CODEWITHSUNDEEP
x86x86-16
samina
samina

Reputation: 11

LEA operation with 8086 assembly

can anyone help me understanding the following instructions-

LES SI,DATA1
MOV DI,OFFSET DATA2
MOV BX,[SI]
MOV CX,[DI]
MOV[SI],CX

Upvotes: 1

Views: 1612

Answers (3)

lijie
lijie

Reputation: 4871

LES is not LEA. LES x, y interprets y as a far pointer and loads its data into ES (a segment register) and x.

The instruction sequence as given is strange though, because ES is not actually used.

Anyway, the instruction sequence (if the [SI] are changed to ES:[SI]) is: given a far pointer (DATA1) and a variable (DATA2), move the contents pointed to by DATA1 into BX and replace them by what is stored currently at DATA2 (which will also be in CX).

Upvotes: 3

Loduwijk
Loduwijk

Reputation: 1977

You can just look up an assembly instruction page, and it would explain what each of those instructions do in a simple to understand way; in many ways assembly in its basic, low-level simplicity is easier to understand a single line than in higher level languages.

I can never remember which x86 assembly syntax orders things which way if I haven't used it in a while (the two major x86 assembly syntaxes order the operands in the opposite order), so I won't say the exact result here.

For the first one, I assume that's a typo and you meant LEA instead of LES? LEA stands for "load effective address" if I recall correctly. Its main purpose it to calculate the memory address of something when you want to actually know the address instead of just use the address.

(edit) I had not used LES before, and Google wanted to redirect me to LEA, hence my above statement. I will leave the above though so you can benefit from that too. (/edit)

MOV moves data from one place to the other. The operands in the MOV instructions that are surrounded by [] square brackets mean you want that memory address instead, so MOV CX,[DI] would be "move the contents of the CX register into the memory location at the address held in the DI register" (or the other way around, [DI] into CX, see above statement about operand order).

I'm not sure on the "OFFSET DATA2" as I don't recall an offset keyword.

Upvotes: 1

BatchyX
BatchyX

Reputation: 5114

corresponds to the following C code :

// DATA1 and DATA2 are 'far' pointers to word of 16 bits, let's say they are 'short'
short bx = *DATA1;
*DATA1 = *DATA2;
// the old value of *DATA1 is still available in the BX register.
// and the value of *DATA2 is still available in the CX register.
// also, ES is set to the segment where DATA1 and DATA2 resides.

Upvotes: 0

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