char vs int

[Merrill & Michele]

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(void){
char *p;
p=malloc(4);
strcpy(p, "tja");
printf("%s\n", p);
free(p);
return 0;
}

Why is p a char and not an int? MPJ

[Michael Mair]

Merrill & Michele wrote:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(void){
char *p;
p=malloc(4);
Check the return value of malloc. Always.
strcpy(p, "tja");
To be on the safe side, you could for example do
*p = '\0';
strncat(p,"tja" ,4-1);
printf("%s\n", p);
free(p);
return 0;
}

Why is p a char and not an int? MPJ

p is a char *. It is definitely not a char or int.
Please follow the simple rule of rereading your message
before you send it as we otherwise have to guess what
you could have meant apart from the on average high
level of weirdness by which many of your messages are
permeated.

My guess is:
"character constants are of type int and many functions
return int/have int parameters when we are dealing with
single characters -- why is it then that we store strings
in char arrays or memory pointed to by char * variables?"

Is that your question?
For that I have an answer.
-Michael
--
E-Mail: Mine is a gmx dot de address.

[dandelion]

"Merrill & Michele" <be********@com cast.net> wrote in message
news:yr******** ************@co mcast.com...

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(void){
char *p;
p=malloc(4);
strcpy(p, "tja");
printf("%s\n", p);
free(p);
return 0;
}

Why is p a char and not an int? MPJ

Becuase you are copying a char-array into it using a function which requires
a char*?

[Nils O. Selåsdal]

Merrill & Michele wrote:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(void){
char *p;
p=malloc(4);
strcpy(p, "tja");
printf("%s\n", p);
free(p);
return 0;
}

Why is p a char and not an int? MPJ

Cause you, the programmer, made it a char pointer.
Which seems good, as you're coping a C string to it,
and strings are usually best manipulated as sequences
of chars, it is also what "%s" in your printf statement
would expect..

[Robert Gamble]

On Wed, 17 Nov 2004 08:40:20 -0600, Merrill & Michele wrote:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(void){
char *p;
p=malloc(4);
strcpy(p, "tja");
printf("%s\n", p);
free(p);
return 0;
}

Why is p a char and not an int? MPJ

p is a pointer to char, not a char, and this is because that is how you
defined it, why in the world would you expect it to be an int??

Rob Gamble

[Thomas Stegen]

Merrill & Michele wrote:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(void){
char *p;
p=malloc(4);
strcpy(p, "tja");
printf("%s\n", p);
free(p);
return 0;
}

Why is p a char and not an int? MPJ

Are you asking why chars are stored in variables of type char?
And p is not a char.

Honestly, I usually try to give the questioner the benefit of the doubt
and I apply a good amount of charity of interpretation, but your
question does not make sense to me. I cannot even tell if you saying
that p is of type char is a mistake or ignorance on your part.

Please tell us why you think p could have been an int (or probably
pointer to int) instead of a pointer to char. Are you by any chance
referring to the fact that getchar and getc returns ints and not chars?
(So you see, I might be beginning to understand your question after all
though I didn't when I wrote the previous paragraph. But still I need
you to ask it.)

--
Thomas.

[J.L.Cooper]

"Merrill & Michele" <be********@com cast.net> wrote:
<Snip>

int main(void){
char *p;
p=malloc(4);
strcpy(p, "tja"); <Snip>
Why is p a char and not an int? MPJ

I am going to assume that you are asking the following question.

Why do we store strings in arrays of char data types instead of arrays of
int data types?

Well I would answer that int data types while capable of holding the same
information as a char data type while doing so they are wasting X bytes of
space (size of X depends on the number of bytes in an int on the system
though on most 32 bit machines it is 4 bytes in length so there are 3 bytes
wasted.)

While this doesn't seem like much how about if you have a string buffer that
needs to store a 1024 characters. This would take up 1Kb of space using an
array of char data types but 4Kb of space (on most 32 bit machines again)
if it was using int data types.

If that is not your question then please repost with a clarified version and
I will have another go.
--------------
Jason Cooper

[Merrill & Michele]

"J.L.Cooper " <[e-mail]> wrote in message
news:cn******** **@sun-cc204.lut.ac.uk ...

"Merrill & Michele" <be********@com cast.net> wrote:
<Snip>

int main(void){
char *p;
p=malloc(4);
strcpy(p, "tja"); <Snip>

Why is p a char and not an int? MPJ

I am going to assume that you are asking the following question.

Why do we store strings in arrays of char data types instead of arrays of
int data types?

Well I would answer that int data types while capable of holding the same
information as a char data type while doing so they are wasting X bytes of
space (size of X depends on the number of bytes in an int on the system
though on most 32 bit machines it is 4 bytes in length so there are 3

bytes wasted.)

While this doesn't seem like much how about if you have a string buffer that needs to store a 1024 characters. This would take up 1Kb of space using an array of char data types but 4Kb of space (on most 32 bit machines again)
if it was using int data types.

If that is not your question then please repost with a clarified version and I will have another go.
--------------
Jason Cooper

Thanks all for replies. That I thought I had partitioned the choice of data
type in this question with int and char shows exactly what I don't
understand. I'll hit K&R and the FAQs and get back to you. MPJ

[Dan Pop]

In <30************ *@uni-berlin.de> Michael Mair <Mi**********@i nvalid.invalid> writes:

Merrill & Michele wrote:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(void){
char *p;
p=malloc(4);

Check the return value of malloc. Always.

That's good advice for code that is meant to be executed. It's bad advice
for code that is meant to merely illustrate a point, because the error
checking code distracts the reader from the real purpose of the code.

strcpy(p, "tja");

To be on the safe side, you could for example do
*p = '\0';
strncat(p,"tja" ,4-1);

It's not any safer than the original, but it is less readable and has
higher maintenance overhead (the magical constant 4 appears in two places
instead of one).

Even in more complex examples, it is open to debate whether this is the
right thing to do: silently generating the *wrong* result is seldom a
better option than reporting an error. That's why strncat is seldom used
as a "safe" strcpy.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de
Currently looking for a job in the European Union

[Merrill & Michele]

> >[OP: Why is a char * a char and not an int]

Michael Mair: [insult snipped] My guess is:
"character constants are of type int and many functions
return int/have int parameters when we are dealing with
single characters -- why is it then that we store strings
in char arrays or memory pointed to by char * variables?"
Is that your question?
For that I have an answer I would imagine that your answer is similar to K&R §5.5 . But since I don't
get it, I would welcome the answer to that question.
Nils Selasdal:
Cause you, the programmer, made it a char pointer.
Which seems good, as you're cop[y]ing a C string to it,
and strings are usually best manipulated as sequences
of chars, it is also what "%s" in your printf statement
would expect..

K&R §2.2: "There are only a few basic data types in C: char, int, float,
double."
Is char * a data type? How about int **? MPJ