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Returning structs containing arrays

P: n/a
It's the first time I try using structs, and I'm getting confused with
it and can't make it work properly

I firstly define the structure by this : typedef struct { char *l1; int
*l2; int Nval; } *arrays;

It's supposed to be a structure containing an array of chars, an array
of ints and an int.

I declare functions like this : arrays *parseline(char *line, int N)

and I point to the arrays I want to return like this :

out.l1=displayln;
out.l2=sumline;
out.Nval=Nval;
return out;

When compiling it gives me an error, such as "error: request of member
<< l1 >> in something that's not a structure or an union". What's wrong
with the way I do it and how should I fix it?

Nov 20 '05 #1
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3 Replies


P: n/a
Michel Rouzic wrote:

It's the first time I try using structs,
and I'm getting confused with
it and can't make it work properly

I firstly define the structure by this :
typedef struct { char *l1; int *l2; int Nval; } *arrays;

It's supposed to be a structure containing an array of chars,
an array of ints and an int. I declare functions like this : arrays *parseline(char *line, int N) out.l1=displayln;
out.l2=sumline;
out.Nval=Nval;
return out; and how should I fix it?


struct arrays {
char c_array[l1];
int i_array[l2];
int Nval;
};

#include <string.h>

struct arrays out;

memcpy(out.c_array, displayln, sizeof out.c_array);
memcpy(out.i_array, sumline, sizeof out.i_array);
out.Nval = Nval;
return out;
.... or sizeof the source object, whichever is smaller.
If displayln is the name of a smaller array then

memcpy(out.c_array, displayln, sizeof displayln);

If displayln is the name of a pointer
to the first element of a smaller array
then you have figure something out.
Likewise for out.i_array.

--
pete
Nov 20 '05 #2

P: n/a
Michel Rouzic wrote:
It's the first time I try using structs, and I'm getting confused with
it and can't make it work properly

I firstly define the structure by this : typedef struct { char *l1; int
*l2; int Nval; } *arrays; ^^^
The * says pointer, why put it there?

Also, l1 and l2 are pore choices for names, since they look too much
like 11 and 12.
It's supposed to be a structure containing an array of chars, an array
of ints and an int.
Well, what you have specified is pointer to char and pointer to int. I
suggest you reread what your text books say about arrays, pointers and
declarations of them.
I declare functions like this : arrays *parseline(char *line, int N)

and I point to the arrays I want to return like this :

out.l1=displayln;
out.l2=sumline;
out.Nval=Nval;
return out;

When compiling it gives me an error, such as "error: request of member
<< l1 >> in something that's not a structure or an union". What's wrong
with the way I do it and how should I fix it?


Quick answer, remove the extraneous *. Although I'll bet that leaves you
with lots of other problems. See my comments above.
--
Flash Gordon
Living in interesting times.
Although my email address says spam, it is real and I read it.
Nov 20 '05 #3

P: n/a
"Michel Rouzic" <Mi********@yahoo.fr> writes:
It's the first time I try using structs, and I'm getting confused with
it and can't make it work properly

I firstly define the structure by this : typedef struct { char *l1; int
*l2; int Nval; } *arrays;
Which is much more legible like this:

typedef struct {
char *l1;
int *l2;
int Nval;
} *arrays;
It's supposed to be a structure containing an array of chars, an array
of ints and an int.
Your structure type doesn't *contain* an array of anything. It
contains two pointers and an integer. The pointers might point to
arrays if you initialize them to point to allocated memory.

The typedef "arrays" doesn't name a structure type; it names a
pointer-to-structure type. You haven't provided a name for the
structure type.

You should consider dropping the typedef altogether and just using
the structure tag directly, something like:

struct arrays {
char *l1;
int *l2;
int Nval;
};

You would then refer to the type as "struct arrays" rather than
"arrays". The typedef saves you the effort of typing "struct", but
that's really not much of a benefit; your code is clearer if it's
obvious that the type is a structure. (There are rare cases where you
might want to hide the nature of the type; this isn't likely to be one
of them.)

You should also choose better names. Flash Gordon has already
mentioned that l1 and l2 are easily confused with 11 and 12. The name
"arrays" could also be improved. Using a plural name for a single
object causes confusion; would an array of them be called "arrayses"?
And it's a structure, not an array. Pick a name that reflects what
it's used for, perhaps "struct line" or "struct line_info".
I declare functions like this : arrays *parseline(char *line, int N)
Given your original declaration, that would return a
pointer-to-pointer-to-structure. You seem to be throwing in '*'s
almost at random. Don't make something a pointer unless you have a
specific reason. (In fact, don't do *anything* unless you have a
specific reason.)

You can just return a structure directly:
arrays parseline(char *line, int N);
or
struct line_info parse_line(char *line, int n);
All-caps names are usually used for macros. Underscores are usually
used to separate words in identifiers; "parseline" is too easy to read
as a nonexistent 3-syllable word.
and I point to the arrays I want to return like this :

out.l1=displayln;
out.l2=sumline;
out.Nval=Nval;
return out;


struct line_info parse_line(char *line, int n)
{
struct line_info result;
...
result.l1 = something;
result.l2 = something_else;
result.Nval = something_else_again;
...
return result;
}

--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.
Nov 20 '05 #4

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