new c programmer question

[John Gerrard]

Hi, my name is John and this is my first posting to comp.lang.c. I am
learning C. I am a high school student with interest in programming. I
have a copy of "The C programming language second edition by Brian W
Kernighan and Dennis M Ritchie". I am using gcc version 4.1 on SuSE 9.

Here is my simple question. I want to set a matrix to contain some
values in my program. I want to pass in the matrix array by address in
a function. The compiler is saying:

hello.c:15: warning: assignment from incompatible pointer type

I can see what is wrong but do not have the ability yet to fix it. Here
is the small code where the error is. I would appreciate some advice on
the syntax to use.

Thank you
John

void setupmatrix(int *m[])
{
int x[4][4]=
{
{ 1, 2, 3, 4},
{ 5, 6, 7, 8},
{ 9,10,11,12},
{13,14,15,16}
};
*m=x;
}

[Frederick Gotham]

Frederick Gotham posted:

system("PAUSE") ;

Sorry, I meant to remove that before posting.

--

Frederick Gotham

[Flash Gordon]

John Gerrard wrote:

Okay thank you for your prompt replies everyone. I had forgotton how
painfully flow learning a new language can be (and humbleing).

I am now compiling my code with

gcc hello.c -ansi -std=c99 -pedantic -Wall -o hello

<OT>
Add a -O which will allow it to detect some other warnings. In this case
it does not show anything, but sometimes it will.
</OT>

I have modified the code. It now compiles compiles but gives this
output at runtime.

jad@dhome2:~/matrix$ ./hello
setupmatrix
exit setupmatrix
printmatrix

Segmentation fault

<OT>
Since you are using Linux you might want to try valgrind. It can help a
lot with problems like this.
</OT>

jad@dhome2:~/matrix$
jad@dhome2:~/matrix$
I have included the COMPLETE program at the end of the posting.

Thank you. This makes it a lot easier you show you what you are doing wrong.

It
doesn't work but it does represent what I am trying to do. I sort of
know what is wrong but don't have the understanding to correct it yet.
This is what I want to do: 1) allocate memory for a n*n 2D array,

OK, you need to read question 6.16 of the comp.lang.c FAQ at
http://c-faq.com/ since it specifically addresses this problem. It
describes in a lot of detail, including some pictures, three methods of
achieving this.

2)
fill it with numbers, 3) print it out, 4) and free up the memory at the
end of it all. I want to allocate the array size at runtime as I will
be, next month at this rate, be multiplying 2-3 matrices together.
Although I would like to say I don't want you to do this for me, I'm
really hoping you don't take that too seriously because I need
something working to see where I should be going.

Thank you all for your help in advance.
John

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

/*setup a n*n matrix and fill it with 1, 2, 3, 4*/
/*============== =============== =============== ==*/
int** setupmatrix(int nSize)
{
printf("setupma trix\r\n");

Why the \r? Doing a \n on its own will take you to the start of the next
line.

int nByNArraySize=n Size*nSize*size of(int);
int *arr=malloc(nBy NArraySize);
for (int i=0;i<nByNArray Size;i++)

nByNArraySize is the number of *bytes* allocated, *not* how many ints
you have space for. It would be closer (but still not a 2D array) if you
did:
int nByNArraySize=n Size*nSize;
int *arr=malloc(nBy NArraySize * sizeof *arr);
for (int i=0;i<nByNArray Size;i++)

Note the way sizeof is being used in the malloc call. No need to specify
the type there, just that it is what arr points to.

arr[i]=0;
printf("exit setupmatrix\r\n ");
return (int **)arr;

Any time you are using a cast, your first thought should be that you are
doing something wrong. There are a few occasions where a cast is needed,
but far more situations where it is harmful because it stops the
compiler from telling you that you have an error. In this case you do
have a serious error and the cast just tells the compiler not to point
it out to you.

int** means pointer to pointer. I.e. if you have:
int **arr2d;
Then arr2d[0] is a *pointer* to a row. You have not set this to point
anywhere instead you have written the integer value 0. See the FAQ
question I pointed out for the correct ways to build your int** allocation.

}

<snip>

If you have problems understanding question 6.16 then tell us what you
don't understand and we will try to help you.

I also suggest you rest of section 6 and the other question I pointed
out in my previous post.
--
Flash Gordon.

[Barry Schwarz]

On Sun, 03 Sep 2006 18:11:30 GMT, Frederick Gotham
<fg*******@SPAM .comwrote:

>John Gerrard posted:

>void setupmatrix(int *m[])
{
int x[4][4]=
{
{ 1, 2, 3, 4},
{ 5, 6, 7, 8},
{ 9,10,11,12},
{13,14,15,16}
};
*m=x;
}

You can't assign one array to another (even if their dimensions are
identical), e.g.:

Since m is an array of pointers (actually it's an int** due the way
arrays are passed to functions), *m is not an array and your
statement, while correct, is not related to this code. *m has type
int* while x in this expression evaluates to type int (*)[4], pointer
to array of four int. The two types are incompatible and there is no
implicit conversion between them. That is the syntax error in the
statement.
Remove del for email

[Keith Thompson]

"osmium" <r1********@com cast.netwrites:
[...]

auto is a "storage class specifier" and if you do not provide one for a
variable, it will implicitly be treated as auto.

.... if and only if the variable is declared within a function
definition.

--
Keith Thompson (The_Other_Keit h) 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.

[Michael Mair]

lovecreatesbea. [e-mail] schrieb:

Michael Mair wrote:

> void setupmatrix(int *m)
or
void setupmatrix(int (*m)[4])

<snip>

> static int x[4][4]

<snip>

>>or even using
memcpy(&m[0], &x[0][0], sizeof x);
or
memcpy(&m[0][0], &x[0][0], sizeof x);

Will the second memcpy fail if the argument to 'void setupmatrix(int
(*m)[4])' is 'int (*m1)[4];'? Because the structures of m and x are
different. The elements in x are stored continuously while the elements
in m are not. And I think it works if the argument is 'int m2[4][4]'.
Am I right?

No.
With "int (*m1)[4]", you have a (hopefully valid) pointer to the
first of an yet unknown number of "arrays 4 of int" which are stored
contiguously, i.e. in memory you can expect
|(*m1)[0]|....|(*m1)[3]|(*(m1+1))[0]|....
or
|m1[0][0]|....|m1[0][3]|m1[1][0]|....

What you probably mean is "int **m1" or even an array of pointers
to arrays 4 of int, i.e. "int (**m1)[4]" or "int (*m1[4])[4]".
In these cases, m1[0] (or (*m1[0])) not necessarily points to
the same object as m1[1] (or (*m1[1])).

Cheers
Michael
--
E-Mail: Mine is an /at/ gmx /dot/ de address.