memcpy? lol wut.

[master131backup]

Hey guys, it's kinda wierd for me to be here but I was wondering what this does.

memcpy((PBYTE)addSuperBullets, "\x33\xC0\x90", 3);

What does "\x33\xC0\x90" do? Like I know it's a 3 byte array and memcpy copies an array of bytes to an address but why those values?

[Ch40zz-C0d3r]

These r the default bytes i think (im EU and idk which addy u mean).
When u use memcpy to HACK than u usually use the bytes x90. This is called NOP which stands for NoOPeration. This means when the Bullet is hit the wall do NOTHING. So u shoot trhough. I hope i could help u.

[master131backup]

I know what NOP means. I have done reverse engineering before :3 Alrighties, thanks.

But I still don't understand this:
memcpy((PBYTE)addSuperBullets, "\x33\xC0\x90", 3); //ON
memcpy((PBYTE)addSuperBullets, "\x0F\x94\xC0", 3); //OFF

The byte arrays are completely different values.

[Ch40zz-C0d3r]

Than its adifferent addy than mine in EU sry IDK. This works different.

[master131backup]

Hmmm, alright thanks for helping anyway. *cough* Where is ᴺᴼᴼᴮ when you need him.

[NOOB]

Ask @mmbob, He's the one who found super bullets. The fact that you can NOP the three bytes and still have the same result is just by chance.

[mmbob]

\x33\xC0\x90 are the bytes for "xor eax, eax; nop"
You need to return 0 from the intersect callback function to tell the engine that the object it gave you is not one that you want to hit. xor eax, eax makes eax (the return value) zero.

[Ch40zz-C0d3r]

\x33\xC0\x90 are the bytes for "xor eax, eax; nop"
You need to return 0 from the intersect callback function to tell the engine that the object it gave you is not one that you want to hit. xor eax, eax makes eax (the return value) zero.

So when u NOP it it must return 0 too or?

[[MPGH]master131]

Oh so they're assembly bytes. I was thinking that they might've been when I woke up this morning. :3 Thanks mmbob.