Function Pointer

**ScriptBlue** · 11-21-2005

Hello,
I'm a pickle at the moment. I have been playing around with import table hijacking, well now I want to rewrite a function and to avoid having the program malfunction, I need to use the old function in the new function; that's besides the point.
I have something like this, which would pointer function
typedef int (__cdecl * oldfunct)(SOCKET s, const char *buf, int len, int flags);
I however do not know how to assign oldfunct a value, in other words, I don't know how to cast it;
oldfunct = pThunk2->Function; //pThunk2->Function == the pointer
doesn't work and neither do
oldfunct = (int __cdecl) pThunk2->Function;
oldfunct = (int __cdecl *)(SOCKET , const char *, int , int ) pThunk2->Function;
oldfunct = (int __cdecl) pThunk2->Function;
oldfunct = (int __cdecl *) (SOCKET , const char *, int , int )pThunk2->Function;
Any advice is appreciated, thanks.

**7stud** · 11-21-2005

I however do not know how to assign oldfunct a value,

oldfunct = someFunctionName;

**7stud** · 11-21-2005

Here's an example:

Code:

#include<iostream>
#include<string>
using namespace std;

int oldfunct(int n, double d)
{
	cout<<n<<endl<<d<<endl;
	
	return 1;
}

void newfunct(int n, double d, string name, int(*pfunc)(int, double) )
{
	pfunc(n, d);
	cout<<name<<endl;
}


int main()
{
	newfunct(10, 12.5, "Betty", oldfunct);

	return 0;
}

**ScriptBlue** · 11-22-2005

Oh sorry for the ambiguity, but the new value is a DWORD. This really shouldn't be C++, but I'm using MS VC++ 2003. Ok, I'm trying this with a simpler situation
typedef int (__cdecl * oldfunct)(const char * ez, ...);
oldfunct = &printf;
(*oldfunct)("test %d %d %d", 1, 2, 3);
yet when I do that, I get an error
projectile.cpp(46): error C2513: 'int (__cdecl *)(const char *,...)' : no variable declared before '='

**7stud** · 11-22-2005

Oh sorry for the ambiguity, but the new value is a DWORD.

What the heck does that mean? I don't have any idea what "the new value" is.

yet when I do that, I get an error
projectile.cpp(46): error C2513: 'int (__cdecl *)(const char *,...)' : no variable declared before '='

You are getting that error because a typedef creates a synonym for a type, i.e. something that is easier to type in than the actual type. So, oldfunct is a synonym for a type, and you can't do this:

Code:

int = 6;

which is equivalent to what you are doing, and which I did in my first post.

Instead, you have to create a variable of type oldfunct and then assign it a value:

Code:

oldfunct pfunc;
pfunc = someFunc;

//or combining the statements:

oldfunct pfunc = someFunc;

Note the name of a function is an address, so you don't use the address-of operator on the function name. To call the function, you would do this:

Code:

pfunc(p1, p2...);

Here's a complete example:

Code:

#include<iostream>
using namespace std;

int someFunc(int n, double d)
{
	cout<<n<<endl<<d;
	
	return 10;
}

typedef int(*IntIntDp)(int, double);

int main()
{
	IntIntDp pfunc = someFunc;

	int result = pfunc(3, 10.1);
	cout<<endl;
	

	return 0;
}

**dwks** · 11-23-2005

Code:

cout<<endl;

Did you mean

Code:

cout << result << endl;

? Otherwise result is used only once.

**7stud** · 11-23-2005

Did you mean

cout << result << endl;

?

No.

Otherwise result is used only once.

Do you think the example was a demonstration of how to display an int returned by a function?

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