# Hamming code error-correction in ARM assembly homework help

July 10, 2024
🇺🇸 United States
Programming
Dr. Madison Davidson, an accomplished programming expert, earned her Ph.D. from the University of York, United Kingdom. With 15 years of experience, she excels in solving intricate programming assignments with finesse and expertise.
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• Correcting Hamming Code in ARM Assembly
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The homework deals with correct errors in a word encoded using Hamming code with 16 information bits, 5 parity bits, and a total parity bit. The implementation detects and can correct single-bit errors and can detect double-bit errors and other errors. The program uses a predefined word and tries to detect and correct the single-bit errors using the parity bits. Go through the solution provided below by our ARM assembly homework help solvers and learn how to correct a hamming code error.

## Correcting Hamming Code in ARM Assembly

```AREA program, CODE ENTRY Main LDR R11,=2_0000000000010011011111 ; word encoded in hamming code as 16 bits data, 5 parity, 1 Pall AND R0,R11,#0x3F MOV R0,R0,LSR #1 ; move the 5 parity bits to R0 MOV R1,R11,LSR #6 ; move the 16 data bits to R1 ; build the Hamming encoding inserting the parity bits at 2^n positions LDR R2,=1 ; current power of 2 LDR R3,=1 ; bit counter LDR R4,=0 ; final hamming encoding AND R6,R11,#1 ; get Pall in R6 LDR R7,=1 ; current power of 2 for parity bloop CMP R3,R7 BNE cpydat ; if the current bit is not a power of 2, skip AND R5,R0,R2 ; get current parity bit ORR R4,R4,R5 ; put bit in current bit position MOV R1,R1,LSL #1 ; shift data bits MOV R7,R7,LSL #1 ; get next power of 2 for parity B next cpydat AND R5,R1,R2 ; get current data bit ORR R4,R4,R5 ; put bit in current bit position MOV R0,R0,LSL #1 ; shift parity bits next MOV R6,R6,LSL #1 ; advance Pall to msb MOV R2,R2,LSL #1 ; get next power of 2 ADD R3,R3,#1 ; increment number of bits CMP R3,#21 ; continue while we haven't built the 21 bit encoding BLE bloop ORR R4,R4,R6 ; add Pall to encoded word MOV R0,R4 BL Parity ; calculate total parity MOV R1,R0 ; R1=total parity LDR R2,=0 ; R2 will contain the 5 parity bits LDR R7,=4 ; calculate the 5 parity bits LDR R5,=P0 ; load start address masks in R5 ploop LDR R6,[R5,R7,LSL #2] ; load bit mask for calculating current P AND R0,R4,R6 ; select the bits to use for calculating parity BL Parity ; calculate parity MOV R2,R2,LSL #1 ; shift old parity bits to the left ORR R2,R2,R0 ; add current parity bit SUBS R7,R7,#1 BGE ploop CMP R2,#0 BEQ NOERR CMP R1,#0 BEQ DBLERR ONEERR LDR R0,=1 shl SUBS R2,R2,#1 BEQ endshl MOV R0,R0,LSL #1 ; use R2 to get the erroneous position B shl endshl EOR R4,R4,R0 ; invert erroneous bit LDR R0,=msg3 ; print single error SWI &2 MOV R0,R4 ; set corrected encoding in R0 B exit DBLERR LDR R0,=msg2 ; print double error SWI &2 B exit NOERR1 CMP R1,#0 ; BEQ NOERR ERR22 ; error in parity bit LDR R0,=1 MOV R0,R0,LSL #21 EOR R4,R4,R0 ; invert bit LDR R0,=msg1 ; print error Pall, corrected SWI &2 MOV R0,R4 ; set corrected encoding in R0 B exit NOERR LDR R0,=msg0 ; print no error SWI &2 MOV R0,R4 ; set corrected encoding in R0 exit SWI &11 ; exit program ; calculate the parity of the word in R0 which is assumed to be 22 bits Parity STMIA SP!, {R4,R5,R14} LDR R4,=0 LDR R5,=22 ; number of bits in the word l1 ; loop to calculate total parity, saving it in R4 MOVS R0,R0,LSR #1 ; move the lsb to the carry EORHS R4,R4,#1 ; XOR the current bit with the carry SUBS R5,R5,#1 ; decrement number of bits to process BNE l1 ; continue for all bits in word MOV R0,R4 ; return parity LDMDB SP!, {R4,R5,PC} ; restore registers and return ; positions in encoded word for calculating the parity bits P0 DCD 2_101010101010101010101 P1 DCD 2_001100110011001100110 P2 DCD 2_110000111100001111000 P3 DCD 2_000000111111110000000 P4 DCD 2_111111000000000000000 msg0 DCB "No error detected.",10,0 msg1 DCB "Error in Pall detected and corrected.",10,0 msg2 DCB "Double error detected, can't be corrected.",10,0 msg3 DCB "Single error detected and corrected.",10,0 END ```

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