Printf, array of structures and pointers...

[Eddahbi Karim]

Hi,

I have a pointer *pos, an array of structure "valid_param_s[]" and a
member name.

(void) fprintf(stderr, "Invalid parameter %s\n"
"Type --help for more "
"informations\n"
,valid_params_s[*pos].name);

This code doesn't work, so I want to know how make it work :-).
I could create another temporary variable to store *pos and do :
(void) fprintf(stderr, "Invalid parameter %s\n"
"Type --help for more "
"informations\n"
,valid_params_s[variable].name);

But I would do it without temporary variable, if it's possible ;).

--
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ThE_TemPLaR

[T.M. Sommers]

Eddahbi Karim wrote:

Hi,

I have a pointer *pos, an array of structure "valid_param_s[]" and a
member name.

(void) fprintf(stderr, "Invalid parameter %s\n"
"Type --help for more "
"informations\n"
,valid_params_s[*pos].name);

Assuming pos is pointing to the array element that you want to print,
the last line should be:

,pos->name);

[Eddahbi Karim]

On Fri, 17 Oct 2003 09:16:41 GMT
"T.M. Sommers" <tm**@mail.ptd.net> wrote:

Assuming pos is pointing to the array element that you want to print,
the last line should be:

,pos->name);

pos is the position in the array.

Maybe http://nopaste.php-q.net/17283 would help you ?
Line 153 ;)

(I prefer to paste a link, because there's no need to flood here :))

--
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ThE_TemPLaR

[Thes]

Eddahbi Karim wrote:

Hi,

I have a pointer *pos, an array of structure "valid_param_s[]" and a
member name.

(void) fprintf(stderr, "Invalid parameter %s\n"
"Type --help for more "
"informations\n"
,valid_params_s[*pos].name);

This code doesn't work, so I want to know how make it work :-).
I could create another temporary variable to store *pos and do :
(void) fprintf(stderr, "Invalid parameter %s\n"
"Type --help for more "
"informations\n"
,valid_params_s[variable].name);

But I would do it without temporary variable, if it's possible ;).

If pos is a pointer to a particular member of your valid_param_s[]
array, then just access the member directly through the pointer:

(void) fprintf(stderr, "Invalid parameter %s\n"
"Type --help for more "
"informations\n"
,pos->name);

An array subscript (that is inside the []'s) must be an integer
expression - you can't use the pointer like that.

[Eddahbi Karim]

On Fri, 17 Oct 2003 10:28:22 +0100
Thes <no*@ctually.myaddress.com> wrote:

If pos is a pointer to a particular member of your valid_param_s[]
array, then just access the member directly through the pointer:

(void) fprintf(stderr, "Invalid parameter %s\n"
"Type --help for more "
"informations\n"
,pos->name);

An array subscript (that is inside the []'s) must be an integer
expression - you can't use the pointer like that.

Ok, So I'll do without pointers ;)

Thank you.
--
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ThE_TemPLaR

[Ed Morton]

Eddahbi Karim wrote:

On Fri, 17 Oct 2003 10:28:22 +0100
Thes <no*@ctually.myaddress.com> wrote:

If pos is a pointer to a particular member of your valid_param_s[]
array, then just access the member directly through the pointer:

(void) fprintf(stderr, "Invalid parameter %s\n"
"Type --help for more "
"informations\n"
,pos->name);

An array subscript (that is inside the []'s) must be an integer
expression - you can't use the pointer like that.
Ok, So I'll do without pointers ;)

No need. The code segment you posted was fine, it just wasn't clear that
*pos was an integer.

I looked at your code posting at http://nopaste.php-q.net/17283 and the
line you say is a problem does not match the code you posted in your
article. The code at line 153 on your web page DOES have a problem. It is:

(void) fprintf(stderr, "Invalid parameter %s\n"
"Type --help for more "
"informations\n"
,valid_params_s.name);
^NOTE!

i.e. it's missing the index ([*pos]) to valid_params_s.

Ed.
Thank you.

[Bertrand Mollinier Toublet]

Eddahbi Karim wrote:

On Fri, 17 Oct 2003 10:28:22 +0100
Thes <no*@ctually.myaddress.com> wrote:

If pos is a pointer to a particular member of your valid_param_s[]
array, then just access the member directly through the pointer:

(void) fprintf(stderr, "Invalid parameter %s\n"
"Type --help for more "
"informations\n"
,pos->name);

An array subscript (that is inside the []'s) must be an integer
expression - you can't use the pointer like that.

Ok, So I'll do without pointers ;)

For us to find out what is wrong with your code, we need to see a
definition of all the involved elements. In particular, we do not have a
definition of struct parameters. You don't tell us what exactly your
problem is either (does not compiler ? gives you a warning ? crashes at
runtime ? demons flying out of your nose ? random cities blasted to
smitherens ?). Be more specific and we should be able to help you...

Were you tired of your co-nationals that you prefered c.l.c to f.c.l.c :-) ?

--
Bertrand Mollinier Toublet
int main(){char*strchr();int j=1234;char t[]=":@abcdefghij-lmnopqrstuv"
"wxyz.\n",*i="iqgbgxmbbla.llsvoaz:zdxylaxejivnidhd @ttopnjeftuh-i";while
(*i){j+=strchr(t,*i++)-t;j%=sizeof t-1;putchar(t[j]);}return 0;}

[Eddahbi Karim]

On Fri, 17 Oct 2003 08:05:51 -0700
Bertrand Mollinier Toublet
<be*****************************@enst-bretagne.fr> wrote:

For us to find out what is wrong with your code, we need to see a
definition of all the involved elements. In particular, we do not have
a definition of struct parameters. You don't tell us what exactly your

problem is either (does not compiler ? gives you a warning ? crashes
at runtime ? demons flying out of your nose ? random cities blasted to

smitherens ?). Be more specific and we should be able to help you...

Ok ok, I'll do it next time :)
Were you tired of your co-nationals that you prefered c.l.c to f.c.l.c
:-) ?

Not really :-D
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ThE_TemPLaR