This is my first time to leaned Assemblr programming. Professor gave us this file and teach. and my problem is, what is this for? I learned c++, and that was for calculating something, etc. what the heck is this? please teach me this as easiest as possible

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;Example of using the add instruction

;add instruction - ; ; General form: ; ADD operand1,operand2 ; Description: ; operand1 = operand1 + operand2 ; Allowable operands: ; reg,reg - reg,mem - reg,immed ; mem,reg - mem,immed ; Flags changed: ; O,S,Z,A,P,C ; Note: ; Both operands must be of the same size. ; Can be unsigned or signed binary integers.

.386 ;identifies minimum CPU for this program

.MODEL flat,stdcall ;flat - protected mode program ;stdcall - enables calling of MS_windows programs

;allocate memory for stack ;(default stack size for 32 bit implementation is 1MB without .STACK directive ; - default works for most situations)

.STACK 4096 ;allocate 4096 bytes (1000h) for stack

;*************************PROTOTYPES*****************************

ExitProcess PROTO,dwExitCode:DWORD ;from Win32 api not Irvine

ReadChar PROTO ;Irvine code for getting a single char from keyboard ;Character is stored in the al register. ;Can be used to pause program execution until key is hit.

WriteHex PROTO ;Irvine function to write a hex number in EAX to the console

;************************DATA SEGMENT***************************

.data

aNum byte 7h bNum byte 1h

;************************CODE SEGMENT****************************

.code

main PROC

mov ebx, 0C456CCh ;ebx = 0C456CCh mov eax, 9ABC12h ;eax = 9ABC12h mov edx, 45234Bh ;edx = 45234Bh mov bl, 10000b ;bl = 10000b = 10h = 16 (base 10) ;add bl to eax

;add eax, bl ; illegal. both operands must be same size

;add eax, ebx is legal but ebx does not contain correct number to add ;solution: use the movzx instruction ;movzx allows you to move a smaller operand into a larger one ;and zeros out the upper part of the larger operand

movzx edx, bl ;copy bl to dl and zero ; out upper part of edx add eax, edx ;add edx to eax

call WriteHex ;Print number in eax to the console in hex

;add ax, bl ; illegal. both operands must be same size ;add eax, bx ; illegal. both operands must be same size ;Problem if sum will not fit into destination mov bx, 0ffffh ;bx = fffffh. ffffh is the largest 2 byte number mov dx, 1ch ;dx = 1ch add dx,bx ;error: sum will not fit into dx (FFFF+1C = 1001Bh but edx contains only 1bh) ;solution if sum will not fit into destination: copy both operands to a larger type using movzx ;then do addition mov dx, 1ch ;reinitialize dx to 1ch movzx ecx, dx ;copy dx to cx and zero out the upper part of ecx movzx esi, bx ;copy bx to si and zero out the upper part of esi add ecx,esi ;add esi to ecx: ffffh+1cH =1001bh

;add 2 variables mov ebx, offset bNum;get address of bNum to look at bNum in memory window ;add aNum, bNum; illegal. no memory to memory operations

;to add one variable to another first copy one of them to a register

mov al, aNum ; copy aNum to a register add bNum, al ; add al to bNum and store answer back in bNum

call ReadChar ; Pause program execution while user inputs a non-displayed char INVOKE ExitProcess,0 ;exit to dos: like C++ exit(0)

main ENDP END main