Write a program using the LOOP instruction with indirect addressing that copies

Question

Write a program using the LOOP instruction with indirect addressing that copies a string from source to target, reversing the character order in the process. Use the folling variables: source BYTE "This is the source string", 0 target BYTE SIZEOF source DUP('#') Insert the following statements immediately after the loop to display the hexadecimal contents of the target string: mov esi,OFFSET target ;offset of variable mov ebx, 1 ; byte format mov ecx, SIZEOF target ; counter call DumpMem If your program works correctly, it will display the following sequence of hexadecimal bytes: 67 6E 69 72 74 73 20 65 63 72 75 6F 73 20 65 68 74 20 73 69 2073 69 68 54

Explanation / Answer

Indirect addressing uses a register as a pointer to hold the offset (address) of the data to be manipulated. This is ideal for array processing because each element of the list is simply an addition to the value stored in the register. You can think of this in the same way C++ uses a pointer to access element of an array.

The LOOP instruction repeats a block of statements a specific number of times. The ECX register is automatically used as a counter and is decremented each time the loop repeats.

The loop destination must be within -128 to + 127 bytes if the current location counter.

A nested loop is a loop inside a loop. When creating a nested loop you must make sure that you save off the value of the ECX register before moving to the nested loop and back

.data

source BYTE "This is the source string",0

target BYTE SIZEOF source DUP('#')

.code

reversecopy PROC

mov ESI,0 ; source register

mov EDI,SIZEOF source-2D ; destination register

mov ECX,SIZEOF source ; loop counter

Copy:

mov DL,source[ESI] ; get a character from source

mov target[EDI],DL ; store it in the target

inc ESI ; move to next character - increment the source index

dec EDI ; decrement the destination index

loop Copy ; repeat for entire string;

; Outputresults;

mov ESI,OFFSET target ;offset of variable

mov EBX,1 ;byte format

mov ECX,SIZEOF target ;counter

call DumpMem;

; Our results should be: 67 6E 69 72 74 73 20 65 63 72 75 6F 73 20 65 68

; 74 20 73 69 20 73 69 68 54;

;Return to operating system;

; exit

reversecopy ENDP

END reversecopy

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.