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Hamming code error-correction in ARM assembly homework help

The assignment 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 assignment helpers and learn how to correct a hamming code error.

Correcting Hamming Code in ARM Assembly

Hamming code error-correction in ARM assembly homework help

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