CODEWITHSUNDEEP
assemblysubtraction
mahmood
mahmood

Reputation: 24745

Sub instruction in assembly and the effect on CF and AF

It seems that CF (carry flag) and AF (auxiliary flag) are tricky when using sub command. For example (first case is correct and easy to understand):

 AL=4CH, BL=29H          (C>9 , 4>2)
 SUB AL,BL     ; AL=23H       CF=0   AF=0

 AL=44H, BL=29H          (4<9 , 4>2)
 SUB AL,BL     ; AL=1BH       CF=0   AF=1

 AL=1CH, BL=29H          (c>9 , 1<2)
 SUB AL,BL     ; AL=F3H       CF=1   AF=0   ALSO SF=1

 AL=13H, BL=29H          (3<9 , 1<2)
 SUB AL,BL     ; AL=F3H       CF=1   AF=1   ALSO SF=1

Now take the second case for more details

 0100 0100
 0010 1001 -
------------
 0001 1011

There is no borrowing from AH to AL. So why the second case results in AF=1?

Regarding the third case, we borrow from AH. So in my opinion, CF=0 and AF=1. But the result is different from what I understand. Why?

Upvotes: 0

Views: 1707

Answers (1)

owacoder
owacoder

Reputation: 4873

The Intel Developer's Manual gives some insight on the AF flag. It is the "Auxiliary Carry," used for BCD arithmetic. The manual says about it:

AF — Auxiliary Carry flag — Set if an arithmetic operation generates a carry or a borrow out of bit 3 of the result; cleared otherwise. This flag is used in binary-coded decimal (BCD) arithmetic.

So in your example, there is no borrowing from AH to AL, but there is borrowing from the low 4-bit nibble in AL to the high 4-bit nibble in AL:

0100 0100
0010 1001 -
------------
0001 1011
     ^
     Borrow needed in low nibble because '1001' (9) is greater than '0100' (4)

Unless you are using BCD arithmetic, you shouldn't have to use the AF flag.

CF, on the other hand, will be set if there is a carry or borrow from the most significant bit of the result, whether the instruction uses 8, 16, or 32 bit operands.

The Intel Developer's Manual says about CF:

CF — Carry flag — Set if an arithmetic operation generates a carry or a borrow out of the most significant bit of the result; cleared otherwise. This flag indicates an overflow condition for unsigned-integer arithmetic. It is also used in multiple-precision arithmetic.

Upvotes: 1

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