Does anyone know how to make a no recoil hack in C++?

[moeezk]

All I really ever want in a hack is no recoil lol. I was using [B]oss's no recoil hack for crossfire, and I really want to know how to make one my self. I know you need basic knowledge of C++ and olly but if any has any tutorials out there, it'd be greatly appreciated

[Biesi]

for(int i=0; i<670; i++)
{
	if((*(DWORD*)((*(DWORD*)(CShell+WeaponMgr))+(4*i)) ) != NULL)
	{
		for (int j=0; j<9 ; j++)
		{
			*(float*)((*(DWORD*)((*(DWORD*)(CShell + WeaponMgr)) +(4*i))) + 0x????? +j*4) = 0.0f;
		}
	}
}

[StupidFuck]

for(int i=0; i<670; i++)
{
	if((*(DWORD*)((*(DWORD*)(CShell+WeaponMgr))+(4*i)) ) != NULL)
	{
		for (int j=0; j<9 ; j++)
		{
			*(float*)((*(DWORD*)((*(DWORD*)(CShell + WeaponMgr)) +(4*i))) + 0x????? +j*4) = 0.0f;
		}
	}
}

As far as I know, for the time being, without an xtrap bypass, it is patched.

[3D]

for(int i=0; i<670; i++)
{
	if((*(DWORD*)((*(DWORD*)(CShell+WeaponMgr))+(4*i)) ) != NULL)
	{
		for (int j=0; j<9 ; j++)
		{
			*(float*)((*(DWORD*)((*(DWORD*)(CShell + WeaponMgr)) +(4*i))) + 0x????? +j*4) = 0.0f;
		}
	}
}

this patched.?

[[H]aaBX]

this patched.?

Yep ..
U get a Client Error (28_3)
Btw You can bypass the client error

[Major_Jake]

Yep ..
U get a Client Error (28_3)
Btw You can bypass the client error

It's useless when you say "you can bypass the client error" when your not give the bypass source codes. Just like any other pro-coders here

[Biesi]

It's useless when you say "you can bypass the client error" when your not give the bypass source codes. Just like any other pro-coders here

Nobody will give a bypass source to leechers
It would probably get detected.

/closed.