6. Implementation of REP MOVS instruction using C++ and verify the functioning of the instruction. (25 points) a) Write a C++ /Python code the emulate the function of REP MOVS instruction in X86 and verify its function. You may decide the length of the string used. b) Implement REP MOVS instruction using the RSIC ARM7 Assembly language programing and verify using Keil micro-vision. c) Make sure you comment your C++/Python and AM7 code well. REP MOVS (DEST SRC) Destination - Source; if(IsByteMove()){ if(DF 0) { IF AddressSize = 16 THEN (E)SI - (E)SI + 1; (E )DI - (E)DI + 1; Use CX for CountReg; ELSE IF AddressSize = 64 and REX.W used THEN Use RCX for CountReg; Fl; ELSE else { (E)SI - (E)SI - 1; (E)DI - (E)DI - 1; Use ECX for CountReg; else if(IswordMove()) { if(DF 0) { Fl; WHILE CountReg # 0 (E)SI - (E)SI + 2; (E)DI - (E)DI + 2; DO else ( Service pending interrupts (if anyk Execute associated string instruction CountReg + (CountReg - 1); IF CountReg = 0 THEN exit WHILE loop; Fl; IF (Repeat prefix is REPZ or REPE) and (ZF = 0) or (Repeat prefix is REPNZ or REPNE) and (ZF = 1) THEN exit WHILE loop; Fl; (E)SI - (E)SI 2; (E) DI - (E)DI 2; else ( //doubleword move if(DF 0) ( (E)SI - (E)SI + 4; (E)DI - (E)DI + 4; else { (E)SI - (E)SI 4; OD; (E) DI - (E)DI 43;
6. Implementation of REP MOVS instruction using C++ and verify the functioning of the instruction. (25 points) a) Write a C++ /Python code the emulate the function of REP MOVS instruction in X86 and verify its function. You may decide the length of the string used. b) Implement REP MOVS instruction using the RSIC ARM7 Assembly language programing and verify using Keil micro-vision. c) Make sure you comment your C++/Python and AM7 code well. REP MOVS (DEST SRC) Destination - Source; if(IsByteMove()){ if(DF 0) { IF AddressSize = 16 THEN (E)SI - (E)SI + 1; (E )DI - (E)DI + 1; Use CX for CountReg; ELSE IF AddressSize = 64 and REX.W used THEN Use RCX for CountReg; Fl; ELSE else { (E)SI - (E)SI - 1; (E)DI - (E)DI - 1; Use ECX for CountReg; else if(IswordMove()) { if(DF 0) { Fl; WHILE CountReg # 0 (E)SI - (E)SI + 2; (E)DI - (E)DI + 2; DO else ( Service pending interrupts (if anyk Execute associated string instruction CountReg + (CountReg - 1); IF CountReg = 0 THEN exit WHILE loop; Fl; IF (Repeat prefix is REPZ or REPE) and (ZF = 0) or (Repeat prefix is REPNZ or REPNE) and (ZF = 1) THEN exit WHILE loop; Fl; (E)SI - (E)SI 2; (E) DI - (E)DI 2; else ( //doubleword move if(DF 0) ( (E)SI - (E)SI + 4; (E)DI - (E)DI + 4; else { (E)SI - (E)SI 4; OD; (E) DI - (E)DI 43;
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Please answer as detailed with all the steps
Transcribed Image Text:6. Implementation of REP MOVS instruction using C++ and verify the
functioning of the instruction. (25 points)
a) Write a C++ /Python code the emulate the function of REP MOVS
instruction in X86 and verify its function. You may decide the length
of the string used.
b) Implement REP MOVS instruction using the RSIC ARM7 Assembly
language programing and verify using Keil micro-vision.
c) Make sure you comment your C++/Python and AM7 code well.
REP MOVS (DEST SRC)
Destination = Source;
if(IsByteMove ()) {
if (DF 0) {
IF AddressSize = 16
(E)SI = (E)SI + 1;
(E)DI = (E)DI + 1;
THEN
Use CX for CountReg;
ELSE IF AddressSize = 64 and REX.W used
THEN Use RCX for CountReg; Fl;
else {
(E)SI - (E)SI - 1;
(E)DI = (E)DI - 1;
ELSE
Use ECX for CountReg;
}
else if(IswordMove ()) {
Fl;
if (DF 0) {
WHILE CountReg = 0
DO
(E)SI = (E)SI + 2;
(E) DI = (E)DI + 2;
else (
Service pending interrupts (if anyk
Execute associated string instruction:
CountReg + (CountReg - 1);
IF CountReg = 0
THEN exit WHILE loop; Fl;
IF (Repeat prefix is REPZ or REPE) and (ZF = 0)
or (Repeat prefix is REPNZ or REPNE) and (ZF = 1)\
THEN exit WHILE loop; Fl;
(E)SI = (E)SI - 2;
(E)DI = (E)DI - 2;
}
}
else { //doubleword move
if (DF 0) {
(E)SI = (E)SI + 4;
(E) DI - (E) DI + 4;
else {
(E)SI = (E)SI - 4;
(E) DI = (E)DI
OD;
4;
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