C# int query

[pigmonkeys21]

Hi guys, I'm going through Microsoft's Developer Step By Step Visual C# 2013 and in chapter 16 I have come across a bit of explanation that i'm not quite understanding, I've taken a photo:

https://dl.dropboxusercontent.com/u/...307_150640.jpg

I was fine up til the last sentence in the Note box "which is actually the int representation of -205 in C#"

I understand that in binary, 204 is:

00000000000000000000000011001100

I understand that ~204 (NOT 204) must be:

11111111111111111111111100110011 - just flip the 1s and 0s

However, I cannot understand how:

11111111111111111111111100110011 = the int representation of -205 in C#.

I would have said that 11111111111111111111111100110011 (using a decimal calculator) was 4,294,967,091. If it actually is -205, then what is the binary for 4,294,967,091 in C#?

I must be missing something fundamental here, can someone explain?

thanks!

[willmotil]

remember that a int and a uint (unsigned) are not the same

the maximum positive value in binary for a int is 0111-1111-1111-1111-1111-1111-1111-1111 = 2147483647
but that's Not the maximum value for a unsigned integer

so ask what is that final bit for in a int ?
yep that's the sign flag it counts as -1

hence -1
1111-1111-1111-1111-1111-1111-1111-1111
Note there is no 0 or significant inverse digit however the sign bit here denotes it is -1 in decimal
i.e. for the decimal value of zero we can say it is considered positive
essentially the sign bit is not on for a decimal value of zero
however as a unsigned integer the above binary is the maximum value of a uint equal to 4294967295

to represent -2 we flip on the sign bit and mark the 1's place inversely significant
1111-1111-1111-1111-1111-1111-1111-1110

for -3 we have with the sign bit
1111-1111-1111-1111-1111-1111-1111-1101
the inverse significant digit is in the twos place
with the additional -1 sign bit implicitly added

and for -4 you can guess
1111-1111-1111-1111-1111-1111-1111-1100
e.g.
if this were positive it would be +1 +2 = 3
but since its in the inverse you count like so
you go (-1)+-1+-2 = -4


I don't normally think of negatives as inverse
i think about if the sign bit is on or not
In fact i had forgotten how it worked i had to think about it.

you might experiment by setting individual bit positions on or off

        public static int SetIntsBit(int theinteger, int bitposition, bool true_or_false)
        {
            int bitvalue = 1 << (bitposition);
            if ((theinteger & bitvalue) > 0)
            {
                if (true_or_false == false)
                {
                    theinteger = theinteger ^ bitvalue;
                }
            }
            else
            {
                if (true_or_false == true)
                {
                    theinteger = theinteger | bitvalue;
                }
            }
            return theinteger;
        }

[pigmonkeys21]

Ah thanks for the explanation, i'd covered int types but nothing that went into the binary representations of the different types etc.

So I guess an unsigned int gets that extra range because it stops using the leading bit as a potential indicator of a negative number, so it gets to actually use it. But with a signed int the leading bit is given up for this use, and only 31 bits can actually be used for the number.

It does seem to get a bit unintuitive though looking at the binary versions of negative numbers, I can see why you had to think about it. I hope I wont have much dealings with them in the future!

[TheCPUWizard]

Actually there is no "sign bit" with 2s complement (which is that is almost universally used). With a sign bit notation (still used in a few places) the pattern would be 10000....0001 (first bit is the sign, then the remaining bits are the absolute value. To understand 2s complement, think about this (I am going to use a much shorted word length)

If you have 11111 and Add 0001, but are limited to 5 bits, what do you get? 00000 !!... ZERO... But what number can you add 1 to and end up with zero? -1..
If you have 11110 and Add 0010, but are limited to 5 bits, what do you get? 00000 !!... ZERO... But what number can you add 1 to and end up with zero? -2..

And so on...

Make sense??

[pigmonkeys21]

i'm struggling, but i'm watching some youtube videos that seem to explain it well. hopefully it will sink in !

From what I understand, aren't those examples overflow situations which simply indicates that the calculation is unsupported ?

What I also wonder is, given no other information than a binary number like 1101, exactly where is the indicator of whether the number is signed or unsigned ? If there never is any such indicator, isn't that somewhat problematic ?

[willmotil]

sorry he's right, i probably said that poorly, though it still denotes the sign but it is a value as well
to say no (positive signed integer) number in either in will have that first bit set true

see the chart on the page in the link below, it will probably make what i said above more clear

this link has two charts that shows all the values of a signed nibble for both types
0 to F hex, or -8 to 7 decimal, or 0000 to 1111 binary
http://www.exploringbinary.com/twos-...ent-converter/

now as far as overflow underflow im not sure you would have to look that up, i would guess
that probably is handled in machine code at the register level on the cpu with specific rules
such things only may occur when a operation is performed such as addition subtraction ect...

there are quite a few sites you can simply google as well that may click a bit better
http://academic.evergreen.edu/projec...am/2s_comp.htm