Re: How to define a pointer to multiple-subscripted arrays

Adrian Hawryluk <adrian.hawryluk-at-gmail.com@xxxxxxxxxx> writes:

Zhou Yan wrote:
Yes, by this changing, it work properly. But I still do not
understand. Could you kindly give a brief explanation?

Also I tried to print all the 4 numbers and I followed this way, I
find I also need:

nPtr[ 0 ] = array[ 0 ];

and I checked array[ 0 ] is just the same as &array[ 0 ][ 0 ] by

printf( "%p\n%p\n", array[ 0 ], &array[ 0 ][ 0 ] );
printf( "%p\n%p\n", array[ 1 ], &array[ 1 ][ 0 ] );

Could you please kindly give me a explanation?
What is array[ 0 ] actually is? Is it an array which contain the the
first row of this array? Or is it a pointer, with the address of array[
0 ][ 0 ]. Or internally it is both?

The full explanation is below.

If I define such an array of pointers, nPtr[ 2 ], I need to initialize
all the pointers in this array,right?

Right.

Is it possible to define a single array which can point to any
address of a 2-D array?

Well, you got one answer on how to get a segment of the array, now for
the full explanation:

An 2D array (as you have allocated) is not an array of pointers to an
array of elements, it is in fact a bunch of elements allocated in one
chunk.

array is of type int[][2] and is convertible to a pointer to an array of
two elements (int (*)[2]). You can declare a variables of this type
like this:

// Get from zeroth 'row' on down
int (*var0)[2] = array; // or
int (*var0)[2] = array+0; // or
int (*var0)[2] = &array[0];
// Get from first 'row' on down
int (*var1)[2] = array+1; // or
int (*var1)[2] = &array[1];

Note: not int *var[2] as this is an array of int pointers of two
elements. Note also that you cannot cast to or define this type this
directly, something in the language/parser prevents this. To
cast/return, you need to generate a typedef like so:

typedef (*ptrToTwoElementArray_t)[2];

(something that I just figured out now, thanks to you. That has
been bugging me forever, as I've never been able to return or cast
to something like this. Thanks very much!)

array[0] is of type int[] and is convertible to a pointer to an array of
elements (int *). You can also declared a variable of this type
like this:

// Get zeroth 'row'
int *var0 = *(array);
int *var0 = *(array+0);
int *var0 = array[0];
// Get first 'row'
int *var1 = *(array);
int *var1 = *(array+1);
int *var1 = array[1];

This is why your code:

printf( "%p\n%p\n", array[ 0 ], &array[ 0 ][ 0 ] );
printf( "%p\n%p\n", array[ 1 ], &array[ 1 ][ 0 ] );

gave you the results that you found.

array[0][0] is of type int and is convertible to an int. You can also
declare a variable of this type like this:

int var = array[0][0];

This should be obvious.

When you dereference an array (declared as you described) using the []
operators, the compiler knows that this is not an array of pointers to
an array of ints, but that it is a big blob of ints and that it can
calculate the offset to the ints via the formula x+y*MaxX.

(I need to ask such silly quetions...but I really know nothing and
cannot find a book talk about these)

Don't think your question silly. I've been trying to figure this out
for 20 years. :) The only silly question is the unasked one. I
certainly learned something from answering this question, I hope that
I explained it sufficiently for you to too.


Adrian

Thank you very much for you explanation. I will print this message
out to read a few times to get the whole idea...

Thanks!
.

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