(*pMyFunction)(int*);" ? why I don't need static in the very beginning ? Thanks !

meow.
0
 
LVL 5

Assisted Solution

by:migoEX
migoEX earned 20 total points
ID: 9741331
'static' is a definition of a SCOPE of a function, not a TYPE, which is parameters and return value.

Exactly as:

class A{
   public:
   static int s_i;
};

the type of s_i is 'int' and not 'static int'
0
 
LVL 10

Expert Comment

by:Sys_Prog
ID: 9745451
Try the following code snippet

int f () {
}
static int f () {
}


This would give u a compile error saying "redefinition of `int f()' "

This demonstrates that static is not at all related to the function signature

Now, technically, static is a type of storage class, thus it only decides what storage class the function belongs to and the behaviour of the compiler gets affected in the corresponding way

When u are declaring a pointer to a function, u just need the function signature and not the storage class of the function

That's the reason for your code to work

HTH

Amit
0
 
LVL 9

Expert Comment

by:tinchos
ID: 10249111
No comment has been added lately, so it's time to clean up this TA.
I will leave the following recommendation for this question in the Cleanup topic area:

Split: rstaveley {http:#9737786} & migoEX {http:#9741331}

Please leave any comments here within the next seven days.
PLEASE DO NOT ACCEPT THIS COMMENT AS AN ANSWER!

Tinchos
EE Cleanup Volunteer
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