Help/Support Need a little help with Obj-C

May 7, 2015

I'm currently trying to make a score multiplier hack for Run Terrio Run, but for some reason whenever I input my own value for the multiplier variable it always stays at zero. I've tested the hack this by hardcoding 20 into the multiplier variable just to make sure that it works, and it does, the score counts by twenties.

Code:

#include <substrate.h>
#include "writeData.h"

int multiplier;
int nothing;

static int hackScore(int);

%hook UnityAppController

- (BOOL)application:(id)fp8 didFinishLaunchingWithOptions:(id)fp12
{
UIAlertView *alert = [[UIAlertView alloc]initWithTitle:@"Run Terrio Run Score Multiplier Modifier" message:@"How much do you want the score to be multiplied by?" delegate:self cancelButtonTitle:@"Cancel" otherButtonTitles:@"Set Multiplier",nil];
[alert setAlertViewStyle:UIAlertViewStylePlainTextInput];
UITextField *inputField = [alert textFieldAtIndex:0];
[inputField setKeyboardType:UIKeyboardTypeNumberPad];
inputField.text = [NSString stringWithFormat:@"%i",multiplier];

multiplier = [inputField.text intValue];

UIAlertView *scoreAlert = [[UIAlertView alloc]initWithTitle:@"Multiplier Modifier" message:[NSString stringWithFormat:@"Score is now going to be multiplied by %i",multiplier] delegate:nil cancelButtonTitle:@"Good shit" otherButtonTitles:nil];
     [scoreAlert show];
     
if(hackScore(multiplier)==255){
writeData(0xHIDDEN, 0xFF0080E2);
}

[alert show];
return %orig;
}
%end

static int hackScore(int multiplier){
   if(multiplier>0){
     return multiplier;
   }
return multiplier;
}

Thanks

Updated May 7, 2015 by Guest

ImChaotic · May 7, 2015

i well try to help

6 God · May 7, 2015

try using code x8486375737488348349354 for it the multiplier variable shouldn't stay at 0

AnotherLurker · May 7, 2015

This gave me eye cancer

alphaMatterr · May 7, 2015

I'm currently trying to make a score multiplier hack for Run Terrio Run, but for some reason whenever I input my own value for the multiplier variable it always stays at zero.

for starters, your method, hackScore, is not doing anything. What your saying is:

"If multiplier is greater than 0, return multiplier"

"If multiplier is equal or less than 0, return multiplier"

since (multiplier ∈ R), what your essentially saying is, if multiplier is a number then return the value for multiplier (in other words, this method is USELESS and does nothing)

secondly, your integer, nothing, is really living up to its name....... by doing nothing

Moreover, from your code:

if(hackScore(multiplier)==255){
writeData(0xHIDDEN, 0xFF0080E2);
}

what your saying is if the user chooses their multiplier to be set as 255, THEN you will write the hex 0xFF... to the given address, however, if they do not choose to have their multiplier set as 255, then they will always have the value of the multiplier which was originally given in the game.

To fix this you have three options

1. write a RIDICULOUSLY HUGE if else OR switch statement containing every integer value from 0 - 255.

This is the most retarded and inefficient way of doing what you want, however it will technically get it done (please dont do it this way if you choose to fix the mistake)

2. You can create a method (or just use hackScore for something useful ) which will:

- find the value inputted for multiplier (already done)

- take that value and convert it to hex

- take your new hex to be written, in this case XX0080E2, and combine it with the value u converted from the previous step

So if they choose a multiplier value of 250 then first you would find that value

multiplier = 250;

then convert it to hex

multiplier = 0xFA;

and then combine your offset with the hex, for example (this is NOT how it would work, just trying to show u what would happen):

multiplier + XX0080E2 = FA + XX0080E2 => FA0080E2

with 0xFA0080E2 being your final hex.

this method will get what you want done fairly efficiently, however it involves having basic, basic, basic programming skills as well as users will only be allowed to enter integers

3. Now the third method, and probably THE BEST METHOD:

just use MSHook and save yourself the trouble.. With MSHook you can accomplish your goal in under 20 lines of code (i didnt actually count but mshook will be 100000000X faster and easier)

..

There are other errors in the code but if you use mshook, most of them will go away because you'll throw the bad code away

Updated May 7, 2015 by alphaMatterr

May 7, 2015

This gave me eye cancer

its my first code, give me a break

-snip-

First of all, thank you for such an informative reply, really means alot

Also, I only put 255 there for testing purposes, I know that if you put in any other number it would stay at zero. Also, when I wrote that function I was thinking about just putting a ton of if statements. I know that it just returns the multiplier, and I wanted it to do that. But now that I think about it I don't even need it, I could have just done "if(multiplier==255)" or something like that. The "nothing" integer I had accidentally left in there, I was only testing with it, I only forgot to take it out.

