character conversion

[mdh]

May I ask the learned board this. Is it correct to assume that a
character (in this case "c") is always converted to an integer, so
that line 2 below ( specifically c - '0' ) will always give the
value as an integer of it's digit representation? (Snippet from page
97 K&R). If the question does not make too much sense, it is because I
am missing something. :-)
f(int *ptr){...

int c;

....

for ( *ptr = 0; isdigit(c); c = getch() ) /* getch gets the next
character from the input */
*ptr = *ptr * 10 + (c - '0'); <----Line 2

....}

[Ben Bacarisse]

mdh <md**@comcast.netwrites:

May I ask the learned board this. Is it correct to assume that a
character (in this case "c") is always converted to an integer, so
that line 2 below ( specifically c - '0' ) will always give the
value as an integer of it's digit representation?

In your case, c is already an int and character constants like '0'
are also ints. When getchar() returns a character, it does so as an
int, so you wording a little odd, but the answer is that the code:

int c;
...
for ( *ptr = 0; isdigit(c); c = getch() ) /* getch gets the next
character from the input */
*ptr = *ptr * 10 + (c - '0'); <----Line 2

is fine. If isdigit(c) is true, c - '0' will evaluate to 0 when c ==
'0', to 1 when c == '1' and so on up to 9 when c == '9'. Is that
answer enough?

--
Ben.

[mdh]

On Apr 21, 5:08*pm, [e-mail] (Walter Roberson)
wrote:

>
In C, characters are one of the integral types, so individual
characters are -already- an integer, rather than being "converted"
.......

Thanks...much cleaner thinking for me that way.
.

>
If the variable c is the encoding of a character, ....... which by definition in C will work
out to the numeric value that the digit would be read as in base 10 notation.

OK...thank you Walter.
Michael.

[mdh]

On Apr 21, 5:09*pm, Ben Bacarisse <[e-mail]:

In your case, c is already an int and character constants like '0'
are also ints. *........ *Is that answer enough?

Yes...it is. Thank you.

[Richard Heathfield]

Ben Bacarisse said:

mdh <md**@comcast.netwrites:

>May I ask the learned board this. Is it correct to assume that a
character (in this case "c") is always converted to an integer, so
that line 2 below ( specifically c - '0' ) will always give the
value as an integer of it's digit representation?

In your case, c is already an int

To be excruciatingly pedantic, c /would/ be an int if he'd written it as
'c'. What he actually wrote - "c" - is a you-must-not-modify-this array of
two char with static storage duration.

--
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

[Keith Thompson]

Richard Heathfield <rj*@see.sig.invalidwrites:

Ben Bacarisse said:

>mdh <md**@comcast.netwrites:

>>May I ask the learned board this. Is it correct to assume that a
character (in this case "c") is always converted to an integer, so
that line 2 below ( specifically c - '0' ) will always give the
value as an integer of it's digit representation?

In your case, c is already an int

To be excruciatingly pedantic, c /would/ be an int if he'd written it as
'c'. What he actually wrote - "c" - is a you-must-not-modify-this array of
two char with static storage duration.

To be slightly less excruciatingly pedantic but slightly more
excruciatingly correct, the code fragment being discussed, and quoted
in the article to which you responded, included the following
declaration:

int c;

The quotation marks in
(in this case "c")
were ordinary natural-language punctuation, not C string literal
delimiters.

(This is why I usually use ``this form'' when I want to quote C code,
unless I can use "quotation marks" without ambiguity.)

Let me think, what's the phrase? Oh, yes. Nyaah nyaah. Terribly
sorry, but you left me with no other choice.

--
Keith Thompson (The_Other_Keith) <ks***@mib.org>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[Richard Heathfield]

Keith Thompson said:

Richard Heathfield <rj*@see.sig.invalidwrites:

>Ben Bacarisse said:

>>mdh <md**@comcast.netwrites:

May I ask the learned board this. Is it correct to assume that a
character (in this case "c") is always converted to an integer, so
that line 2 below ( specifically c - '0' ) will always give the
value as an integer of it's digit representation?

In your case, c is already an int

To be excruciatingly pedantic, c /would/ be an int if he'd written it as
'c'. What he actually wrote - "c" - is a you-must-not-modify-this array
of two char with static storage duration.

To be slightly less excruciatingly pedantic but slightly more
excruciatingly correct,

Ah, it seems that I took it out of context (unwittingly). My mistake.
Apologies.

--
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