Re: FAQ 20-12

**Richard Heathfield** · Jul 4 '08, 05:35 AM

Re: FAQ 20-12

mdh said:

On Jul 3, 6:51 pm, viza <tom.v...@gm-il.com.obviousc hange.invalid>
wrote:
>

do {
*--p = "0123456789abcd ef"[num % base]; << -------This

>>

>line has me confused? If someone could help, I will appreciate it.

>>
>Let me space it out a bit:
>>
>do {
>char hexidecimal_dig its[]= {'0','1','2', ...etc };
>>
>p= p - 1;
>>
>*p= hexidecimal_dig its[ num % base ];
>>
>The only bit of maths that you might not understand is that ( num %
>base ) means the remainder when num is divided by base.

>
>
I have just never seen ( not that that necessarily means anything) a
char array initialized like that. I am curious as to how you get from
this ("0123456789abc def"[num % base];) to this (char
hexidecimal_dig its[]= {'0','1','2', ...etc };).

Come on Michael, he's just spreading it out a little to help you see the
logic, that's all. There is no hexadecimal_dig its array, because it isn't
necessary - the string literal will do just as well. But what viza is
telling you is that we could write code that has the same effect, using a
temporary array instead of a string literal, and you seem to agree that
this equivalent would be easy to understand.

The trick is that there is no trick. "A"[0] is 'A'. "AB"[0] is also 'A',
and "AB"[1] is 'B'. "ABC"[0] is 'A', "ABC"[1] is 'B', "ABC"[2] is 'C'.

"0123456789abcd ef"[0] is '0'.
"0123456789abcd ef"[1] is '1'.
"0123456789abcd ef"[9] is '9'.
"0123456789abcd ef"[10] is 'a'.
"0123456789abcd ef"[11] is 'b'.

Clear now?

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

**CBFalconer** · Jul 4 '08, 06:45 AM

Re: FAQ 20-12

mdh wrote:

>

.... snip ...

>
I have just never seen ( not that that necessarily means anything)
a char array initialized like that. I am curious as to how you get
from this ("0123456789abc def"[num % base];) to this (char
hexidecimal_dig its[]= {'0','1','2', ...etc };). ( The math is fine
(viz the modulo operator). Thank you for your help.

Why the difficulty? "012...def" is a char array. num % base is a
number in the range 0 through 15 as long as num and base are
positive, and base is in the range 1 through 16. Therefore
"0123...def "[num % base] selects one of those characters.

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home .att.net>
Try the download section.

**mdh** · Jul 4 '08, 01:45 PM

Re: FAQ 20-12

On Jul 3, 10:34 pm, Richard Heathfield <r...@see.sig.i nvalidwrote:

mdh said:

Come on Michael, he's just spreading it out a little to help you see the
logic, that's all.

Sounds perfectly logical to me!

There is no hexadecimal_dig its array, because it isn't

necessary - the string literal will do just as well.

....snip....

The trick is that there is no trick. "A"[0] is 'A'. "AB"[0] is also 'A',
and "AB"[1] is 'B'. "ABC"[0] is 'A', "ABC"[1] is 'B', "ABC"[2] is 'C'.
>
"0123456789abcd ef"[0] is '0'.
"0123456789abcd ef"[1] is '1'.
"0123456789abcd ef"[9] is '9'.
"0123456789abcd ef"[10] is 'a'.
"0123456789abcd ef"[11] is 'b'.
>
Clear now?
>

Yep...I have gone back and looked at K&R and agree it is a very neat
way of doing this. What I was trying to say was that even though the
construct "I am a string literal" is well explained in K&R, I was not
aware that one could refer to each character in the constant by using
the operator [] immediately following it. All I know is that when I
have simply assumed things in the past, it has often gotten me into
hot water...so I know I can ask here and all my assumptions will be
rapidly corrected!!! :-)
Thank you Richard.

**mdh** · Jul 4 '08, 01:55 PM

Re: FAQ 20-12

On Jul 3, 7:58 pm, CBFalconer <cbfalco...@yah oo.comwrote:

>
Why the difficulty? "012...def" is a char array. num % base is a
number in the range 0 through 15 as long as num and base are
positive, and base is in the range 1 through 16. Therefore
"0123...def "[num % base] selects one of those characters.
>
--

Hi Chuck...I think I made it more difficult than it is.

One last question. The line

p = &retbuf[sizeof(retbuf)-1];

where retbuf[] is declared as a static char array

uses the ampersand when assigning the address to the pointer 'p'. Is
this necessary because using the name alone would assign the address
of retbuf[0] which is not what we want ie retbuf[x] by itself would
not do it? Sorry if this is another obvious question, but that's the
way I learn!!!

Thanks for taking the time to reply.

Thank you for your help.

