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strcpy - my implementation

I have created my own implementation of strcpy library function. I would
like to have comments for improvements:
/* My version of "strcpy - a C Library Function */

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

enum { ARRSIZE = 101 };

char* my_strcpy( char*, char* );

int main( int argc, char** argv )
{
char* pc;

char arr_in[ARRSIZE];
char arr_out[ARRSIZE];

memset( arr_in, '\0', ARRSIZE );
memset( arr_out, '\0', ARRSIZE );
if( 2 != argc )
{
perror("USAGE: ./exec \" your input \"\n");
exit( EXIT_FAILURE );
}
else
{
strcpy( arr_in , argv[1] );
}

pc = my_strcpy( arr_out, arr_in );

while( *pc )
{
printf("*pc = %c\n", *pc++);
}

return EXIT_SUCCESS;
}

char* my_strcpy( char* arr_out, char* arr_in )
{
char* pc;

pc = arr_out;

while( (*arr_out++ = *arr_in++) ) ;

return pc;
}
=============== OUTPUT =============== =======

[arnuld@dune ztest]$ gcc -ansi -pedantic -Wall -Wextra check_STRCPY.c
[arnuld@dune ztest]$ ./a.out like
*pc = l
*pc = i
*pc = k
*pc = e
[arnuld@dune ztest]$

It works fine without troubles. Now if you change the last return call in
my_strcpy from "return pc" to return "return arr_out", then while loop in
main() will not print anything at all. I really did not understand it.
Using thr array name will give a pointer to its 1st element but int htis
case it is giving a pointer to its last element. Why ? Thats why I
introduced the extra "char* pc" in first place.


--
www.lispmachine.wordpress.com
my email is @ the above blog.
Google Groups is Blocked. Reason: Excessive Spamming

Sep 8 '08 #1
77 8329
Hi

On Mon, 08 Sep 2008 11:51:45 +0500, arnuld wrote:
/* My version of "strcpy - a C Library Function */
char* my_strcpy( char* arr_out, char* arr_in ) {
That should be: char *strcpy(char *dest, const char *src);
char* pc;
pc = arr_out;
while( (*arr_out++ = *arr_in++) ) ;
return pc;
}
It works fine without troubles. Now if you change the last return call
in my_strcpy from "return pc" to return "return arr_out", then while
loop in main() will not print anything at all. I really did not
understand it. Using thr array name will give a pointer to its 1st
element but int htis case it is giving a pointer to its last element.
Why ? Thats why I introduced the extra "char* pc" in first place.
arr_in is not an array, it is a pointer. If you increment the pointer
repeatedly it will be a pointer to a later address in memory. In this
case, at the end of the function it will point past the end of the string.

Your implementation is much slower than what that would be used in almost
any real c library, because it copies one byte at a time. If you know
something more about the way the target processor works than the C
standard tells you, then you can copy several bytes at a time using (for
example) int.

HTH
viza
Sep 8 '08 #2
arnuld said:
I have created my own implementation of strcpy library function. I would
like to have comments for improvements:
/* My version of "strcpy - a C Library Function */

#include <stdio.h>
#include <stdlib.h>
I understand that you might need these for testing purposes...
#include <unistd.h>
....but why on earth do you need this non-standard header for testing an
implementation of a standard and rather simple C function?
#include <string.h>
Strange that you should need this, but okay.
enum { ARRSIZE = 101 };

char* my_strcpy( char*, char* );
Better: char *my_strcpy(char *, const char *);
>
int main( int argc, char** argv )
{
char* pc;

char arr_in[ARRSIZE];
char arr_out[ARRSIZE];

memset( arr_in, '\0', ARRSIZE );
memset( arr_out, '\0', ARRSIZE );
Or just:

char arr_in[ARRSIZE] = {0};
char arr_out[ARRSIZE] = {0};

which saves you two memset calls.
if( 2 != argc )
{
perror("USAGE: ./exec \" your input \"\n");
exit( EXIT_FAILURE );
}
else
{
strcpy( arr_in , argv[1] );
What if argv[1] is longer than 100 bytes? You should test for this.

