2

I wrote a function that returns a pointer to a 1D array. The function prototype is:

double* NameFunction (int,double*,double*)

The function seems to work fine with 1D array.

Here's the question: I'd like to use the function to fill one row of a 2D array.

How do I write in the calling function a pointer to the row of the 2D array to fill?

Can I use the same pointer structure to indicate a row of a 2D array as argument?

Lucky Murari
  • 12,672
  • 5
  • 22
  • 43
Giovanni
  • 33
  • 1
  • 5

3 Answers3

2

Yes just pass the name of the 2D array with the first index filled in:

NameFunction (12121212,&array[0],some pointer here) // for example if you want to pass the first row
Pepe
  • 6,360
  • 5
  • 27
  • 29
1

A variable pointing to an array in C always contains just an address location i.e., the pointer to the starting of the array. So, whatever the array type, it just needs a pointer which is a *.So this function should work as it is now.

But you may need to change some programming logic.

Lucky Murari
  • 12,672
  • 5
  • 22
  • 43
0

You can take double foo[20] as 1×20 array, or you can take it as two-dimensional 4×5 array. Then you have the coordinates like this:

foo    foo+1  foo+2  foo+3  foo+4
foo+5  foo+6  foo+7  foo+8  foo+9
foo+10 foo+11 foo+12 foo+13 foo+14
foo+15 foo+16 foo+17 foo+18 foo+19

So if you have a function that returns double *, you can pass it (or make it return) foo + 5*n to point at row n.

#define ROW_LEN 5
#define ROWS 4

void fillRow(double * row)
{
     int i;
     for (i = 0; i < ROW_LEN; i++)
     {
          row[i] = 12;
          (row + i) = 12; // this is the same thing written differently
     }
}

double arr[ROW_LEN * ROWS];
fillRow(arr + 2 * ROW_LEN); // fill third row

Note that C does not do any range checking at all, so if you are not careful and accidentally to something like arr[627] = 553 somewhere in your code, it's going to blindly overwrite everything that is at the computed address in the memory.

che
  • 12,097
  • 7
  • 42
  • 71