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qazm
#1 / 23
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 char --->int char c = '0x12' ;
int i = ????
I want i to be assigned the number 0x12. How do you do that? Should I use atoi()?
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| Tue, 30 Nov 2004 13:37:19 GMT |
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Joona I Palast
#2 / 23
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char --->int
Quote:
>> char c = '0x12' ;
>> int i = ????
>> I want i to be assigned the number 0x12. How do you do that? Should I use atoi()?
> You cannot assign '0x12' to a char, because '0x12' is four characters.
> But you don't need your char variable anyway. To assign 0x12 to an int
> variable, use:
> int i = 0x12;
More on this topic:
If you meant, how can you change the character whose code is 0x12 to an
int, you can just assign it:
char c = 0x12; /* or char c = '\x12'; */
int i = c;
If you meant, how can you change the string "0x12" into an int value,
you can use strtol:
char *c = "0x12";
int i = strtol(c, NULL, 16);
--
| Kingpriest of "The Flying Lemon Tree" G++ FR FW+ M- #108 D+ ADA N+++|
| http://www.helsinki.fi/~palaste W++ B OP+ |
\----------------------------------------- Finland rules! ------------/
"I am lying."
- Anon
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| Tue, 30 Nov 2004 14:36:52 GMT |
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Zoran Cutur
#3 / 23
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char --->int
Quote: > char c = '0x12' ; > int i = ???? > I want i to be assigned the number 0x12. How do you do that? Should I use
You are mixing up things! '0x12' isn't a string. It isn't a
literal character either. You probably meant char *c="0x12"?
In this case you can use atoi(), but even better would be
strtol() to convert the string into an integer value.
--
"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
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| Tue, 30 Nov 2004 14:45:07 GMT |
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Martin Ambuh
#4 / 23
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char --->int Quote:
> char c = '0x12' ;
> int i = ????
> I want i to be assigned the number 0x12. How do you do that?
i = c;
Quote: > Should I use atoi()?
The variable c is not a string: it has the numeric value 0x12.
atoi() is not only pointless, but is meaningless when called for
anything not a string.
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| Tue, 30 Nov 2004 15:42:13 GMT |
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Richard B
#5 / 23
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char --->int Quote:
> char c = '0x12' ; > int i = ???? > I want i to be assigned the number 0x12. How do you do that? Should I use atoi()?
This'll do it:
int i=0x12;
Your char initialisation, however, is broken. It assigns something to c
whose value is implementation-defined. You probably meant either
char c='\x12';
or
char c=0x12;
which are equivalent, and both assign the number 0x12, alias 18, to c.
Richard
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| Tue, 30 Nov 2004 16:40:22 GMT |
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CBFalcone
#6 / 23
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char --->int Quote:
> > char c = '0x12' ;
> > int i = ????
> > I want i to be assigned the number 0x12. How do you do that?
> > Should I use atoi()?
> You are mixing up things! '0x12' isn't a string. It isn't a
> literal character either. You probably meant char *c="0x12"?
> In this case you can use atoi(), but even better would be
> strtol() to convert the string into an integer value.
Harrumph, I would expect atoi to return zero from that string.
--
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home.att.net> USE worldnet address!
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| Tue, 30 Nov 2004 19:56:46 GMT |
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Thomas Stege
#7 / 23
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char --->int Quote:
>>char c = '0x12' ;
>>int i = ????
>>I want i to be assigned the number 0x12. How do you do that? Should I use atoi()?
> You cannot assign '0x12' to a char...
You can, but the result is implementation defined.
--
Thomas Stegen
http://www.geocities.com/thinkoidz
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| Tue, 30 Nov 2004 21:07:35 GMT |
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Richard B
#8 / 23
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char --->int
Quote:
> > char c = '0x12' ; > > int i = ???? > > I want i to be assigned the number 0x12. How do you do that? Should I use atoi()?
> You cannot assign '0x12' to a char, because '0x12' is four characters.
Yes, you can, and no, it isn't. "0x12" is four characters, plus a null
terminator. '0x12' is _one_ character whose value is implementation-
dependent. IOW, you can assign '0x12' to a char variable, but ISO C
doesn't tell you anything about which character it will turn out to be.
Richard
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| Tue, 30 Nov 2004 20:09:53 GMT |
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Dan P
#9 / 23
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char --->int
Quote: >> You are mixing up things! '0x12' isn't a string. It isn't a
>> literal character either. ^^^^^^^^^^
^^^^^^^^^^^^^^^^^^^^^^^^
Why? Chapter and verse, please.
Dan
--
Dan Pop
DESY Zeuthen, RZ group
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| Tue, 30 Nov 2004 23:33:06 GMT |
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Dave Vandervi
#10 / 23
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char --->int
Quote:
>> You cannot assign '0x12' to a char, because '0x12' is four characters. >Yes, you can, and no, it isn't. "0x12" is four characters, plus a null
>terminator. '0x12' is _one_ character whose value is implementation-
>dependent.
Isn't it an int?
dave
--
Because C needs to know how long the array elements are, otherwise
pointer arithmetic goes all woozy (don't you just love a precise,
detailed technical term?). --Joona I Palaste in comp.lang.c
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| Wed, 01 Dec 2004 02:26:45 GMT |
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Zoran Cutur
#11 / 23
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char --->int
Quote:
>>> You are mixing up things! '0x12' isn't a string. It isn't a
>>> literal character either. ^^^^^^^^^^
> ^^^^^^^^^^^^^^^^^^^^^^^^
> Why? Chapter and verse, please.
Which literal character is it?
--
"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
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| Fri, 03 Dec 2004 17:19:15 GMT |
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Zoran Cutur
#12 / 23
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char --->int
Quote:
>> > char c = '0x12' ;
>> > int i = ????
>> > I want i to be assigned the number 0x12. How do you do that?
>> > Should I use atoi()?
>> You are mixing up things! '0x12' isn't a string. It isn't a
>> literal character either. You probably meant char *c="0x12"?
>> In this case you can use atoi(), but even better would be
>> strtol() to convert the string into an integer value.
> Harrumph, I would expect atoi to return zero from that string.
Ahh, I see, I thought atoi() was equivalent to strtol with base
0 (which would mean octal, decimal, or hexadecimal) but it acts
as with base 10. My fault. Actually I never really used atoi on
my own (something close to an excuse but it really isn't).
--
"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
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| Fri, 03 Dec 2004 17:28:23 GMT |
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Richard B
#13 / 23
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char --->int Quote:
> >> You cannot assign '0x12' to a char, because '0x12' is four characters.
> >Yes, you can, and no, it isn't. "0x12" is four characters, plus a null
> >terminator. '0x12' is _one_ character whose value is implementation-
> >dependent.
> Isn't it an int?
Erm, yes, of course you're right. Worse, it could be an int that isn't a
valid char value. You can still assign it to a char, however, although
its value is still (even more!) implementation-dependent.
Richard
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| Fri, 03 Dec 2004 18:30:38 GMT |
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Thomas Stege
#14 / 23
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char --->int Quote:
>>>>You are mixing up things! '0x12' isn't a string. It isn't a
>>>>literal character either. ^^^^^^^^^^
>> ^^^^^^^^^^^^^^^^^^^^^^^^
>>Why? Chapter and verse, please.
> Which literal character is it?
An implementation defined one.
--
Thomas Stegen
http://www.geocities.com/thinkoidz
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| Fri, 03 Dec 2004 20:22:48 GMT |
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