string copy

[bml]

char s1 = "this string";

char *s2;
s2 = (char *)malloc(30);
memset(s2, 0, 30);

/* this supposes to copy string s1 to s2 */
while((*s2++ = *s1++) != '\0');

/* Why empty content of s2 print out here*/

Thanks a lot!

[Artie Gold]

bml wrote:

char s1 = "this string";

char *s2;
s2 = (char *)malloc(30);
memset(s2, 0, 30);

/* this supposes to copy string s1 to s2 */
while((*s2++ = *s1++) != '\0');

/* Why empty content of s2 print out here*/

Hint: What is s2 now pointing to?

[BTW -- in the future, post *real* code (the most minimal possible
runnable snippet); otherwise we're just guessing.]

HTH,
--ag

--
Artie Gold -- Austin, Texas

[Martin Ambuhl]

bml wrote:

char s1 = "this string"; ^^ missing '*'
char *s2;
s2 = (char *)malloc(30); ^^^^^^^ silly cast
No error check memset(s2, 0, 30); ^^^^^^^^^^^^^^^ ^^
useless function call
/* this supposes to copy string s1 to s2 */
while((*s2++ = *s1++) != '\0');

/* Why empty content of s2 print out here*/ You were unlucky; your computer should have exploded.
Because s2 points beyond the end of the string. What did you think was
happening when you incremented it?
Thanks a lot!

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main(void)
{
char *s1 = "this string";
char *t1 = s1;
char *s2;
char *t2;
if (!(s2 = malloc(strlen(s 1) + 1))) {
fprintf(stderr, "malloc failed for s1\n");
exit(EXIT_FAILU RE);
}
t2 = s2;
while ((*t2++ = *t1++)) ;
printf("s1: \"%s\"\ns2: \"%s\"\n", s1, s2);
return 0;
}

--
Martin Ambuhl

[Thomas Pfaff]

"bml" <le*****@yahoo. com> writes:

char s1 = "this string";
I assume this is a typo and you meant to write

char *s1 = "this string";
char *s2;
s2 = (char *)malloc(30);
Don't cast malloc. It is not necessarry and may hide a serious error
if you forget to include <stdlib.h> (which provide the prototype for
malloc).
memset(s2, 0, 30);
Make sure malloc succeeded before accessing s2.

if (s2 != NULL)
/* use s2 here */

If you simply want to "empty" the string then *s2 = '\0' will do.
/* this supposes to copy string s1 to s2 */
while((*s2++ = *s1++) != '\0');
Why not simply use strcpy?
/* Why empty content of s2 print out here*/

Where does s2 point after the loop completes? You'll need to remember
the original pointer,

char *t = s2;
while (*s2++ = *s1++);
s2 = t;
printf ("%s\n", s2);

[Jarno A Wuolijoki]

On 1 Feb 2004, Thomas Pfaff wrote:

Where does s2 point after the loop completes? You'll need to remember
the original pointer,

char *t = s2;
while (*s2++ = *s1++);
s2 = t;

char *t = s2;
while (*t++ = *s1++) ;

If this seems like a mindless nit I'm probably just overly biased against
the idiom of modifying a variable and then popping it back..

[Thomas Matthews]

bml wrote:

char s1 = "this string";
const char * s1 = "this string";
Always treat string literals as constants. Thus
s1 is declared as a pointer to constant data.
Constant meaning it won't change.

char *s2;
s2 = (char *)malloc(30);
memset(s2, 0, 30);
char * s2;
if (strlen(s1) < 30)
{
s2 = malloc(30);
}
else
{
s2 = malloc(strlen(s 1) + 1);
}
if (s2 == NULL)
{
/* error processing */
}


/* this supposes to copy string s1 to s2 */
while((*s2++ = *s1++) != '\0');
strcpy(s2, s1); /* copy string s1 to s2 */


/* Why empty content of s2 print out here*/
puts("String s2:");
puts(s2);

Thanks a lot!

--
Thomas Matthews

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http://www.slack.net/~shiva/welcome.txt
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Other sites:
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http://www.sgi.com/tech/stl -- Standard Template Library

[Emmanuel Delahaye]

In 'comp.lang.c', "bml" <le*****@yahoo. com> wrote:

char s1 = "this string";

char *s2;
s2 = (char *)malloc(30);
memset(s2, 0, 30);

/* this supposes to copy string s1 to s2 */
while((*s2++ = *s1++) != '\0');

/* Why empty content of s2 print out here*/

Empty ? I'm not sure. I would have said 'undeterminated ', due to the lack of
final 0.

add :
*s2 = 0;

but you are making a design mistake. You have changed the value of s2, hence
you have lost it's initial value:

- The string copy was somehow correct, but you don't know any more where does
the copied string begin.
- You are now unable to free the allocated block which has created a memory
leak. (A Bad Thing (c))

If you want to make a copy of a string, I recommend to desing your own
'strdup()' function:

#include <string.h>
#include <stdlib.h>

char *str_dup (char const * const s)
{
char *sdup;

if (s != NULL)
{
size_t size = strlen (s) + 1;

sdup = malloc (size);

if (sdup != NULL)
{
memcpy (sdup, s, size);
}
}
return sdup;
}

--
-ed- em**********@no os.fr [remove YOURBRA before answering me]
The C-language FAQ: http://www.eskimo.com/~scs/C-faq/top.html
C-reference: http://www.dinkumware.com/manuals/reader.aspx?lib=cpp
FAQ de f.c.l.c : http://www.isty-info.uvsq.fr/~rumeau/fclc/

[Al Bowers]

Emmanuel Delahaye wrote:

If you want to make a copy of a string, I recommend to desing your own
'strdup()' function:

#include <string.h>
#include <stdlib.h>

char *str_dup (char const * const s)
{
char *sdup;
To correct the flaw in the return value should the argument, s,
be a NULL value, make this:

char *sdup = NULL;

if (s != NULL)
{
size_t size = strlen (s) + 1;

sdup = malloc (size);

if (sdup != NULL)
{
memcpy (sdup, s, size);
}
}
return sdup;
}

--
Al Bowers
Tampa, Fl USA
mailto: xa******@myrapi dsys.com (remove the x to send email)
http://www.geocities.com/abowers822/

[Peter Pichler]

"Emmanuel Delahaye" <em**********@n oos.fr> wrote in message
news:Xn******** *************** ****@213.228.0. 196...

In 'comp.lang.c', "bml" <le*****@yahoo. com> wrote:

char s1 = "this string";

char *s2;
s2 = (char *)malloc(30);
memset(s2, 0, 30);

/* this supposes to copy string s1 to s2 */
while((*s2++ = *s1++) != '\0');

/* Why empty content of s2 print out here*/
Empty ? I'm not sure. I would have said 'undeterminated ', due to the lack

of final 0.

That's probably why he added the otherwise useless memset() in the first
place ;-)

[Peter Pichler]

"Thomas Matthews" <Th************ *************@s bcglobal.net> wrote in
message news:40******** ******@sbcgloba l.net...

char * s2;
if (strlen(s1) < 30)
{
s2 = malloc(30);
}
else
{
s2 = malloc(strlen(s 1) + 1);
}

Why do you need the first branch?