<snip>
char* my_strcpy( char* arr_out, char* arr_in )
Better: char *my_strcpy(char * arr_out, const char *arr_in)
{
char* pc;

pc = arr_out;

while( (*arr_out++ = *arr_in++) ) ;

return pc;
}
How is this *your* implementation? It isn't significantly different from
the implementation on p105 of K&R2, with names changed to protect the
innocent and a return value added to get a closer match to ISO strcpy.

<snip>
It works fine without troubles. Now if you change the last return call in
my_strcpy from "return pc" to return "return arr_out", then while loop in
main() will not print anything at all.
arr_out is badly named. In my_strcpy, it's not an array, but a pointer.

Consider what value that pointer has at the time you've completed the loop.
To which byte is it pointing? What value does that byte have? And
consequently, if you return it to main and capture that pointer value in
pc = my_strcpy(arr_o ut, arr_in), to what byte is pc pointing?

You would find it easier to learn and we would find it easier to explain if
you chose different names in called functions from those in calling
functions. When discussing your program, I can't refer to 'pc' without
saying which pc I mean. Nor can I refer to arr_in or arr_out without
saying which I mean. Had you used different names in my_strcpy, it would
have made discussion easier.
I really did not understand it.
Using thr array name will give a pointer to its 1st element
Yes. And that's what my_strcpy's arr_in is - a copy of a pointer to the
first element of an array. But then *you* change the value of that
pointer.
but int htis
case it is giving a pointer to its last element. Why ?
If you change an object's value, you should not be surprised if that
object's value changes.

Thats why I introduced the extra "char* pc" in first place.
Yes. It's necessary in this case. Well, there is another way round,
actually:

char *my_strcpy(char *target, const char *source)
{
size_t i = 0;
while(target[i] = source[i])
{
++i;
}

return target;
}

which avoids the extra pointer, but only at the expense of an extra size_t.

--
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
Sep 8 '08 #3
arnuld wrote:
>
char* my_strcpy( char* arr_out, char* arr_in )
{
char* pc;

pc = arr_out;

while( (*arr_out++ = *arr_in++) ) ;

return pc;
}
As has been pointed out elsewhere, the source pointer should be const
and the names don't make sense. The inner parentheses in the while loop
are superfluous.
>
It works fine without troubles. Now if you change the last return call in
my_strcpy from "return pc" to return "return arr_out", then while loop in
main() will not print anything at all. I really did not understand it.
Using thr array name will give a pointer to its 1st element but int htis
case it is giving a pointer to its last element. Why ? Thats why I
introduced the extra "char* pc" in first place.
Because you have incremented it to point the the '\0' at the end of the
string.

--
Ian Collins.
Sep 8 '08 #4
viza <to******@gm-il.com.obviousc hange.invalidwr ites:
Hi

On Mon, 08 Sep 2008 11:51:45 +0500, arnuld wrote:
>/* My version of "strcpy - a C Library Function */
>char* my_strcpy( char* arr_out, char* arr_in ) {

That should be: char *strcpy(char *dest, const char *src);
Oh for goodness sake.
>
> char* pc;
pc = arr_out;
while( (*arr_out++ = *arr_in++) ) ;
return pc;
}
>It works fine without troubles. Now if you change the last return call
in my_strcpy from "return pc" to return "return arr_out", then while
loop in main() will not print anything at all. I really did not
understand it. Using thr array name will give a pointer to its 1st
element but int htis case it is giving a pointer to its last element.
Why ? Thats why I introduced the extra "char* pc" in first place.

arr_in is not an array, it is a pointer. If you increment the pointer
repeatedly it will be a pointer to a later address in memory. In this
If you increment it once it will be a pointer to a later address in
memory...
case, at the end of the function it will point past the end of the
string.
So what?
>
Your implementation is much slower than what that would be used in
almost
Much slower?
any real c library, because it copies one byte at a time. If you know
something more about the way the target processor works than the C
standard tells you, then you can copy several bytes at a time using (for
example) int.

