I'm having a very hard time trying to figure out RCR reg,reg . Please help ?

What it does or how to use?

If you what to use a register as src it only can be CL.

I don't understand that picture lol...... I want to know what the opcode does. Please explain in words lol

Rotate Right through Carry.
For example, when the instruction RCR al,1 is executed the low order bit (bit0) of al is copied to the carry flag. The carry flag is copied to the high order bit (bit7) of al. Bit 7 copies to bit 6 etc. So all the 8 register bits plus the carry bit play musical chairs and shift one bit (or chair) in a circle.

If the instruction is RCR al,cl, for example, the number of times the shift takes place for the one instruction is the value that is in the cl register. For example, if you preceed the rcr al,cl instruction with mov cl,2, then two shifts will take place.

Chris

Do you mean the high order bit ( bit 7 ) by eax ?
eax can hold 8 bits .... in the way you explain it from 0-7

I still dont understand what happens here :

mov al, 36h
rcr al,2h

al is equal to "Dh" after the execution . How is the low order bit --- 6h in this case copied into the carry flag ? The carry flag can't become 6 ....

However, al can only hold a byte
So for example

mov al,36h

which al in bits is 110110

There are only 6 binary numbers there.... and the carry flag would count as the 7th. Why do you say 8 then ? Is there something I'm missing ?

> Is there something I'm missing ?

YES. After 2 years and 233 posts

> So for example
> mov al,36h
> which al in bits is 110110

00110110

> There are only 6 binary numbers there

NO. always 8

> the carry flag would count as the 7th.

NO.

> Why do you say 8 then ?

Always 8.



BTW, C must have been zero before your RCR

Lol I know I don't have all the basics down..

> Do you mean the high order bit ( bit 7 ) by eax ?
> eax can hold 8 bits .... in the way you explain it from 0-7

The example was using an 8 bit register. If using 16 bit register then high order bit will be 15, and 31 if using a 32 bit register.

> al is equal to "Dh" after the execution . How is the low order bit --- 6h in > this case copied into the carry flag ? The carry flag can't become 6 ....

The low order bit is 1h. You are confusing a hex nibble with a bit. Only one bit is copied to the carry flag, not the four bits of a hex symbol.


mov al,36h gives 00110110 in binary. You left off the top two bits. So there are 8 bits and the carry flag could count as the ninth bit during the rotate.

Well thank you everyone. I understand now