**santosh** · Jul 4 '08, 02:15 PM

Re: FAQ 20-12

mdh wrote:

On Jul 3, 7:58 pm, CBFalconer <cbfalco...@yah oo.comwrote:

>>
>Why the difficulty? "012...def" is a char array. num % base is a
>number in the range 0 through 15 as long as num and base are
>positive, and base is in the range 1 through 16. Therefore
>"0123...def "[num % base] selects one of those characters.
>>
>--

>
>
Hi Chuck...I think I made it more difficult than it is.
>
>
One last question. The line
>
p = &retbuf[sizeof(retbuf)-1];
>
where retbuf[] is declared as a static char array
>
uses the ampersand when assigning the address to the pointer 'p'. Is
this necessary because using the name alone would assign the address
of retbuf[0] which is not what we want ie retbuf[x] by itself would
not do it? Sorry if this is another obvious question, but that's the
way I learn!!!

You're right. Just doing:

p = retbuf;

will assign the address of the first element of the array retbuf to p,
which must be of the appropriate pointer type. Doing:

p = &retbuf;

will assign the address of the array 'retbuf' to p, which must be of the
appropriate pointer to array type. Doing:

p = retbuf[x];

will assign the _value_ of element 'x' of retbuf to p, which is probably
not what you want. Doing:

p = &retbuf[x];

will assign the address of element 'x' of retbuf to p, which is what we
want in this code posted up-thread.

<snip>

**mdh** · Jul 4 '08, 02:25 PM

Re: FAQ 20-12

On Jul 4, 7:05 am, santosh <santosh....@gm ail.comwrote:

mdh wrote:

On Jul 3, 7:58 pm, CBFalconer <cbfalco...@yah oo.comwrote:

>

Why the difficulty? "012...def" is a char array. num % base is a
number in the range 0 through 15 as long as num and base are
positive, and base is in the range 1 through 16. Therefore
"0123...def "[num % base] selects one of those characters.

>

--

>

Hi Chuck...I think I made it more difficult than it is.

>

One last question. The line

>

p = &retbuf[sizeof(retbuf)-1];

>

where retbuf[] is declared as a static char array

>

uses the ampersand when assigning the address to the pointer 'p'. Is
this necessary because using the name alone would assign the address
of retbuf[0] which is not what we want ie retbuf[x] by itself would
not do it? Sorry if this is another obvious question, but that's the
way I learn!!!

>
You're right. Just doing:
>
p = retbuf;
>
will assign the address of the first element of the array retbuf to p,
which must be of the appropriate pointer type. Doing:
>
p = &retbuf;
>
will assign the address of the array 'retbuf' to p, which must be of the
appropriate pointer to array type. Doing:
>
p = retbuf[x];
>
will assign the _value_ of element 'x' of retbuf to p, which is probably
not what you want. Doing:
>
p = &retbuf[x];
>
will assign the address of element 'x' of retbuf to p, which is what we
want in this code posted up-thread.
>
<snip>

Thanks Santosh.

**Keith Thompson** · Jul 4 '08, 07:25 PM

Re: FAQ 20-12

santosh <santosh.k83@gm ail.comwrites:
[...]

You're right. Just doing:
>
p = retbuf;
>
will assign the address of the first element of the array retbuf to p,
which must be of the appropriate pointer type. Doing:
>
p = &retbuf;
>
will assign the address of the array 'retbuf' to p, which must be of the
appropriate pointer to array type. Doing:
>
p = retbuf[x];
>
will assign the _value_ of element 'x' of retbuf to p, which is probably
not what you want. Doing:
>
p = &retbuf[x];
>
will assign the address of element 'x' of retbuf to p, which is what we
want in this code posted up-thread.

[...]

And the last is equivalent to

p = retbuf + x;

where the "+" is doing pointer arithmetic. But

p = &retbuf[x];

is arguably clearer.

--
Keith Thompson (The_Other_Keit h) kst-u@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

**CBFalconer** · Jul 5 '08, 12:15 AM

Re: FAQ 20-12

mdh wrote:

>

.... snip ...

>
One last question. The line
>
p = &retbuf[sizeof(retbuf)-1];
>
where retbuf[] is declared as a static char array

I have no idea. Not enough code is shown. Bear in mind that
pointers to local data are invalid outside of the function (barring
static data).

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home .att.net>
Try the download section.

**Richard Heathfield** · Jul 5 '08, 12:35 AM

Re: FAQ 20-12

CBFalconer said:

mdh wrote:

>>

... snip ...

>>
>One last question. The line
>>
> p = &retbuf[sizeof(retbuf)-1];
>>
>where retbuf[] is declared as a static char array

>
I have no idea. Not enough code is shown.

Looks like plenty to me.

Bear in mind that
pointers to local data are invalid outside of the function

He did say it's static.

(barring static data).

Right.

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

**Richard** · Jul 5 '08, 09:05 AM

Re: FAQ 20-12

CBFalconer <cbfalconer@yah oo.comwrites:

mdh wrote:

>>

... snip ...

>>
>One last question. The line
>>
> p = &retbuf[sizeof(retbuf)-1];
>>
>where retbuf[] is declared as a static char array

>
I have no idea. Not enough code is shown. Bear in mind that
pointers to local data are invalid outside of the function (barring
static data).

There is more than enough information for the question asked.

Re: FAQ 20-12

Re: FAQ 20-12

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