pointer to pointer as parameter of func

Question

 void print_first_n_row(double **matrix, int n, int row) {
   int i,j;
   double x;
   for(i=0;i<n;i++){
     for(j=0;j<row;j++){
       x=*(*(matrix+i)+j);
       printf("%lf",x); 
    }
    printf("\n");
  } 
}

I am not getting output with this func. What can be reason?

Answer 1

In C, arrays are arranged in memory consecutively, row by row, referred to as row major for example, an array created like this:

int array[4][3] = {{1,2,3},{4,5,6},{7,8,9},{10,11,12}};//4 rows of 3 columns

...is stored in memory as:

|1|2|3|4|5|6|7|8|9|10|11|12|
|row 1|row 2|row 3| row 4  |

So your function, as it is written , should be passing n as the number of rows to print in the second argument (as it does), but pass how many columns there are in the 3rd argument.

Change this:

void print_first_n_row(double **matrix, int n, int row)  

to this:

void print_first_n_row(double **matrix, int n, int col)    

I tested this change by creating [4][3] index accessible memory locations using allocated memory for: double **matrix; (4 pointers to double, each pointing to 3 double locations.) then populated each location with a series of numbers,and passed appropriate arguments:

int n = 4;
int col = 3;
void print_first_n_row(matrix, n, col);  

I saw this:

When passed:

int n = 2;
int col = 3;
void print_first_n_row(matrix, n, col);  

I saw this:

One other suggestion, as stated in comments:

This:

x = matrix[i][j];  

Is much more readable than:

x=*(*(matrix+i)+j);

Answer 2

"I am not getting output."

If you ain´t got output, either n or row (or even both) shall be 0 or smaller than 0 because if it would be different you shall get at least one output.

Check the values you pass as arguments to n and row in the calling function.

My personally assumption is that the newline form printf("\n"); is getting printed, but you didn´t noticed the output of the line feed.