How is it working? Am I doing it right?

[Benoit Lefebvre]

Weird subject you will say.. I didn't know what to put in there..

Long story short, I'm changing the value of a variable from a
function.. Nothing unusual there..

How come the value of the "word" variable change when using the
testfunct function? Am I sending a pointer to the word variable to
this function or whatever ??

I know it's a weird question.. especially that this is doing exactly
what I need.. but I want to know if I'm doing it right and if not, how
to make it work correctly.

The program where this is going is checking a configuration text file
and puts all the lines in a 2d char array for comparison later. Here
is a simplified "extraction " of the code.

------
#include <stdio.h>
#include <strings.h>

int testfunct (char *myword)
{
strcpy(myword," hello");

return 0;
}

int main (int argc, char *argv[])
{
char word[100];

strcpy(word,"we eee");
printf("WORD: %s\n", word);
testfunct(word) ;
printf("WORD: %s\n", word);

return 0;
}
------

[Walter Roberson]

Re: How is it working? Am I doing it right?

In article <d009b12e-13ff-44bd-a75b-b66a58854ca0@27 g2000hsf.google groups.com>,
Benoit Lefebvre <benoit.lefebvr e@gmail.comwrot e:

>How come the value of the "word" variable change when using the
>testfunct function? Am I sending a pointer to the word variable to
>this function or whatever ??

>int testfunct (char *myword)
>{
strcpy(myword," hello");
>
return 0;
>}

>int main (int argc, char *argv[])
>{
char word[100];
>
strcpy(word,"we eee");
printf("WORD: %s\n", word);
testfunct(word) ;

When you pass an array name as a function parameter, it gets silently
converted into a pointer to the first element of the array.

printf("WORD: %s\n", word);
>
return 0;
>}

So yes, the above is fine / legal / well-defined, as long as testfunct()
does not write more characters into myword than the object is defined
to hold.
--
"The art of storytelling is reaching its end because the epic
side of truth, wisdom, is dying out." -- Walter Benjamin

[Keith Thompson]

Re: How is it working? Am I doing it right?

Benoit Lefebvre <benoit.lefebvr e@gmail.comwrit es:

Weird subject you will say.. I didn't know what to put in there..
>
Long story short, I'm changing the value of a variable from a
function.. Nothing unusual there..
>
How come the value of the "word" variable change when using the
testfunct function? Am I sending a pointer to the word variable to
this function or whatever ??
>
I know it's a weird question.. especially that this is doing exactly
what I need.. but I want to know if I'm doing it right and if not, how
to make it work correctly.
>
The program where this is going is checking a configuration text file
and puts all the lines in a 2d char array for comparison later. Here
is a simplified "extraction " of the code.
>
------
#include <stdio.h>
#include <strings.h>
>
int testfunct (char *myword)
{
strcpy(myword," hello");
>
return 0;
}
>
int main (int argc, char *argv[])
{
char word[100];
>
strcpy(word,"we eee");
printf("WORD: %s\n", word);
testfunct(word) ;
printf("WORD: %s\n", word);
>
return 0;
}

Yes, you're sending a pointer to the function.

``word'' is an array object, so the argument to testfunct is an
expression of array type, which is implicitly converted to a pointer
to the array's first element. testfunct() then modifies the data that
its argument points to.

Read section 6 of the comp.lang.c FAQ, <http://www.c-faq.com/>.

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

[Keith Thompson]

Re: How is it working? Am I doing it right?

roberson@ibd.nr c-cnrc.gc.ca (Walter Roberson) writes:
[...]

When you pass an array name as a function parameter, it gets silently
converted into a pointer to the first element of the array.

[...]

Yes. This is, of course, just one instance of a more general rule
having nothing to do with functions.

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

[Jim Langston]

Re: How is it working? Am I doing it right?

Benoit Lefebvre wrote:

Weird subject you will say.. I didn't know what to put in there..
>
Long story short, I'm changing the value of a variable from a
function.. Nothing unusual there..
>
How come the value of the "word" variable change when using the
testfunct function? Am I sending a pointer to the word variable to
this function or whatever ??
>
I know it's a weird question.. especially that this is doing exactly
what I need.. but I want to know if I'm doing it right and if not, how
to make it work correctly.
>
The program where this is going is checking a configuration text file
and puts all the lines in a 2d char array for comparison later. Here
is a simplified "extraction " of the code.
>
------
#include <stdio.h>
#include <strings.h>
>
int testfunct (char *myword)

testfunct takes as a parameter a character pointer. A pointer to a
character or array of characters depending on how you use it.

{
strcpy(myword," hello");

strcpy will copy into the location of the character pointer the contents of
the 2nd parameter.

>
return 0;
}
>
int main (int argc, char *argv[])
{
char word[100];

word is defined as an array of 100 characters. &word[0] is the address of
the first character, but so is word by itself. It gets silently converted.

>
strcpy(word,"we eee");

here word is silently converted to a char pointer, the same as &word[0]
which is where strcpy will start copying the contents of the 2nd parameter.

printf("WORD: %s\n", word);
testfunct(word) ;

here testfunct takes a char pointer so the compiler silently converts word
to a char pointer, the same as &word[0]

printf("WORD: %s\n", word);
>
return 0;
}
------

It is a well formed program.

Note, however, that you may wish to pass a 2nd parameter to your function,
the size of the buffer.

If you had done:
char word[2];
testfunct(word) ;

The compiler would accept it and when you ran it you would get a buffer
overflow because strcpy doesn't know how long the buffer the char pointer is
pointing to, it has no way to know. So it mindlessly copies until it
reaches the terminating null character in the copy from buffer, effectively
trying to stuff 10 pounds of stuff in a 5 pound bag.


--
Jim Langston
tazmaster@rocke tmail.com