http://www.neilstuff.com/guide_to_cpp/notes/Multi%20Dimension%20Arrays%20and%20Pointer%20Pointers.htm
According to this site, I should be able to use the following code:
double stuff[3][3];
double **p_stuff;
p_stuff = stuff;
But I get a complaint that the conversion is not allowed by assignment.
Am I doing something wrong?
I have an extern "C" type function that I want to pass this double stuff[3][3] to. So I think i need to make it a pointer, right?
Regarding the edit: to pass this double stuff[3][3]
to a C function, you could
1) pass a pointer to the whole 2D array:
void dostuff(double (*a)[3][3])
{
// access them as (*a)[0][0] .. (*a)[2][2]
}
int main()
{
double stuff[3][3];
double (*p_stuff)[3][3] = &stuff;
dostuff(p_stuff);
}
2) pass a pointer to the first 1D array (first row) and the number of rows
void dostuff(double a[][3], int rows)
{
// access them as a[0][0] .. a[2][2]
}
int main()
{
double stuff[3][3];
double (*p_stuff)[3] = stuff;
dostuff(p_stuff, 3);
}
3) pass a pointer to the first value in the first row and the number of both columns and rows
void dostuff(double a[], int rows, int cols)
{
// access them as a[0] .. a[8];
}
int main()
{
double stuff[3][3];
double *p_stuff = stuff[0];
dostuff(p_stuff, 3, 3);
}
(that this last option is not strictly standards-compliant since it advances a pointer to an element of a 1D array (the first row) past the end of that array)
If that wasn't a C function, there'd be a few more options!
Your assigned is flawed. p_stuff;
is pointer to pointer to double
whereas stuff
is two dimensional array( array of arrays)
A single dimension array decays to the pointer to its first element. A 2 dimensional array decays to pointer to a single dimension array.
Try this
double stuff[3][3];
double (*p_stuff)[3]; // pointer to array of 3 int
p_stuff = stuff;
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