ARM assembler for ARM learning

ARM assembly language program structure

In ARM (Thumb) assembly language program, the code is organized by program segments. Segments can be divided into code sections and data sections. A assembler program should have at least one code segment. When the program is long, it can be divided into multiple code segments and data segments. When the program is compiled and linked, multiple segments eventually form an executable image file.
The composition of the executable image file:


The linker arranges the segments in the memory according to the system default or user set rules. Therefore, the relative position between the segments of the source program and the executable image file is generally not the same.

Assembly language subroutine call

In ARM assembly language program, the call of subroutine is usually realized by BL instruction.

bl DELAY           ;DELAY Subroutine name
...
ldr r0 ,= 0xffff
DELAY:	
	sub r0,r0,#1    ;r0 = r0 - 1
	cmp r0,#0; compare if the value of r0 is equal to 0
	bne DELAY       ;If it's not equal, it's back DELAY implement
ldr r1 ,= 0x08
...

(1) when the subprogram is executed, the return address of the subprogram should be saved in the connection register LR, and the program counter PC points to the entry point of the subprogram.
(2) the subroutine needs to be returned after execution, only the return address stored in LR needs to be copied back to the program counter PC.
(3) at the same time of calling subroutine, it can also transfer parameters and return operation results from subroutine. Generally, it can be completed by register R0~R3

Compilation example

Assembly implementation string copy:
Compilation environment: Keil4

	area init,code,readonly  
	entry
	 
	ldr r0,= datablock1   ;Get the address of the string and save it to r0 in
	ldr r2,= datablock2   ;Get the address of the allocated memory unit with the size of 100 and save it to r2 in
cat
	ldrb r1,[r0],#1; take out 1 byte of data and store it in r1, r0 address plus 1
	strb r1,[r2],#1. Save the 1 byte data to r2, r2 address plus 1
	cmp r1, #0; compare whether the extracted string character is' \ 0 '
	bne cat              ;If not, continue copying

	b .
	
datablock1
	dcb "hello young man!",0

datablock2
	space 100           ;Allocate 100 size memory units

	end                      ;Marking the end of the program

Get student's highest score:

;init.s
	area init,code,readonly	 ;The definition has a name init Code segment(code) Property is read-only
	entry

	import getmax
	import scores
	import maxscore
	import numofstudent

;	mov r0,#1
;	mov r1,#2
;	cmp r1,r0
;	movhi r0,r1


	ldr r2,=scores
	ldr r4,=numofstudent
	ldrb r5,[r4]	 ;Ten students
	ldrb r0,[r2]   ;Take out the scores of the first student
	add r2,r2,#1
	sub r5,r5,#1; take out the first student, minus 1

	bl getmax

	ldr r4,= maxscore
	strb r0,[r4]

 	b .

	end
;maxoftwo.s
;Two numbers to compare are r0,r1 in
;The maximum number of comparisons is r0 Among

	area max,code,readonly
	import scores              ;statement scores From outside

	export getmax             ;statement getmax Can be called externally


getmax   
	 ldrb r1,[r2],#1. Take out the second student's score
	 sub r5,r5,#Take out another student's score, and reduce 1
	 cmp r1,r0
	 movhi r0,r1  ; Compare, put the high scores in r0 in
	 cmp r5,#0; judge whether ten students have finished
	 bne getmax	  ;Not to continue
	 bx lr
	 end
;score.s

	export scores                   ;Declaration may scores Called externally        
	export numofstudent
	export maxscore

	area score ,data,readwrite  ;Define data segment, readable and writable

scores 
	dcb 65,78,92,47,77,83,59,93,82,97        ;Student achievement 
numofstudent
	dcb 10                ; Number of students
maxscore
	dcb 0;//Store maximum score
	
	end

Tags: Assembly Language Linker

Posted on Mon, 18 Nov 2019 09:55:47 -0500 by CrazeD