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pete <> wrote in message news:<>...
> Alan Gillespie wrote:
> >
> > > > > unsigned char *src, *end;
> > > > > size_t bufsiz;
> > > > > ...
> > :
> > :
> > :
> > >
> > > If bufsiz is of type ptrdiff_t, then the expression
> > > ((end - src) > bufsiz)
> > > is valid as long as end and src point to the same object,
> > > or one byte past.
> > >
> >
> > A pedant would say you mean one "char" past.
> > The standard leaves the precise
> > size of a char to the implementation.
>
> Pedant that I am,
> I say that you don't know what you're talking about.
>
> N869
> 6.5.3.4 The sizeof operator
> Semantics
> [#3]
> When applied to an operand that has type char, unsigned char,
> or signed char, (or a qualified version thereof) the result is 1.

That is defining the sizeof operator, not the size of a char.

The sizeof operator gives the size of its operand in chars, not bytes.
Standard C has no notion of a byte.

Now who doesn't know what he's talking about?

Micah Cowan <> wrote in message news:<>...
> "Alan Gillespie" <> writes:
>
> > > > > unsigned char *src, *end;
> > > > > size_t bufsiz;
> > > > > ...
> > :
> > :
> > :
> > >
> > > If bufsiz is of type ptrdiff_t, then the expression
> > > ((end - src) > bufsiz)
> > > is valid as long as end and src point to the same object,
> > > or one byte past.
> > >
> >
> > A pedant would say you mean one "char" past. The standard leaves the precise
> > size of a char to the implementation.
>
> Yes, the precise size of a char... in *bits*. A char (or unsigned
> char) is *always* one byte in size, regardless of how many bits may be
> in a byte.
>
> -Micah

Not quite. sizeof(char) always gives 1 because the sizeof operator is
defined that way - it gives the size of its operand in chars.

A byte is /always/ 8 bits, but that has nothing to do with C. C has no
notion of a byte.

> My opinion, is that a warning about signed/unsigned comparison
> is a good warning, and I wouldn't lower my compiler's warning
> level to avoid it.

You're right not to lower the warning threshold. I wish more people
thought like that.

Alan Gillespie wrote:
>
> pete <> wrote in message news:<>...
> > Alan Gillespie wrote:
> > >
> > > > > > unsigned char *src, *end;
> > > > > > size_t bufsiz;
> > > > > > ...
> > > :
> > > :
> > > :
> > > >
> > > > If bufsiz is of type ptrdiff_t, then the expression
> > > > ((end - src) > bufsiz)
> > > > is valid as long as end and src point to the same object,
> > > > or one byte past.
> > > >
> > >
> > > A pedant would say you mean one "char" past.
> > > The standard leaves the precise
> > > size of a char to the implementation.
> >
> > Pedant that I am,
> > I say that you don't know what you're talking about.
> >
> > N869
> > 6.5.3.4 The sizeof operator
> > Semantics
> > [#3]
> > When applied to an operand that has type char, unsigned char,
> > or signed char, (or a qualified version thereof) the result is 1.
>
> That is defining the sizeof operator, not the size of a char.
>
> The sizeof operator gives the size of its operand in chars, not bytes.
> Standard C has no notion of a byte.
>
> Now who doesn't know what he's talking about?

You, and worse today than yesterday.

Richard Bos made a post 5 minutes ago,
in response to your post in the
Re:size of array??
thread, quoting N869 and explaining what Micah Cowan has already
explained to you in this thread.

I always avoid making posts which make it appear as though
it is substantially easier for me to post than to:
1 access a copy of the standard
2 acess a compiler

http://anubis.dkuug.dk/jtc1/sc22/wg14/www/docs/n869/

Alan Gillespie <> wrote:
> Micah Cowan <> wrote in message news:<>...
>> "Alan Gillespie" <> writes:
>>
>> > > > > unsigned char *src, *end;
>> > > > > size_t bufsiz;
>> > > > > ...
>> > :
>> > :
>> > :
>> > >
>> > > If bufsiz is of type ptrdiff_t, then the expression
>> > > ((end - src) > bufsiz)
>> > > is valid as long as end and src point to the same object,
>> > > or one byte past.
>> > >
>> >
>> > A pedant would say you mean one "char" past. The standard leaves the precise
>> > size of a char to the implementation.
>>
>> Yes, the precise size of a char... in *bits*. A char (or unsigned
>> char) is *always* one byte in size, regardless of how many bits may be
>> in a byte.
>>
>> -Micah
>
> Not quite. sizeof(char) always gives 1 because the sizeof operator is
> defined that way - it gives the size of its operand in chars.
>
> A byte is /always/ 8 bits, but that has nothing to do with C. C has no
> notion of a byte.

