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MASM Tutorial

Follow this tutorial step by step:

- You can use almost any text editor to create an assembly program. In
this example, we will use Microsofts EDIT. Type edit example1.asm on the command prompt and enter the text of the program. Save the file by Alt-F,Alt+S. Exit Alt-F,Alt-X

- Compile and link the assembly file by issuing ml /Zi example1.asm

- Now lets start and configure the Code View debugger.

o Type cv example1.exe at the command prompt.Enter Alt-W and make sure that you have the following windows on the screen: Code1 Registers Memory 1 Press Alt-F5 to arrange the windows on the screen.

- Set the options. Alt-O -> Preferences. Set the options as shown and
click ok.

- Again, Alt-O -> Source 1 window

- Alt-O - >Memory 1 window

The configuration is now complete.

Lets look at the program.

- step through the program and observe execution of each instruction.

o Press F10. o The debugger will show execution of the first line of the prolog. o Press F10 until instruction MOV AX,0 is highlighted. This is the first instruction of your program.

Observe the value in the register EAX. Register AX contains number 09DFH.

Now press F10. The debugger will execute the highlighted instruction. Note the change in the content of EAX and the fact that the register has been highlighted by the debugger, indicating the change.

The highlighting the code window moved to the next instruction.

Note that the line of the source code MOV AL, VAR1 became MOV AL, [000C] where 000CH is the actual offset of VAR1 in the data segment. You can check that this is true by checking the content of memory location DS:000CH in the data window. Now execute this instruction by pressing F10. Content of the register AL changed, taking the value from the VAR1.

The next instruction is MOV BX, OFFSET VAR2. VAR2 follows VAR1 in memory and has offset of 000DH. This is the value that will be placed into the BX upon execution of this instruction. Press F10 to execute.

The following instruction MOV [BX], AL will copy the content of AL into the memory location pointed by BX within the data segment. After the previous instruction BX contains the offset of the first byte of VAR2 or 000DH. That is where the data from AL will appear. Press F10 to execute. Note the debugger also highlighted changes in the data window.

Instruction MOV [BX+1], AL will copy the content of the register AL into the memory location with offset equal whatever the number is in BX plus 1. In our case BX=000DH, then the offset is 000DH+0001H=000EH. That is the second byte of the VAR2. Press F10 to execute. Note the change in the memory content.

Instruction MOV EAX, 12345678H will place number 12345678H into the register EAX. Press F10 to execute.

The instruction MOV VAR3, EAX became MOV DWORD PTR [000F], EAX.

VAR3 has been replaced by the actual offset (000FH) of VAR3 in the data memory. This instruction will take the content of the EAX and place into the four consecutive bytes of memory (a 32-bit variable) starting with the offset 000FH. Press F10 to execute.

That was the last instruction of the user program. The remaining instructions are generated by the .EXIT directive and serve to terminate the program. Press F10 until the process terminates.

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