HTH
viza
I would be interested to hear how you would do this in standard C.
Sep 8 '08 #5
Richard wrote:
viza <to******@gm-il.com.obviousc hange.invalidwr ites:
>Hi

On Mon, 08 Sep 2008 11:51:45 +0500, arnuld wrote:
>>/* My version of "strcpy - a C Library Function */
char* my_strcpy( char* arr_out, char* arr_in ) {
>That should be: char *strcpy(char *dest, const char *src);

Oh for goodness sake.
Now you are being an arse, the correction is correct. The second
pointer points to an invariant.
>>element but int htis case it is giving a pointer to its last element.
Why ? Thats why I introduced the extra "char* pc" in first place.
>arr_in is not an array, it is a pointer. If you increment the pointer
repeatedly it will be a pointer to a later address in memory. In this
case, at the end of the function it will point past the end of the
string.

So what?
He's answering the question, boy you are dense tonight..

--
Ian Collins.
Sep 8 '08 #6
Ian Collins <ia******@hotma il.comwrites:
Richard wrote:
>viza <to******@gm-il.com.obviousc hange.invalidwr ites:
>>Hi

On Mon, 08 Sep 2008 11:51:45 +0500, arnuld wrote:
/* My version of "strcpy - a C Library Function */
char* my_strcpy( char* arr_out, char* arr_in ) {
>>That should be: char *strcpy(char *dest, const char *src);

Oh for goodness sake.
Now you are being an arse, the correction is correct. The second
pointer points to an invariant.
I was referring to the name change.
>
>>>element but int htis case it is giving a pointer to its last element.
Why ? Thats why I introduced the extra "char* pc" in first place.
>>arr_in is not an array, it is a pointer. If you increment the pointer
repeatedly it will be a pointer to a later address in memory. In this
case, at the end of the function it will point past the end of the
string.

So what?
He's answering the question, boy you are dense tonight..
I am pointing out that it has zero impact. So nothing to worry
about. But I could have, and should have, been more explicit.
Sep 8 '08 #7
On Sep 8, 11:21 am, Ian Collins <ian-n...@hotmail.co mwrote:
Richard wrote:
viza <tom.v...@gm-il.com.obviousc hange.invalidwr ites:
Hi
On Mon, 08 Sep 2008 11:51:45 +0500, arnuld wrote:
/* My version of "strcpy - a C Library Function */
char* my_strcpy( char* arr_out, char* arr_in ) {
That should be: char *strcpy(char *dest, const char *src);
Oh for goodness sake.

Now you are being an arse, the correction is correct. The second
pointer points to an invariant.
No, it's wrong.
It's C code, not a C implementation, therefore it shouldn't have the
name strcpy. (a hint that it's C code is that he includes several
standard header files, ignoring <unistd.hand that he has main there)
Sep 8 '08 #8

"Ian Collins" <ia******@hotma il.comwrote in message
news:6i******** ****@mid.indivi dual.net...
arnuld wrote:
>>
char* my_strcpy( char* arr_out, char* arr_in )
{
char* pc;

pc = arr_out;

while( (*arr_out++ = *arr_in++) ) ;

return pc;
}
The inner parentheses in the while loop
are superfluous.
I thought this was sometimes used to stop the compiler warning about using
"=" instead of "==".

--
Bartc

Sep 8 '08 #9
"arnuld" <su*****@invali d.addresswrote in message
news:pa******** *************** *****@invalid.a ddress...
>I have created my own implementation of strcpy library function. I would
like to have comments for improvements:
char* my_strcpy( char* arr_out, char* arr_in )
{
char* pc;
if (arr_out==NULL| |arr_in==NULL)r eturn "";

Testing for null strings might be useful; either return empty string or
abort. But not everyone agrees.
pc = arr_out;

while( (*arr_out++ = *arr_in++) ) ;
If you're interested in speed you might like to test your version against
the standard strcpy, for various lengths of strings and copying a few
million times.

You might then also like to test a version using strlen() and memcpy().

--
Bartc

Sep 8 '08 #10

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