Sorry Alan,

but you're wrong. Look what ISO/IEC 9899:1999 has to say about this:

3.6
1 *byte* addressable unit of data storage large enough to hold any
member of the basic character set of the execution environment
2 NOTE 1 It is possible to express the address of each individual byte
of an object uniquely.
3 NOTE 2 A byte is composed of a contiguous sequence of bits, the
number of which is implementationdefined. The least significant bit
is called the low-order bit; the most significant bit is called the
high-order bit.

6.5.3.4 The sizeof operator

2 The sizeof operator yields the size (in bytes) of its operand, ...

3 When applied to an operand that has type char, unsigned char, or
signed char, (or a qualified version thereof) the result is 1.

which implies that "a char is a byte is a char"!


--
Z ()
"LISP is worth learning for the profound enlightenment experience
you will have when you finally get it; that experience will make you
a better programmer for the rest of your days." -- Eric S. Raymond

In article < >,
(Alan Gillespie) wrote:

> pete <> wrote in message
> news:<>...
> > Alan Gillespie wrote:
> > >
> > > > > > unsigned char *src, *end;
> > > > > > size_t bufsiz;
> > > > > > ...
> > > :
> > > :
> > > :
> > > >
> > > > If bufsiz is of type ptrdiff_t, then the expression
> > > > ((end - src) > bufsiz)
> > > > is valid as long as end and src point to the same object,
> > > > or one byte past.
> > > >
> > >
> > > A pedant would say you mean one "char" past.
> > > The standard leaves the precise
> > > size of a char to the implementation.
> >
> > Pedant that I am,
> > I say that you don't know what you're talking about.
> >
> > N869
> > 6.5.3.4 The sizeof operator
> > Semantics
> > [#3]
> > When applied to an operand that has type char, unsigned char,
> > or signed char, (or a qualified version thereof) the result is 1.
>
> That is defining the sizeof operator, not the size of a char.
>
> The sizeof operator gives the size of its operand in chars, not bytes.
> Standard C has no notion of a byte.
>
> Now who doesn't know what he's talking about?

From the C99 Standard:

3.4: byte
addressable unit of data storage large enough to hold any member of the
basic character set of the execution environment

In article < >,
(Alan Gillespie) wrote:

> Micah Cowan <> wrote in message
> news:<>...
> > "Alan Gillespie" <> writes:
> >
> > > > > > unsigned char *src, *end;
> > > > > > size_t bufsiz;
> > > > > > ...
> > > :
> > > :
> > > :
> > > >
> > > > If bufsiz is of type ptrdiff_t, then the expression
> > > > ((end - src) > bufsiz)
> > > > is valid as long as end and src point to the same object,
> > > > or one byte past.
> > > >
> > >
> > > A pedant would say you mean one "char" past. The standard leaves the
> > > precise
> > > size of a char to the implementation.
> >
> > Yes, the precise size of a char... in *bits*. A char (or unsigned
> > char) is *always* one byte in size, regardless of how many bits may be
> > in a byte.
> >
> > -Micah
>
> Not quite. sizeof(char) always gives 1 because the sizeof operator is
> defined that way - it gives the size of its operand in chars.
>
> A byte is /always/ 8 bits, but that has nothing to do with C. C has no
> notion of a byte.

You are confusing bytes with octets.

(Alan Gillespie) writes:

> Micah Cowan <> wrote in message news:<>...
> > "Alan Gillespie" <> writes:
> >
> > > > > > unsigned char *src, *end;
> > > > > > size_t bufsiz;
> > > > > > ...
> > > :
> > > :
> > > :
> > > >
> > > > If bufsiz is of type ptrdiff_t, then the expression
> > > > ((end - src) > bufsiz)
> > > > is valid as long as end and src point to the same object,
> > > > or one byte past.
> > > >
> > >
> > > A pedant would say you mean one "char" past. The standard leaves the precise
> > > size of a char to the implementation.
> >
> > Yes, the precise size of a char... in *bits*. A char (or unsigned
> > char) is *always* one byte in size, regardless of how many bits may be
> > in a byte.
> >
> > -Micah
>
> Not quite. sizeof(char) always gives 1 because the sizeof operator is
> defined that way - it gives the size of its operand in chars.
>
> A byte is /always/ 8 bits, but that has nothing to do with C. C has no
> notion of a byte.

Prepare to be corrected.

Read the C standard (or at least a draft) before correcting me,
please.

-Micah

"Alan Gillespie" <> wrote in
news:bee4oq$m7q$:

> A pedant would say you mean one "char" past. The standard leaves the
> precise size of a char to the implementation.
>

The standard also defines "byte" to mean the size of a char. Therefore, by
definition, a char and a byte are the same size.