How To Change A Value In A Register To A Negative Number

I'thousand on 24-hour interval 3 of Larn ASM in 28 Days and I'g confused when it says this:

Storing negative numbers, however, is legal, but the number volition get "wrapped" to fit. For example, if you assign -1 to A, information technology will actually hold 255. If you assign -2330 to BC, it will really concur 63206. Calculation one plus the maximum value the register will hold gives you the value that will be stored.

I empathise the wrap part, but the last sentence doesn't make any sense. If A tin hold 255, then i more is 256. But if you have -3 for instance, information technology stores 253, which is not 256. Why does it say that?

If you lot were making a CPU, how would you shop negative numbers? The way preferred on the Z80 (and most other CPUs you'll discover today) is called two'southward complement.

The mode the wrapping works, 0 = 256.
So -i = 255, -2 = 254, -iii = 253.

This is obvious if you write them out in binary. 256 = 100000000b, which has the 9 bits, however the register simply has 8 bits, then information technology is truncated to 00000000b.

for 256 wouldn't you need two registers though?
And, unless I'm very confused, I don't think you see what i was inquiring about. Information technology says it wraps but so it contradicts itself and says information technology's always the value a register tin store + 1.

MrIdrik wrote:

for 256 wouldn't y'all demand 2 registers though?

Yes, that's why I said "truncated to 00000000b".

Quote:

And, unless I'yard very confused, I don't retrieve you run across what i was inquiring about. It says it wraps simply then it contradicts itself and says it's always the value a register can store + 1.

The formula for 2s complement negative numbers is that to negate a number, northward, you lot store it is (~north)+1, where ~ is the bitflipping operation. (i.due east. ~1001 = 0110). Everything is modulo 256 when you have an 8 bit register, which is "the largest number information technology tin can store + 1", but modulo that means you lot all numbers must be less than information technology.

It means add together ane plus the max value [to the value yous're storing]. -3+i+255=253.

If it helps, negative numbers in z80 ASM are entirely a function of how you interpret them. $FF can be -i or 255, depending on what you desire information technology to exist.

Yeah, it's confusingly written. It'south saying that if y'all write "LD A,-iii", the value stored in A is -3 plus (255 + 1), which is 253. But this add-on is washed by your assembler, non in the Z80 CPU itself - as far as the Z80 is concerned, -three and 253 are the same number.

Fixing the way that ASM in 28 Days discusses negative numbers is one of the things at the top of my list for if (when?) I ever become around to making a 3.0 of the guide, after fixing the interrupt chapter, adding TI-84+ and TI-84+CSE information, updating the assembler/editor instructions, and tossing in a lot more sample code and programs.

Thanks to everyone for all the replies. It'southward more articulate to me at present. So basically, I don't even have to convert negative numbers to positive myself?

And one more than affair, could someone explain this to me:

It turns out that there are many signed numbering schemes, merely the just i we're interested in is called the 2'due south complement. When we have a signed value in two's complement, the almost significant bit of the number is termed the sign bit and its state determines the sign of the number. The existence of the sign scrap naturally imposes a restriction on the number of $.25 a number may exist composed of. With this, the amount of bits at our disposal to represent the number is reduced by i; for a string of eight $.25, we tin can have a numeric range of -128 to +127. For a string of xvi, it's -32, 768 to 32, 767, etc.

So the almost meaning scrap is the leftmost ane, just I don't come across how the sign bit restricts the states. Does it have the identify of the about significant bit? What fifty-fifty IS a sign fleck (like, what does it expect like). Someone higher up put a ~ to show it, and then LD A,-3 would be 253, which is 11111100, and to show it is signed it would be ~11111100b? And if the assembler does it, when would I ever need to apply information technology anyway?

~ for bitwise logic means invert. When you practice ~1 in 8 bit, you lot get 254 (0b11111110), which is the i'due south complement. Two'southward complement is: If the number is negative, capsize and then add together 1.

-three is 253, but it is not 0b11111100 in 2'due south complement, that is i's complement.

For your confusion on the sign bit, look at a 2's complement viii-bit number:
S|NNNNNNN

You have 7 bits worth to make numbers, essentially. For positive, you lot have 0 (0000000) to 127(1111111). This is when the sign bit is 0.

When the sign bit is 1, and then you accept negative numbers (essentially flips and shifts a number line). -1 is (1111111) (2's complement of -i goes ~0000001 -> 1111110+1 -> 1111111) and -128 (1111111).

<edit> Made post about numbers and not very nice bits Razz

So the sign bit is just a number?
If a 1 is the almost significant bit, it's negative, and when 0 is the most significant scrap, information technology'south positive?

-3 is 253, simply information technology is non 0b11111100 in 2's compliment, that is ane's compliment.

Why is there is a 0 in front, is that the sign bit? Wouldn't the b go at the end? And how is information technology not Ii's Complement? It says this on the site which is what I did:

Calculate null minus the number (like negative numbers in the Existent World). If you're confused how to do this, you tin can consider 0 and 256 (or 65536 if advisable) to be the same number. Therefore, -half dozen would be 256 - half dozen or 250: %11111010.

0b is just notation proverb that you lot take a binary number, just as 0x is for hex and 0 is for octal.

Virtually the 253, the binary number is wrong, 253 is 11111101

Oh, right, and then %11111101 is indeed -3?
And so I accidentally wrote it wrong and it happened to be one's complement?

I'grand all the same confused nearly what a sign bit actually is:

Quote:

So the sign bit is but a number?
If a 1 is the most significant flake, it's negative, and when 0 is the almost significant flake, information technology's positive?

More asm learners! Very Happy

1'due south complement is essentially an "xor $FF", meaning as in the byte is inverted. The i's complement of 1 would be:
%0000 0001
%1111 1111 (XOR)
----------------
%1111 1110

This is equivalent to -ii. Two's complement adds 1 to this value: %1111 1111 (or -1).

Retrieve about it this fashion, an 8-bit number can simply hold values from 0-255. If you're at 0 and decrease one, it will wrap to 255. And so 0-1 = -1=255=%1111 1111.

The sign bit is the leftmost bit. Every negative number (represented in binary) volition have a ane in the very leftmost bit, so negative numbers get from %1111 1111 (-ane aka 255) to %1000 0000 (-128 aka 128). Positive numbers go from %0000 0000 to %0111 1111.

I think I've got it now, thanks everyone.

Take your own thoughts to add to this or whatsoever other topic? Want to inquire a question, offer a suggestion, share your own programs and projects, upload a file to the file archives, get aid with calculator and computer programming, or simply conversation with agreeing coders and tech and figurer enthusiasts via the site-wide AJAX SAX widget? Registration for a free Cemetech business relationship just takes a minute.

» Go to Registration page

You cannot postal service new topics in this forum
You cannot respond to topics in this forum
Yous cannot edit your posts in this forum
Yous cannot delete your posts in this forum
You cannot vote in polls in this forum