Microcomputer principle operation

example 4-12：

;eg0501.asm
	include io32.inc
	.data
string	byte 'Do you have fun with Assembly?',0
	.code
start:
	xor ebx,ebx
again:
	mov al,string[ebx]
	cmp al,0
	jz done
	inc ebx
	jmp again
done:	mov eax,ebx
	call dispuid

	exit 0
	end start

exercises 4-13

;eg0201.asm
	include io32.inc
	.data
string	byte 'Do you have fun with Assembly?',0
space	dword ?
	.code
start:
	mov esi,offset string
	xor ebx,ebx
again:	mov al,[esi]
	cmp al,0
	jz done
	cmp al,20h
	jne next
	inc bx
next:	inc esi
	jmp again
done:	mov space,ebx

	exit 0;
	end start

example 4-10：

include io32.inc
	.data

	.code
start:
	mov edx,1
	mov esi,2
done1:	add edx,esi 	     ; Add natural numbers , Sum up 
	jc done                         ; Highest carry , Jump 
	inc esi                           ; Addend plus one 
	jmp done1                    ; Add natural numbers repeatedly 
done:	sub esi,1     	      ; Plus minus one , Make the cumulative sum effective without carry 	
	mov ebx,1
	mov ecx,2
nom:	add ebx,ecx                   ; Find the effective cumulative sum 
	cmp ecx,esi                    ; Add to N The boundary of stops 
	jz again
	inc ecx                           ; Addend plus one 
	jmp nom                        ; Add natural numbers repeatedly 
again:	mov eax,ebx                  ; Displays the maximum value of the effective cumulative sum 
	call dispuid
	call dispcrlf                     ; Line break 
	mov eax,ecx                   ; Show N The boundaries of 
	call dispuid	
	
	exit 0
	end start

Example 4-15：

;eg0101.asm
	include io32.inc
	.data
regd	byte 'EAX=',8 dup(0),'H',0
	.code
start:
	mov eax,1234abcdh
	xor ebx,ebx
	mov ecx,8
again:	rol eax,4
	push eax
	call htoasc
	mov regd+4[ebx],al
	pop eax
	inc ebx
	dec ecx
	jnz again
	mov eax,offset regd
	call dispmsg
htoasc	proc
	and al,0fh
	or al,30h
	cmp al,39h
	jbe htoend
	add al,7
htoend:	ret
htoasc	endp
	
	exit 0
	end start

exercises 4-29：

;eg0502.asm
	include io32.inc
	.data
var	byte 'This is a test!'
	.code 
start:	mov eax,offset var
	mov ecx,sizeof var
	call dispmem
	exit 0
dispmem	proc
	push ebx
	mov ebx,eax
dispml:	mov al,[ebx]
	call disphb
	mov al,' '
	call dispc
	inc ebx
	loop dispml
	pop ebx
	ret
dispmem	endp
	end start

example 4-18：

 

;eg0101.asm
	include io32.inc
	.data
count	=10
array	dword count dup(0)
temp	dword ?
readbuf	byte 30 dup(0)
	.code
start:
	mov ecx,count
	mov ebx,offset array
again:	call read
	mov eax,temp
	mov [ebx],eax
	add ebx,4
	dec ecx
	jnz again
read 	proc
	push eax
 	push ebx
	push ecx
	push edx
read0:	mov eax,offset readbuf
	call readmsg
	test eax,eax
	jz readerr
	cmp eax,12
	ja readerr
	mov edx,offset readbuf
	xor ebx,ebx
	xor ecx,ecx
	mov al,[edx]
	cmp al,'+'
	jz read1
	cmp al,'-'
	jnz read2
	mov ecx,-1
read1:	inc edx
	mov al,[edx]
	test al,al
	jz read3
read2:	cmp al,'0'
	jb readerr
	cmp al,'9'
	ja readerr
	sub al,30h
	imul ebx,10
	jc readerr
	movzx eax,al
	add ebx,eax
	cmp ebx,80000000h
	jbe read1
readerr:	mov eax,offset errmsg
	call dispmsg
	jmp read0
read3:	test ecx,ecx
	jz read4
	neg ebx
	jmp read5
read4:	cmp ebx,7fffffffh
	ja readerr
read5:	mov temp,ebx
	pop edx
	pop ecx
	pop ebx
	pop eax
	ret
errmsg	byte 'Input error,enter again:',0
read	endp
	
	exit 0
	end start

exercises 7-22：