What is the difference between signed and unsigned char?

[tinesan]

Hello fellow C programmers,

I'm just learning to program with C, and I'm wondering what the
difference between signed and unsigned char is. To me there seems to
be no difference, and the standard doesn't even care what a normal char
is (because signed and unsigned have equal behavior).

For example if someone does this:

unsigned char a = -2; /* or = 254 */
signed char b = -2; /* or = 254 */

putchar(a);
putchar(b); /* both print the same character (ex ascii 254)*/

-------------
It seems to me that it doesn't matter whether char is signed or
unsigned, because the output functions just look at the bit pattern and
deal with it as a positive number.
Also, I assigned a negative number to unsigned char, it wraps around
and creates the same bit pattern as assigning the same negative number
to signed char.

So my question is, what really is the difference between unsigned and
signed char?

Also, for other integral types, are the normal types always equal to
the signed types (int = signed int, long = signed long,etc. etc.)... or
is that implementation defined just like for chars?
Any help will be appreciated.

 

[Kobu]

Hello fellow C programmers,

I'm just learning to program with C, and I'm wondering what the
difference between signed and unsigned char is. To me there seems to
be no difference, and the standard doesn't even care what a normal char
is (because signed and unsigned have equal behavior).

For example if someone does this:

unsigned char a = -2; /* or = 254 */
signed char b = -2; /* or = 254 */

I don't think you can assign a negative initializer to a signed
integer.
Am I right people?

 

[Gregory Dean]

The other way around...
unsigned char does not have a sign extension.

 

[Alex Fraser]

I'm just learning to program with C, and I'm wondering what the
difference between signed and unsigned char is. To me there seems to
be no difference, and the standard doesn't even care what a normal char
is (because signed and unsigned have equal behavior).

For example if someone does this:

unsigned char a = -2; /* or = 254 */

In this, the value -2, of type int, is converted to unsigned char. This
conversion is specified as being equivalent to repeatedly adding or
subtracting UCHAR_MAX + 1 (where UCHAR_MAX is the maximum value an unsigned
char can have; apparently 255 for your compiler) until the result is between
0 and UCHAR_MAX inclusive. So a is assigned the value 254.

signed char b = -2; /* or = 254 */

Here, -2, again of type int, is converted to signed char. Note however that
the effect of assigning 254 to b (which can hold values between SCHAR_MIN
and SCHAR_MAX inclusive, probably -128 and 127 respectively in your case) is
undefined by the standards.

putchar(a);
putchar(b); /* both print the same character (ex ascii 254)*/

The putchar function takes an int, so for both these calls, the argument is
converted to type int; the calls are equivalent to putchar(254) and
putchar(-2) respectively. The putchar function is specified as converting
its parameter to unsigned char, which uses the rule above. Therefore, with
UCHAR_MAX being 255, the second call is equivalent to the first in terms of
the result.

It seems to me that it doesn't matter whether char is signed or
unsigned, because the output functions just look at the bit pattern and
deal with it as a positive number.

See above.

Also, I assigned a negative number to unsigned char, it wraps around
and creates the same bit pattern as assigning the same negative number
to signed char.

See the rule above. The wrapping around is what the standards specify. The
fact it is the same bit pattern is common, because two's complement
representation for signed numbers is common, but two's complement is not
required by the standards.

So my question is, what really is the difference between unsigned and
signed char?

One can represent unsigned values, and the other signed values (obviously).
As indicated above, conversion of out-of-range values to signed types (such
as signed char) is undefined by the standard. Similarly, the result of
arithmetic that produces out-of-range values for the type is undefined. But
the behaviour in both these cases for unsigned types (such as unsigned char)
*is* defined.

Also, for other integral types, are the normal types always equal to
the signed types (int = signed int, long = signed long,etc. etc.)... or
is that implementation defined just like for chars?

char can represent the same range of values as either signed char or
unsigned char, but all three are distinct types. (Similarly, int and long
may be able to represent the same range of values on a given implementation,
but they too are distinct types.)

int, short, long (and long long in C99) are always capable of representing
negative numbers. I think that int and signed int are the same type, and
similarly for short, long and long long. Hopefully someone else can clarify
this point.

HTH,
Alex

 

[Peter Nilsson]

For example if someone does this:

unsigned char a = -2; ...
signed char b = -2; ...

putchar(a);
putchar(b); /* both print the same character (ex ascii 254)*/

Abstractly, putchar() is a wrapper for fputc(), and...

