Chez 0 Report post Posted February 9, 2006 (edited) I'm trying to convert a 32-bit Hex string into 32-bit binary and then back.Whats the theory behind this? I was checking online and some articles said check each char divided against a power of 10. ... ? A bit confused here. Any help appreciated.using SPIMon Fedora Core platform Edited February 10, 2006 by Chez (see edit history) Share this post Link to post Share on other sites
Cerebral Stasis 0 Report post Posted February 9, 2006 I'm assuming that by "Hx" you mean hexadecimal. You may want to try one of the free conversion scripts found here. Share this post Link to post Share on other sites
Chez 0 Report post Posted February 9, 2006 (edited) Yes, hexadecimal. But I was actually looking for code help, not an actual converter. thanks though.I can convert from decimal to binary simply using a recurring function with remainders...recurring function:47 / 2 gives DIV=23 MOD=1, bin string = 123 / 2 gives DIV=11 MOD=1, bin string = 1111/2 gives DIV=5 MOD=1, bin string = 1115/2 gives DIV=2 MOD=1, bin string = 11112/2 gives DIV=1 MOD=0, bin string = 011111/2 gives DIV=0 MOD=1, bin string = 101111but Hex is harder to code because it has letters as well as numbers. So I'm not sure how I can create a function that will do that.not looking for code, just how would you go about converting hex to binary and then back again? Edited February 10, 2006 by Chez (see edit history) Share this post Link to post Share on other sites
Chez 0 Report post Posted February 17, 2006 ok, so I figured it out myself after days of reading up on the complexities of MIPS architecture. I basically IO-ed the HEX input into a register and used 'lb' to grab each character (by cooredinating ascii value). this is the part I had problems with previously: I didn't know the loaded byte would be converted to binary in the register, so I was origionally comparing ascii values. But lo and behold, Assembly language did not completely let me down once I figured you could compare chars as in C or Java simply by bordering your target char with single quotes. Once I figured that out, the whole comparison was a peice of cake. Finding, say, '3' would run through the converter loop and input 0011 to a string in another register, srl 4, increase position, and continue. Fairly easy once I got the jist of it. code follows: # This program inputs a Hex number and# converts it to Binary. Then alters the # Binary number and outputs new Hex number. .dataPROMPT: .asciiz "Enter Hex: "ORI: .asciiz " "RES: .asciiz " "OUT: .asciiz " -> " .textmain:################################### INITIALIZE REGISTERS ################################### li $s0, 0 #INPUT HEX li $s1, 0 #TEMP BYTE STORAGE li $s2, 0 #I li $s3, 0 #BINARY NUMBER li $s4, 0 #TEMP BINARY NUMBER li $s5, 0 #TEMP BINARY CHAR la $s6, ORI #ORIGIONAL HEX la $s7, RES #FINAL HEX################################### MAIN ################################### j INPUT #PROMT FOR HEX AND STORE ITSTART: j HIN #HEX INPUT TO BINARY STRINGL2: j BINC #BINARY CONVERTERL3: add $s7, $s7, 7 j HOUT #HEX OUTPUT FROM BINARY STRINGL4: j PRINT #PRINT AND END#################################### READ AND STORE INPUT HEX ####################################INPUT: la $a0, PROMPT #LOAD THE ADDRESS OF PROMPT FOR HEX li $v0, 4 #SYSTEM CALL FOR THE PRINT STRING syscall #PRINT THE PROMPT li $v0, 8 #SYSTEM CALL TO READ IN A STRING move $a0,$s6 syscall #READ IN HEX move $s0, $v0 #COPY ORIGIONAL HEX VALUE INTO s0 move $s0,$s6 #PUT TEMP HEX VALUE INTO s6 j START #CONVERT TO BINARY STRING#################################### HEX INPUT ####################################HIN: beq $s2,8,L2 #IF END OF HEX INPUT(8), BREAK lb $s1,0($s0) #STORE CURRENT HEX CHAR IN $s1 sll $s3,$s3,4 #SHIFT LEFT 4 BITS FOR NEXT INPLACE beq $s1,'0',IF0 #IF CHAR= 0 beq $s1,'1',IF1 #IF CHAR= 1 beq $s1,'2',IF2 #IF CHAR= 2 beq $s1,'3',IF3 #IF CHAR= 3 beq $s1,'4',IF4 #IF CHAR= 4 beq $s1,'5',IF5 #IF CHAR= 5 beq $s1,'6',IF6 #IF CHAR= 6 beq $s1,'7',IF7 #IF CHAR= 7 beq $s1,'8',IF8 #IF CHAR= 8 beq $s1,'9',IF9 #IF CHAR= 9 beq $s1,'a',IFA #IF CHAR= A beq $s1,'b',IFB #IF CHAR= B beq $s1,'c',IFC #IF CHAR= C beq $s1,'d',IFD #IF CHAR= D beq $s1,'e',IFE #IF CHAR= E beq $s1,'f',IFF #IF CHAR= FIF0: ori $s3,$s3,0 j RUTGER IF1: ori $s3,$s3,1 j RUTGER IF2: ori $s3,$s3,2 j RUTGER IF3: ori $s3,$s3,3 j RUTGER IF4: ori $s3,$s3,4 j RUTGER IF5: ori $s3,$s3,5 j RUTGER IF6: ori $s3,$s3,6 j RUTGER IF7: ori $s3,$s3,7 j RUTGER IF8: ori $s3,$s3,8 j RUTGER IF9: ori $s3,$s3,9 j RUTGER IFA: ori $s3,$s3,10 j RUTGER IFB: ori $s3,$s3,11 j RUTGER IFC: ori $s3,$s3,12 j RUTGER IFD: ori $s3,$s3,13 j RUTGER IFE: ori $s3,$s3,14 j RUTGER IFF: ori $s3,$s3,15 j RUTGERRUTGER: addi $s2,$s2,1 #INCREASE I addi $s0,$s0,1 #INCREASE POSITION j HIN #RESTART LOOP#################################### BINARY CONVERTER ####################################BINC: li $s1,0 #CLEAR TEMP BYTE STORAGE li $s2,0 #CLEAR I srl $s4,$s3,1 #SHIFT RIGHT 1 xor $s3,$s3,$s4 #EXCLUSIVE OR TO CHANGE BINARY STRING j L3 #NEXT STEP#################################### HEX OUTPUT ####################################HOUT: beq $s2,8,L4 #END OF BINARY STRING, BREAK and $s5,$s3,0xf #GET CHAR srl $s3,$s3,4 #'TYPEWRITTER' SHIFT RIGHT beq $s5,0,IF0000 #IF BYTE= 0000 beq $s5,1,IF0001 #IF BYTE= 0001 beq $s5,2,IF0010 #IF BYTE= 0010 beq $s5,3,IF0011 #IF BYTE= 0011 beq $s5,4,IF0100 #IF BYTE= 0100 beq $s5,5,IF0101 #IF BYTE= 0101 beq $s5,6,IF0110 #IF BYTE= 0110 beq $s5,7,IF0111 #IF BYTE= 0111 beq $s5,8,IF1000 #IF BYTE= 1000 beq $s5,9,IF1001 #IF BYTE= 1001 beq $s5,10,IF1010 #IF BYTE= 1010 beq $s5,11,IF1011 #IF BYTE= 1011 beq $s5,12,IF1100 #IF BYTE= 1100 beq $s5,13,IF1101 #IF BYTE= 1101 beq $s5,14,IF1110 #IF BYTE= 1110 beq $s5,15,IF1111 #IF BYTE= 1111 IF0000: li $t0,'0' sb $t0,0($s7) j MUTGERIF0001: li $t0,'1' sb $t0,0($s7) j MUTGER IF0010: li $t0,'2' sb $t0,0($s7) j MUTGER IF0011: li $t0,'3' sb $t0,0($s7) j MUTGER IF0100: li $t0,'4' sb $t0,0($s7) j MUTGER IF0101: li $t0,'5' sb $t0,0($s7) j MUTGER IF0110: li $t0,'6' sb $t0,0($s7) j MUTGER IF0111: li $t0,'7' sb $t0,0($s7) j MUTGER IF1000: li $t0,'8' sb $t0,0($s7) j MUTGER IF1001: li $t0,'9' sb $t0,0($s7) j MUTGER IF1010: li $t0,'a' sb $t0,0($s7) j MUTGER IF1011: li $t0,'b' sb $t0,0($s7) j MUTGER IF1100: li $t0,'c' sb $t0,0($s7) j MUTGERIF1101: li $t0,'d' sb $t0,0($s7) j MUTGER IF1110: li $t0,'e' sb $t0,0($s7) j MUTGER IF1111: li $t0,'f' sb $t0,0($s7) j MUTGERMUTGER: sub $s7,$s7,1 #INCREASE POSITION addi $s2,$s2,1 #INCREASE I j HOUT #RESTART LOOP#################################### PRINT THE OUTPUT ####################################PRINT: sb $0,8($s6) #Seperate the inputs(/n -> 0) la $s7, RES #RESET $s7 li $v0, 4 #LOADS SYSTEM CALL TO PRINT STRING move $a0, $s6 #LOADS VALUE OF ORIGIONAL HEX TO CALL syscall #PRINT MIN la $a0, OUT #LOAD ADDRESS FOR FIRST STRING PART li $v0, 4 #SYSTEM CALL TO PRINT STRING syscall #PRINT STRING (" -> ") li $v0, 4 #LOADS SYSTEM CALL TO PRINT STRING move $a0, $s7 #LOADS VALUE OF FINALIZED HEX TO CALL syscall #PRINT MIN jr $ra #EXIT #END PROGRAM I'm sure it's not as optimimized as I would like, and I'm sure I could drop the number of registers down to around 4, but heck, it works and that'll all I care about.*why Rutger and Mutger... I don't know. don't ask. I was running out of names for methods. Share this post Link to post Share on other sites
DeveloperX 0 Report post Posted April 11, 2006 3 years ago I wrote HexToBin function. function HexToBin(Hexadecimal: string): string;const BCD: array [0..15] of string = ('0000', '0001', '0010', '0011', '0100', '0101', '0110', '0111', '1000', '1001', '1010', '1011', '1100', '1101', '1110', '1111');var i: integer;begin for i := Length(Hexadecimal) downto 1 do Result := BCD[StrToInt('$' + Hexadecimal[i])] + Result;end;procedure TForm1.Button1Click(Sender: TObject);begin ShowMessage(HexToBin('FFA1')); // Returns 1111111110100001end; But for BinToHex(s:string) converting you can use standard BinToHex function in Math module.And such code:BinToHex = DecToHex(BinToDec(Bin)); //maybe works! good luck! Share this post Link to post Share on other sites
