I have to do a simple calculator in assembly using EMU8086, but every time I try to launch it EMU8086 gives this error:

I checked the other stuff, but there were no mistakes:

I cut most of the code segment because it wasn't important here.

After experimenting with the code I found that the problem was related to the lengths of the messages in the data segment. menu1 & menu2 were too long and any message after them can't be printed (msg1 & msg2 are printed, but nothing after them). I checked if I should merge menu1 & menu2, but it didn't help out. Please help me find out what is wrong with it.

The error message means you use int 21h / AH=09h on a string that didn't end with a $ (ASCII 24h). The system-call handler checked 2000 bytes without finding one.

Often, that means your code or data is buggy, e.g. in a fixed string you forgot a $ at the end, or if copying bytes into a buffer then you maybe overwrote or never stored a '$' in the first place.

But in this case, it appears that EMU8086 has a bug assembling push offset msg4. (In a way that truncates the 00B5h 16-bit address to 8-bit, and sign-extends back to 16, creating a wrong pointer that points past where any $ characters are in your data.)

Based on the error message below I know you are using EMU8086 as your development environment.

INT 21h, AH=09h - address: 170B5 byte 24h not found after 2000 bytes. ; correct example of INT 21h/9h: mov dx, offset msg mov ah, 9 int 21h ret msg db "Hello$"

I'm no expert on EMU8086 by any stretch of the imagination. I do know why your offsets don't work. I can't tell you if there is a proper way to resolve this, or if it's an EMU8086 bug. Someone with a better background on this emulator would know.

You have created a data segment with some variables. It seems okay to me (but I may be missing something). I decided to load up EMU8086 to actually try this code. It assembled without error. Using the debugger I single stepped to the push offset msg1 line near the beginning of the program. I knew right away from the instruction encoding what was going on. This is the decoded instruction I saw:

It shows the instruction was encoded as push 0b5h where 0b5h is the offset. The trouble is that it is encoded as a push imm8 . The two highlighted bytes on the left hand pane show it was encoded with these bytes:

If you review an instruction set reference you'll find the encodings for PUSH instruction encoded with 6A is listed as:

You may say that B5 fits within a byte (imm8) so what is the problem? The smallest value that can be pushed onto the stack with push in 16-bit mode is a 16-bit word. Since a byte is smaller than a word, the processor takes the byte and sign extends it to make a 16-bit value. The instruction set reference actually says this:

If the source operand is an immediate of size less than the operand size, a sign-extended value is pushed on the stack

B5 is binary 10110101 . The sign bit is the left most bit. Since it is 1 the upper 8 bits placed onto the stack will be 11111111b (FF). If the sign bit is 0 then then 00000000b is placed in the upper 8 bits. The emulator didn't place 00B5 onto the stack, it placed FFB5. That is incorrect! This can be confirmed if I step through the push 0b5h instruction and review the stack. This is what I saw:

Observe that the value placed on the stack is FFB5. I could not find an appropriate syntax (even using the word modifier) to force EMU8086 to encode this as push imm16. A push imm16 would be able to encode the entire word as push 00b5 which would work.

Two things you can do. You can place 256 bytes of dummy data in your data segment like this:

Why does this work? Every variable defined after the dummy data will be an offset that can't be represented in a single byte. Because of this EMU8086 is forced to encode push offset msg1 as a word push.

The cleaner solution is to use the LEA instruction. This is the load effective address instruction. It takes a memory operand and computes the address (in this case the offset relative to the data segment). You can replace all your code that uses offset with something like:

AX can be any of the general purpose 16-bit registers. Once in a register, push the 16-bit register onto the stack.

Someone may have a better solution for this, or know a way to resolve this. If so please feel free to comment.

Given the information above, you may ask why did it seem to work when you moved the data around? The reason is that the way you reorganized all the strings (placing the long one last) caused all the variables to start with offsets that were less than < 128. Because of this the PUSH of an 8-bit immediate offset sign extended a 0 in the top bits when placed on the stack. The offsets would be correct. Once the offsets are >= 128 (and < 256) the sign bit is 1 and the value placed on the stack sign will have an upper 8 bits of 1 rather than 0.

There are other bugs in your program, I'm concentrating on the issue directly related to the error you are receiving.

I reviewed your code and concentrated on the following sequence of instructions:

So your sequence

essentially means, that you are setting BX to the address of 'choice', then setting 'choice'([BX]) to AL and copying it back to AL.

This is redundant.

After that, you compare that char to '+' and...

No error here.

So your problem with menu1 and menu2 may stem from some escape characters included in your strings like %,/,\. For example, % is a MACRO character in some assemblers which may create problems.

simple solution is that your strings should always end in '$' change DUP(?) to DUP('$') and all other strings end with ,'$'