CODEWITHSUNDEEP
assemblyx86dos
Austin Davis
Austin Davis

Reputation: 3756

What happens in the following assembly commands?

Just a precaution. I'm a complete noob on assembly. My only question is that only the following commands in DOS:

MOV  AH, 09
INT  21

MOV  AX, 4C01
INT  21

Does the second interrupt still have the parameter 09 meaning display screen or does the AX change that in some way?

Upvotes: 4

Views: 604

Answers (2)

paxdiablo
paxdiablo

Reputation: 882336

You need to search for Ralf Brown's Interrupt List, it's the definitive guide to these old DOS/Windows interrupts.

Calling INT 21h with AH = 09H outputs a $-terminated string located at DS:DX to the console.

Calling INT 21H with AH = 4cH exits your program with the return code in AL (01H in this case).

The second call will not have AH set to 09H simply because AH is simply the upper 8 bits of AX. The second MOV sets AH to 4cH as part of setting AX:

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In more detail (from Wikipedia):

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This register-made-from-other-registers has a long history at Intel with:

  • ax being made up of ah and al;
  • eax being made up from 16 bits plus the other 16 from ax;
  • rax being made up of 32 bits plus the other 32 from eax.

along with similar approaches for the other general purpose registers and special registers like index or base registers, and the stack pointer.

Interestingly enough, they didn't duplicate the ability to access the high halves of the expanded registers, such as with eax being made up of eah and eal (both mythical, but the latter being an alias for ax).

That would have given us a large cache of extra small registers in higher modes, though I'm not sure what the cost would have been (no doubt Intel/AMD do know the cost and it was probably deemed too expensive).

Upvotes: 3

Carl Norum
Carl Norum

Reputation: 225142

AH and AL are the high and low halves of AX, respectively. So yes, setting AX changes AH.

4C is "exit program".

Upvotes: 3

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