"The fputc function writes the character specified by c
(converted to an unsigned char)..."

So, the calls will both send the same byte value to stdout.

No, the conversion is _NOT_ specified in terms of bit pattern.

On a signed magnitude machine, the 8-bit representation of -2 is...

10000010

Irrespective of the signed char representation, the conversion
will always yield UCHAR_MAX + 1 - 2.

Also, I assigned a negative number to unsigned char, it wraps
around and creates the same bit pattern as assigning the same
negative number to signed char.

It will do so on two's complement machines. However, the standard
doesn't _require_ two's complement integer representation for
negative signed integers.

So my question is, what really is the difference between unsigned
and signed char?

Also, for other integral types, are the normal types always equal
to the signed types (int = signed int, long = signed long,etc.
etc.)...
Yes.

or is that implementation defined just like for chars?

No.

 

[Peter Nilsson]

The other way around...
unsigned char does not have a sign extension.

Please don't top post in clc.

Apart from the obvious!?

You haven't read the standard, have you?

The value of an unsigned integer cannot be negative.

/* or UCHAR_MAX + 1 - 2 */

Since -2 is in the range -127..127, the minimum range for
signed char, b will always be assigned the value -2 here
on any conforming implementation.

No.

 

[Keith Thompson]

I don't think you can assign a negative initializer to a signed
integer.
Am I right people?

No. Both declarations above are legal.

 

[Chris Croughton]

Hello fellow C programmers,

I'm just learning to program with C, and I'm wondering what the
difference between signed and unsigned char is. To me there seems to
be no difference, and the standard doesn't even care what a normal char
is (because signed and unsigned have equal behavior).

For example if someone does this:

unsigned char a = -2; /* or = 254 */
signed char b = -2; /* or = 254 */

putchar(a);
putchar(b); /* both print the same character (ex ascii 254)*/

On your machine. They won't on a 1-s complement machine. Incidentally,
value 254 (0xFE) is not an ASCII character, ASCII only defines 7 bit
characters in the range 0x00 to 0x7FF (0 to 127).

Usually, yes. Not guaranteed, though (some output libraries will fail
if the parameter is negative).

Also, I assigned a negative number to unsigned char, it wraps around
and creates the same bit pattern as assigning the same negative number
to signed char.

Again, on your machine.

So my question is, what really is the difference between unsigned and
signed char?

int main(void)
{
unsigned char a = -2;
signed char b = -2;

if (a == b)
printf("equal\n");

if (a > 0)
printf("a > 0\n");

if (b > 0)
printf("b > 0\n");
return 0;
}

What, if anything, will be output?

Also, for other integral types, are the normal types always equal to
the signed types (int = signed int, long = signed long,etc. etc.)... or
is that implementation defined just like for chars?

They are defined to take the same amount of storage. It is also defined
that a signed value which is positive can be converted to an unsigned
value of the same type and back again with no change in value. However,
an unsigned value which is greater than the maximum positive value which
a signed version can hold cannot be reliably converted into a signed
value, nor is it defined what value an unsigned version of a negative
value has.

Chris C

 

[CBFalconer]

.... snip on signed/unsigned char ...

They are defined to take the same amount of storage. It is also defined
that a signed value which is positive can be converted to an unsigned
value of the same type and back again with no change in value. However,
an unsigned value which is greater than the maximum positive value which
a signed version can hold cannot be reliably converted into a signed
value, nor is it defined what value an unsigned version of a negative
value has.

That last provision is wrong. There is a specific process for
converting any out-of-range value to an unsigned value. It just
isn't necessarily reversible.

 

[Gregory Dean]

lly pre
Please don't top post in clc.


Apart from the obvious!?


You haven't read the standard, have you?

The value of an unsigned integer cannot be negative.


/* or UCHAR_MAX + 1 - 2 */


Since -2 is in the range -127..127, the minimum range for
signed char, b will always be assigned the value -2 here
on any conforming implementation.


No.

Silly preferences setting in Entourage 2004. Sorry.
-Greg

 

[pete]

On your machine. They won't on a 1-s complement machine.

They won't do what exactly, on a 1-s complement machine?

((unsigned char)-2) may or may not be 254,
but putchar(-2) does the same thing
as putchar((unsigned char)-2), always.

See Alex Fraser's post in this thread for a good explanation.