Bi dimensional table <> double pointer
Hello,
I can't understand why this not works,
int tab1[15][20];
int tab2[20][15];
transpose(tab1,tab2);
void transpose (int **tab1, int **tab2) {
/* cannot convert parameters from 'int [15][20]' to 'int ** ' */
...
}
Because when I try to get an int of my two dimension table with a double pointer it works,
int tab[4][5] = {5,10};
printf("%d\n", **tab);
/* 5 */
printf("%d\n", *(*tab+1));
/* 10 */
If someone know why this, help !
I can't understand why this not works,
int tab1[15][20];
int tab2[20][15];
transpose(tab1,tab2);
void transpose (int **tab1, int **tab2) {
/* cannot convert parameters from 'int [15][20]' to 'int ** ' */
...
}
Because when I try to get an int of my two dimension table with a double pointer it works,
int tab[4][5] = {5,10};
printf("%d\n", **tab);
/* 5 */
printf("%d\n", *(*tab+1));
/* 10 */
If someone know why this, help !
ASKER
I know i can use this
void transpose (int tab1[15][20], int tab2[20][15]) {
or this
void transpose (int tab1[][20], int tab2[][15]) {
or this
void transpose (int (*tab1)[20], int (*tab2)[15]) {
but I don't understand why I can't use this one
void transpose (int **tab1, int **tab2) {
void transpose (int tab1[15][20], int tab2[20][15]) {
or this
void transpose (int tab1[][20], int tab2[][15]) {
or this
void transpose (int (*tab1)[20], int (*tab2)[15]) {
but I don't understand why I can't use this one
void transpose (int **tab1, int **tab2) {
ASKER CERTIFIED SOLUTION
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ASKER
Thanks, so once I put a signature with the number of columns
void transpose (int ptr1[][20], int ptr2[][15]) {
I can use
ptr2++
and
*(*ptr2+1)
*(*(ptr2+1))
Because now it knows how it has to be calculated because
he knows the constant size of the internal arrays
I am right ? :-)
void transpose (int ptr1[][20], int ptr2[][15]) {
I can use
ptr2++
and
*(*ptr2+1)
*(*(ptr2+1))
Because now it knows how it has to be calculated because
he knows the constant size of the internal arrays
I am right ? :-)
Hi,
the more general rule is that if a multidimensional array has to
be passed to a funtion, only the first dimension is free, rest of the dimensions
will have to be specified.
for example, consider a 2 * 3 array, this is how the elemnts are stored in memory,
[0,0] [0,1] [0,2][1,0][1,1][1,2] <---- array A;
if u pass to function as::
void transpose(p)
int **p;
{
..........
Imagine referencing A[1][2]; It wont be possible to reference this element through p
unless it is known that A[1][0] comes after A[0][2]. So u need to tell the second index
size.
hope this helps,
van_dy
the more general rule is that if a multidimensional array has to
be passed to a funtion, only the first dimension is free, rest of the dimensions
will have to be specified.
for example, consider a 2 * 3 array, this is how the elemnts are stored in memory,
[0,0] [0,1] [0,2][1,0][1,1][1,2] <---- array A;
if u pass to function as::
void transpose(p)
int **p;
{
..........
Imagine referencing A[1][2]; It wont be possible to reference this element through p
unless it is known that A[1][0] comes after A[0][2]. So u need to tell the second index
size.
hope this helps,
van_dy
well.... take care with this, as the number of columns of one is the number of rows of the other :-)
what you may want to do is also pass in the number of rows/columns expected, to ensure this is a valid call....
what you may want to do is also pass in the number of rows/columns expected, to ensure this is a valid call....
LOL, seems like it got answered b4 i clicked the submit button.
ASKER
Oki Thanks, i get it ;-)
(sorry van_dy lol)
(sorry van_dy lol)
Hi matthew016,
Two-dimensional array constants are "flat" arrays, whereas double-pointers are ... double pointers.
Or, a bit more technical:
if you have
int a[3][5];
int b[15];
&a[2][2]-&a
is the same as
&b[2*3+2]-&b
Cheers!
Stefan
Two-dimensional array constants are "flat" arrays, whereas double-pointers are ... double pointers.
Or, a bit more technical:
if you have
int a[3][5];
int b[15];
&a[2][2]-&a
is the same as
&b[2*3+2]-&b
Cheers!
Stefan
void transpose( int[][], int[][] );