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## Binary to Integer (Read hex binary file)

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### #1greenpig83  Members

Posted 19 February 2014 - 12:09 PM

I'm trying to read an unknown type file (a save game file, I change the value and then see the file in Hex editor) !

And i want to read the integer values. I know where it is ! but the hex value is quite strange! I use Decimal here for easy understanding!

(byte-byte -> integer )
//128,63 -> 1           128 =  10000000   63=00111111
//0, 64 -> 2(10)            00000000-0100000064=    01000000
//64-64 -> 3(11)            01000000-01000000
//128-64 -> 4(100)          10000000-01000000
//160-64 -> 5(101)      160=10100000-01000000
//192-64 -> 6(110)      192=11000000-01000000
//224-64 -> 7(111)          11100000-01000000
//0-65 -> 8 (1000)          00000000-01000001
//16-65 -> 9(1001)          00010000-01000001
//32-65 ->10(1010)          00100000-01000001
//48-65 ->11(1011)          00110000-01000001
//64-65 ->12(1100)          01000000-01000001
//80-65 ->13(1101)          01010000-01000001
//96-65 ->14(1110)          01100000-01000001
//112-65 ->15(1111)         01110000-01000001
//128-65 ->16(10000)        10000000-01000001

//136-65 ->17(10001)        10001000-01000001
//144-65 ->18(10010)        10010000-01000001
//152-65 ->19(10011)        10011000-01000001

//160-65 ->20(10100)        10100000-01000001
//168-65 ->21(10101)        10101000-01000001
//176-65 ->22(10110)        10110000-01000001

//32-66 ->40(101000)        00100000-01000010

//48,64,80

//160-66 ->80(1010000)      10100000-01000010
//162-66 ->81(1010001)      10100010-01000010
//164-66 ->82(1010010)      10100100-01000010

(bigger values, int,int,int -> byte-byte )
//80,81,82  -> 160-66,162-66,164-66
//180,181,182 -> : 52-67,53-67,54-67
//20,21,22 -> 160 ,168 , 176 -65
//30,40,50 ->  240-65,248-65, 0-66  (3 integer -> 3 (2bytes))

//128-191 -> -1             10000000-10111111
//0-192 -> -2               00000000-11000000
//64-192 -> -3              01000000-11000000

The values is like above. It use 4 byte (but only 2 byte have values) . It seems that the left-most bit of 2nd byte are for Signed!
There is some rule , but i find it hard to figure out !
Thanks !

Edited by greenpig83, 19 February 2014 - 12:57 PM.

### #2tanzanite7  Members

Posted 19 February 2014 - 01:47 PM

Your notation is extremely cryptic and you do not explain anything - i would be amazed if anyone would be able to guess what you are talking about.

Anyway, just in case, this is how values are stored on your typical little-endian format:

(2 byte integer)

original value: 4660 -(hex)-> 0x1234 -(as bytes)-> 0x34, 0x12

original value: -4660 -(hex)-> 0xEDCC -(as bytes)-> 0xCC, 0xED

Note: 0xEDCC = 0x10000 - 0x1234

(4 byte integer)

original value: -123456789 -(hex)-> 0xF8A432EB -(as bytes)-> 0xEB, 0x32, 0xA4, 0xF8

edit:

If i understood your notation - looks like floating point to me. Treat it as a standard 4 byte float - see if that does the trick.

Edited by tanzanite7, 19 February 2014 - 02:04 PM.

### #3tanzanite7  Members

Posted 19 February 2014 - 02:09 PM

Hats off to my binary pattern recognition: I am fairly certain it is float:

http://www.binaryconvert.com/result_float.html?decimal=049

http://www.binaryconvert.com/result_float.html?decimal=056049

etc.

### #4Álvaro  Members

Posted 19 February 2014 - 02:49 PM

It's a 16-bit floating point type.

#include <cstdio>
#include <cmath>

void print_as_tinyfloat(int i) {
int sign = (i < 0);
int exponent = static_cast<int>(std::floor(std::log2(std::abs(i))));
int mantissa = (i << (7 - exponent)) ^ (1 << 7);
unsigned tinyfloat = (sign << 15) + ((exponent + 127) << 7 ) + mantissa;

std::printf("%d -> %d %d\n", i, tinyfloat % 256, tinyfloat / 256);
}

int main() {
for (int i=1; i<20; ++i)
print_as_tinyfloat(i);
}


### #5greenpig83  Members

Posted 19 February 2014 - 04:06 PM

Wow, never think of that. I just think inside the box that it must be integer!

I'll try and see if it's right ! Spend hours to figure a very simple thing

Yeah. that it ! Using C# it's more simple !

byte []t={0,0,32,65};
float myFloat = System.BitConverter.ToSingle(t, 0); //-> return 10

Because the game only let me input integer value, so I think that it's integer !

Edited by greenpig83, 19 February 2014 - 04:43 PM.

### #6tanzanite7  Members

Posted 19 February 2014 - 05:15 PM

It's a 16-bit floating point type.

Half-float does not have an 8-bit exponent . The realization that an exponent could explain the related change in the byte pairs and the one bit crossing the byte border (due to the sign bit) was the smoking gun for me. The exponent encoding works as a fair giveaway too, but missed it at first (64 = 128 >> 1).

Because the game only let me input integer value, so I think that it's integer !

This has been actually very common occurrence i my savegamemucking endeavors. Even with values where having a float makes no sense - ie. the value can logically only be whole numbers (a'la unit count etc).

Edited by tanzanite7, 19 February 2014 - 05:16 PM.

### #7Álvaro  Members

Posted 19 February 2014 - 07:52 PM

I thought for a minute that it was a strange custom 16-bit float with 8-bit exponent and 7-bit mantissa, but what's really going on is that it's a traditional 32-bit float, and the OP only gave us the 2 non-zero bytes (and I missed the comment about the two zero bytes). The other two bytes are always zero because for small integers they represent things after the point.

So just forget the code I posted. The answer is that it's a float, as tanzanite7 pointed out.

Edited by Álvaro, 19 February 2014 - 07:53 PM.

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