*(DWORD*) pointers etc

Hey community!

I saw this in a cf hack and I dont understand why there as example:

if ((*(DWORD*)((*(DWORD*)(CShell+WeaponMgr))+(4*i))) != 0)

are 2 *(DWORD*)

*(DWORD*) is de-referencing that pointer to an actual DWORD value.
and (DWORD*) is a cast to convert a value to a pointer to a DWORD.

But i dont understand why there are 2 from them

Hope you can help me

DWORD CShell = ( DWORD )GetModuleHandleA( "CShell.dll");
DWORD WeaponMgr = *( DWORD *)( CShell + 0x0 );

if( WeaponMgr )
{
for( int i = INT_MAX; i = 0; i++ )
{
DWORD Wep = *( DWORD *)( Wep + ( i * 4 ) );

if( Wep )
{
//hacks
}
}

More clean code than yours, if you want really understand just learn the ASM logic and C++ logic.

Originally Posted by XarutoUsoCrack

More clean code than yours, if you want really understand just learn the ASM logic and C++ logic.

nice answer...

here a example:

Code:

00725430 . DC6A6900 DD execu.00696ADC *(DWORD *)0x00725430 make a pointer for address 00696ADC ... 00696ADC 80BD4400 DD execu.0044BD80 *(DWORD *)0x00696ADC make a pointer for address 0044BD80 ... 0044BD80 /. 55 PUSH EBP 0044BD81 |. 8BEC MOV EBP,ESP 0044BD83 |. 51 PUSH ECX then *(DWORD *)*(DWORD *) or **(DWORD **)0x00725430 make a pointer directly to the address 0044BD80

Originally Posted by luizimloko

Nice Tut!

Originally Posted by Nik08154

break that line into steps, starting at the inner-most ( )'s

Code:

*(DWORD*)((*(DWORD*)(CShell+WeaponMgr))+(4*i))

1. add CShell+weaponMgr --> we'll store the answer in (a temporary variable) 'aa'
aa = CShell+WeaponMgr;

2. *(DWORD*)(aa)
--read a DWORD (4 bytes) starting at the addr aa --> these 4 bytes will be a memory address. we'll store in (a temporary variable) 'bb'
bb = *(DWORD*)(aa);

3. + (4*i)
bb += 4*i;

4. *(DWORD*)(bb)
--same as step 2. Read 4 bytes starting at 'bb'. we'll store this in (a temporary variable) 'cc'
cc = *(DWORD*)(bb); // cc could be a memory addr -- or regular Int32 value, representing..something.
5. cc != 0 //last item in the expression.

------------------------

Code:

*(DWORD*)(CShell+WeaponMgr)

If we only did one *(DWORD*)() that would read 4 bytes from CShell+WeaponMgr
it would return an address like 0x11223344.
If we then read a DWORD from (0x11223344 + (4*i)), we get some useful number from the game like gun ammo - or possibly* a pointer to some object in the game.

-----------------------------

Originally Posted by XarutoUsoCrack

More clean code than yours, if you want really understand just learn the ASM logic and C++ logic.

Troll?

DWORD WeaponMgr = *( DWORD *)( CShell + 0x0 ); // +0 means nothing. ?

if( WeaponMgr )
{
for( int i = INT_MAX; i = 0; i++ ) //This loop will never execute its body... Why is i being set =0, not checked against ==0? If the value of a condition is 0, it's == false, the body won't execute.
{

DWORD Wep = *( DWORD *)( Wep + ( i * 4 ) ); 'Wep' used in it's own definition. Wild goose chase, or compiler error?
//Wep + ( i * 4 ); has 'undefined' value.

test:

Code:

#include <climits>
#include <Windows.h>


int main()
{

 for (int i = INT_MAX; i = 0; i++) // error in condition; same as above.
 {
	 MessageBoxA(0,"Annoying yet?","Nope. This code never runs.",0);// never executed
	 bool _breakPoint = false; // never executed
 }

 MessageBoxA(0,"How many times did that loop run?","Zero???",0); // executed
 bool _done = true; // executed

Originally Posted by abuckau907

break that line into steps, starting at the inner-most ( )'s

Code:

*(DWORD*)((*(DWORD*)(CShell+WeaponMgr))+(4*i))

Code:

*(DWORD*)(CShell+WeaponMgr)

Code:

#include <climits>
#include <Windows.h>


int main()
{

 for (int i = INT_MAX; i = 0; i++) // error in condition; same as above.
 {
	 MessageBoxA(0,"Annoying yet?","Nope. This code never runs.",0);// never executed
	 bool _breakPoint = false; // never executed
 }

 MessageBoxA(0,"How many times did that loop run?","Zero???",0); // executed
 bool _done = true; // executed

i just show to him a easy way to create same shit whitout this 2 yo method un-clean, whit my code you can easy understand the code.

Originally Posted by XarutoUsoCrack

i just show to him a easy way to create same shit whitout this 2 yo method un-clean, whit my code you can easy understand the code.

Uh, your code would never run???

Code:

 for (int i = INT_MAX; i = 0; i++)

middle statement is "i = 0" ...not "i == 0".

for (initialize; test; update)

your test condition would always be 0 (ie. false), and therefore never execute its block of code.

?????????????????????????????????????????????????? ??????????????????

And I don't think the line

Code:

DWORD Wep = *( DWORD *)( Wep + ( i * 4 ) );

would even compile?

bump.
@XarutoUsoCrack

??

noob proof protection, just understand dat.

lol bro, just stop being a idiot, i just showed a EXAMPLE to this guy, just a little research in crossfire source section will find a lot of information.

edit: I'm not even sure...this shit is bananas. Thanks for helping : /

Originally Posted by luizimloko

here a example:

Code:

00725430 . DC6A6900 DD execu.00696ADC *(DWORD *)0x00725430 make a pointer for address 00696ADC ... 00696ADC 80BD4400 DD execu.0044BD80 *(DWORD *)0x00696ADC make a pointer for address 0044BD80 ... 0044BD80 /. 55 PUSH EBP 0044BD81 |. 8BEC MOV EBP,ESP 0044BD83 |. 51 PUSH ECX then *(DWORD *)*(DWORD *) or **(DWORD **)0x00725430 make a pointer directly to the address 0044BD80

And whats the advantage of making a direct pointer?

Originally Posted by Nik08154

And whats the advantage of making a direct pointer?

less code lines.

a other example:

Code:

class CRealFunc { public: DWORD stuff; }; class CPointer { public: union { DWORD Address; CRealFunc *Pointer; }; char _0x0000[4]; DWORD stuff; }; class CTest { public: union { DWORD Address; CPointer *Pointer; }; char _0x0000[4]; DWORD stuff; }; //1) CTest *test = (CTest *)0x00725430; test->Pointer->stuff = 1; test->Pointer->Pointer->stuff = 1; //2) CTest *test = *(CTest **)0x00725430; //the address for this func will be: 0x00696ADC

(DWORD) pointers etc

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