But I think I will go to with the second way, I do not know MSHook (only the MSHook that hackers are supposed to know) But I will report back to you And thanks again for such a great reply/explanation

Updated May 7, 2015 by Guest

May 7, 2015

@@alphaMatterr @

Before I did what you suggested, I took out everything and just had the alerts and the multiplier variable, and it is still zero whenever I input my own number. Can you help me fix this? Here is the code:

#include <substrate.h>
#include "writeData.h"

int multiplier;

//static int hackScore(int);

%hook UnityAppController

- (BOOL)application:(id)fp8 didFinishLaunchingWithOptions:(id)fp12
{
UIAlertView *alert = [[UIAlertView alloc]initWithTitle:@"Run Terrio Run Score Multiplier Modifier" message:@"How much do you want the score to be multiplied by?" delegate:self cancelButtonTitle:@"Cancel" otherButtonTitles:@"Set Multiplier",nil];
[alert setAlertViewStyle:UIAlertViewStylePlainTextInput];
UITextField *inputField = [alert textFieldAtIndex:0];
[inputField setKeyboardType:UIKeyboardTypeNumberPad];
inputField.text = [NSString stringWithFormat:@"%i", multiplier];

multiplier = [inputField.text intValue];

UIAlertView *scoreAlert = [[UIAlertView alloc]initWithTitle:@"Multiplier Modifier" message:[NSString stringWithFormat:@"Score is now going to be multiplied by %i",multiplier] delegate:nil cancelButtonTitle:@"Good shit" otherButtonTitles:nil];
[scoreAlert show];
[alert show];
return %orig;
}
%end

Thanks

AnotherLurker · May 7, 2015

How can it work if there is no writeData

@@shmoo

KingRalph · May 8, 2015

try using code x8486375737488348349354 for it the multiplier variable shouldn't stay at 0

O_o bruh

Moreover, from your code:
if(hackScore(multiplier)==255){
writeData(0xHIDDEN, 0xFF0080E2);
}
what your saying is if the user chooses their multiplier to be set as 255, THEN you will write the hex 0xFF... to the given

To fix this you have three options

1. write a RIDICULOUSLY HUGE if else OR switch statement containing every integer value from 0 - 255.

This is the most retarded and inefficient way of doing what you want, however it will technically get it done (please dont do it this way if you choose to fix the mistake)

You don't have to do that. Obj-C has a built in solution.

Change your code from

if(hackScore(multiplier)==255){
writeData(0xHIDDEN, 0xFF0080E2);
}

To:

if(hackScore(multiplier)<256){
writeData(0xHIDDEN, 0xFF0080E2);
}

By doing this, you're saying that if your user chooses their multiplier to be any number that's equal to 255 or less, THAT'S when you'll write the hex to 0xFF...

Then again. I don't see a reason as to why you want to give the user the option to choose a score multiplier.

CSCI ftw

alphaMatterr · May 8, 2015

@KingRalph, the goal is to allow users to enter any integer between 0 and 255 as a score multiplier.. If you only wanted one score multiplier option then you wouldn't need any user input (the UITextField would be useless)...

..

@shmoo, the code you posted accomplishes the first part of the three i mentioned in the post.

- find the value inputted for multiplier

- take that value and convert it to hex

- take your new hex to be written, in this case XX0080E2, and combine it with the value you converted from the previous step

now you have to convert the value of the multiplier to hex... so if

multiplier = 10;

then after converting you'll have:

multiplier = 0x0A;

..

The final step after that will be to take your offset, XX0080E2 and replace the XX with the given multiplier hex value.

Another way you can accomplish steps 2 and 3 (involves a little bit of math but if you understand this method it might be easier / cooler). Check the spoiler below for more info

First were going to derive an "incrementation value", I'll explain more below...

****************************

SHORT DERIVATION

0x000080E2 and 0x010080E2 only differ by 0x01000000, which when converted to decimal is 16777216

additionally, 0x000080E2 converted to decimal is 32994. Thus, we can make a linear function, f(x) = mx + b, where b =32994 and m = 16777216. so if a multiplier of 10 is entered we can just do:

f(10) = 16777216 * 20 + 32994 = 335577314

converting to hex then yield the correct offset of 0x140080E2

NOTE: you don't technically have to convert to decimal, you could just do all of the math through hex, however I thought that may be a bit confusing because its base 16 instead of base 10.

****************************

LONG DERIVATION

Okay, so if you set your multiplier to zero, you get the hex:

0x000080E2

converting this to decimal, you get:

32994

NOW the multiplier can only be an integer values between 0 - 255.

if you choose your multiplier to be 1, you get

0x010080E2

converting this to decimal yields the number:

16810210

lastly, subtract the initial int value of 0x000080E2 from our previous conversion and we get:

16810210 - 32994 = 16777216

16777216 can be treated as our "incrementation value". This is useful and can actually simplify the amount of code used and just use more math instead.

For example! Lets say that the user inputted a multiplier value of 20. Thus we have

multiplier = 20;

now if we want to "create" our new hex with this multiplier value, we can do something like this:

new_hex = 32994 + 16777216 * multiplier = 32994 + 16777216 * 20 = 335577314

now just convert 335577314 back to hex!

(335577314)_10 = (140080E2)_16

i.e. new_hex = 0x140080E2

now to double check we can see that

0x14 is in fact 20 when converted back to decimal and thus our math checks out!

ainet mathee phun!

****************************

Doing it this way means you can skip step 3 entirely and wont have to combine the multiplier hex with the offset via code (instead you do it mathematically)

either method you choose will get you the correct hex value needed. Now all that is left is to use writeData to apply your patch...

Updated May 12, 2015 by alphaMatterr