Thread: Some basic C++ concepts

My professor told me that the concepts is extremely important than programming.So today he shows us:

char c=0;
c=~c;(which is the bitwise)

so what is c=?

Since C is 00000000,and use not(~) so it is 11111111 now,but he said the answer is -1 instead of 255.Because he says it use one's complement or something~~~so,why it is -1?
(I know it can't be 255,since char is 1 byte with sign,which means it is -128~127,so 255 is out of range,so why it is -1?)

Ok,guys I know I am so stupid since for storing int ,we use two's complement notation,so 1111 1111 means -1.
but I want to know We also ues Excess notation for int,so why this case we use two's complement notation instead of Excess notation?which situations we use Excess notation? thx!

and also for this :

we know 0111 1111 is 127,how about 0111 1111+1 if the variable is char(which is -128~127)?
so because it is 1000 0000 after +1,so 1 is use for sign(-),why the answer is -128?


also for this one,we use two's complement so 1000 0000 means the least num!

So I know I am so stupid,since When we study this in CS 2,our teacher just go through it so quickly and we didn't focus on it since he thinks programming is more important.I know it is an excuse and I should be good at it.

> but he said the answer is -1 instead of 255
It's the same bit pattern, do you regard 'char' as being signed or unsigned ?

Oh dear. Your teacher makes a LOT of unwarranted assumptions here. Namely:
1) "C is 00000000"
No. There's no guarantee that char has 8 bits, even though that's the most common number.
2) "the answer is -1 instead of 255"
No. Whether char is signed or unsigned is implementation-defined. That's why char, signed char and unsigned char are distinct types. (Unlike the other integral types, where signed xxx and xxx are the same.) Thus, it is implementation-defined whether the result is 255 (unsigned) or ... something else (signed).
3) "the answer is -1"
No. It is implementation-defined (or even unspecified) what negative representation is used, with the only restriction being that an all-zero bit pattern must represent 0. Thus, 2's complement, 1's complement and sign-value are all valid (and have been used at least once), whereas excess representation is invalid (0 is not all-zero).
4) "so it is 11111111 now"
No. That is, not if char is signed and another common condition holds: then it's undefined behaviour. The reason is that ~ does integral promotion, so the result of ~c is actually an int, and all ones at that. The other common condition mentioned above is that sizeof(int) > sizeof(char). If that's the case, then the assignment back to c constitutes a signed overflow, which is UB.

So ... to answer the original question, "what is c?"
If char is signed, the question cannot be answered, since c = ~c is undefined behaviour.
If char is unsigned, c is CHAR_MAX.

Originally Posted by CornedBee

The other common condition mentioned above is that sizeof(int) > sizeof(char).

I think the C standard mandates that his is true.

QuantumPete

Originally Posted by QuantumPete

I think the C standard mandates that his is true.

QuantumPete

It thought the mandated point was sizeof(int) >= sizeof(char).

As to the point of Cornedbee, the behaviour of c = ~c may well technically be undefined for a signed char, but I have yet to see a system where it's not resulting in a -1, and I have worked on several processor architectures: x86, 68K, 29K, ARM, 6502, Z80, 8031, VAX, PDP-11. All of which show this behaviour. So whilst it's technically correct that the result is undefined, it is also quit unusual that the result is something OTHER than -1.

--
Mats

Originally Posted by robatino

No, it doesn't - it's guaranteed that 1 == sizeof(char) <= sizeof(short) <= sizeof(int) <= sizeof(long),

my mistake

CS1 was a lot of hand holding in my experience, CS2 was when they started teaching real programming, and that is when most students find out they are not cut out for it, after that, it is the second year classes that separate the programmers and the people who are just kinda interested in computers.