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I'm confused about 2 dimensional arrays and there memory storage and
how to handle them using pointers; for example below something is
wrong but can't spot why?

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

double **test ()
{
double **arr =malloc(100);
arr[0][0] = 5.6;
arr[0][1]=17.4;
return arr;
}

int main ()
{
double **myarray;
myarray=test();
printf("%f -- %f", myarray[0][0],myarray[1][1]);
}

// result: 5.600000 -- 17.400000 but expected to give error of
myarray[1][1] not 17.400000

On 02/24/2011 11:35 AM, Francogrex wrote:
> I'm confused about 2 dimensional arrays and there memory storage and
> how to handle them using pointers;

Have you read section 6 of the CLC FAQ? <http://c-faq.com>
You probably should - especially question 6.16.

Francogrex <> writes:

> I'm confused about 2 dimensional arrays and there memory storage and
> how to handle them using pointers; for example below something is
> wrong but can't spot why?
>
> #include <stdio.h>
> #include <stdlib.h>
>
> double **test ()
> {
> double **arr =malloc(100);
> arr[0][0] = 5.6;
> arr[0][1]=17.4;
> return arr;

Arrays are not pointers and pointers are not arrays. Your malloc line
allocates 100 bytes (that's odd in itself, but you're the boss) and
points to it with a pointer that will interpret those 100 bytes as a
sequence of pointers to doubles. Until you set those pointers to point
to something valid, you can't use arr[0][0] at all.

Your best bet is to start with the FAQ. It has a summary of how 2D
arrays can be allocated and used in C:
http://c-faq.com/aryptr/dynmuldimary.html

The FAQ is a little old, and if you have access to a compiler that
supports what are called variably modified arrays, you can often write
much neater 2D manipulation functions.

> }
>
> int main ()
> {
> double **myarray;
> myarray=test();
> printf("%f -- %f", myarray[0][0],myarray[1][1]);
> }
>
> // result: 5.600000 -- 17.400000 but expected to give error of
> myarray[1][1] not 17.400000

--
Ben.

pete <> writes:

> Francogrex wrote:
>> I'm confused about 2 dimensional arrays and there memory storage and
>> how to handle them using pointers; for example below something is
>> wrong but can't spot why?
>>
>> #include <stdio.h>
>> #include <stdlib.h>
>>
>> double **test ()
>> {
>> double **arr =malloc(100);
>> arr[0][0] = 5.6;
>> arr[0][1]=17.4;
>> return arr;
>> }
>
> &(arr[0][1]) - &(arr[0][0]) is equal to one;

No, both &arr[0][1] and &arr[0][0] are undefined. The & and the *
implied by the []s cancel and your get arr[0]+1 - arr[0] and arr[0] is
not been given a value at any point in the program.

I suspect you intended to describe what should be the case in a proper
2D array but the declaration of 'arr' makes it a pointer to the start of
a plain 1D array.

> but there is no way to tell from the above code
> how much &(arr[1][0]) - &(arr[0][0]) is supposed to be,
> which is something that the compiler needs to know
> in order to implement arr as a two dimensional array.

--
Ben.