Thread: Rofl where did I go wrong ? Need Help Plz !

up readup on the usage of printf: it should be

Code:

printf("%f\n",A[i][j]);

cant remember if its %f or somethin else but just have a quick read and you'll be able to find it all, I know float is %f
printf(A[i][j]"\t");
actually thats not the only problem, you are accessing **** too high proram should be somethin like:

Code:

void main(void) 
{ 
  int i,j; 
  double x,y; 
  double A[2][3]; 
  for(i=0;i<2;i++) 
  { 
    for(j=0;j<3;j++) 
    { 
      A[i][j] = 10; 
      printf(A[i][j]"\t"); 
    } 
    printf("\n") 
  } 
  getch(); 
}

because doing:
double A[2][3]; means using index's:
A[0][0];
A[0][1];
A[0][2];
A[1][0];
A[1][1];
A[1][2];

double A[2][3]; means using index's:
A[0][0];
A[0][1];
A[0][2];
A[1][0];
A[1][1];
A[1][2];

I though this would have an array of 3x4 what your saying is my array is now 2x3 ?

When you declare an array like such..

Code:

double A[2][3];

you are declaring an array that has 2 X 3 elements, however when you index those elements it will be indexed as such:

Code:

A[0][0]
A[0][1]
A[0][2]
A[1][0]
A[1][1]
A[1][2]

the numbers you specify when declaring the array is the exact number you can use so as you have
double A[2][3];
it means you have 2 rows, and 3 colums or 6 cells, now when you want to access a cell you get it and minus 1, so if you want to manipulate the first cell you would have:

Code:

A[0][0] = whatever;

may take a little to get used to it but you'll pick it up pretty easy, become second nature soon.

Originally posted by Halo
double A[2][3]; means using index's:
A[0][0];
A[0][1];
A[0][2];
A[1][0];
A[1][1];
A[1][2];

I though this would have an array of 3x4 what your saying is my array is now 2x3 ?

Huh? Are you a Visual Basic programmer or something like that?