   Question using Log
  Hey, I was just wondering if there was a possible log function or any function, or way, that I could do this.I want to get three separate integers:int a = 0x1;int b = 0x0;int c = 0x0d;However, I want to put these integers so the new integer intact is 0x100d.I understand if I have 100d and just doint num1 = (intTemp / 0x10000);int rem1 = (intTemp % 0x10000);int num2 = ( rem1 / 0x1000 );int rem2 = ( rem1 % 0x1000 );int num3 = rem2;It will split it up reverse of how I want it done now.  Is there a way to reverse what I did here^^ ?Thanks.   
  you can use logic operations to do that.& - logical and| - logical or<< bit shift left>> bit shift rightFor instance:0x1 | (0x1 << 1) => 0x11(0x20 & (0xF0) ) >> 8 => 0x2   ;  Alright thanks! I'll see if I can figure it out from that   ;For instance:0x1 | (0x1 << 1) => 0x11(0x20 & (0xF0) ) >> 8 => 0x2Yeah shifting and or'ing is the answer, except you're example's a little bit off 0x1 | (0x1 << 1) => 0x3(0x20 & (0xF0) ) >> 8 => 0x00x1 | (0x1 << 4) => 0x11(0x20 & (0xF0) ) >> 4 => 0x2   ;[quote name='KulSeran' timestamp='1303872992' post='4803360']For instance:0x1 | (0x1 << 1) => 0x11(0x20 & (0xF0) ) >> 8 => 0x2Yeah shifting and or'ing is the answer, except you're example's a little bit off 0x1 | (0x1 << 1) => 0x3(0x20 & (0xF0) ) >> 8 => 0x00x1 | (0x1 << 4) => 0x11(0x20 & (0xF0) ) >> 4 => 0x2[/quote]ah...... derp. thanks hodman.  obviously i was only half thinking when i wrote that.   