Darnell:
I don't understand your solution. Can you explain how you can reference an element in the double_array in your function in the format double_array[idx1][idx2] when you passed it as a double **?
Here's what I understand about the way C stores two-dimension arrays in memory.
At a low level, each location in computer memory has only one address, not a row,column address. C stores the elements of a two dimension array in memory as if the array was a one dimension array, starting with the elements of row 0, and then following them with the elements of row 1, etc.
So if I create a two-dimension array with the following data:
double my2d_array[2][5] = {
{1, 2, 3, 4, 5}, // row 0
{11, 12, 13, 14, 15} // row 1
};
C stores the data in a linear fashion:
1
2
3
4
5
11
12
13
14
15
There is no marker in memory to show where one row stops and the next one begins. C knows where one row stops only because you tell it how many columns are in each row.
If you tell C that you want my2d_array[1][3], it calculates the address by adding the starting address of my2d_array, the desired row number (1) times the number of elements in a row (5) and the column number.
So the address of my2d_array[1][3] = *(my2d_array + 1*5 + 3).
The way C knows how to address the elements of a two-dimension array by the array[row][column] format is by knowing how many elements are in each row. When you pass a 2d array as a double**, it doesn't know how many columns are in each row, so it doesn't know where to go when you show a [row][column] format.
I'd suggest you flatten you 2d array by passing it as a double * (instead of a double **). Since C has no problem passing a 1d array to a function, I'd also suggest you pass your 1d array as a double array ([]) instead as a double*.
Another thing I don't understand about your solution is the following statement:
conv_data[q][u]= temp_data[q][u];
Since your 2d arrays are dimensioned to [q][u], element [q][u] is out of the array bounds. You statement above will compile with no errors, but you will get a run-time error trying to access non-existant array elements. Did you run your program to prove you solution?
Here's how I would suggest reworking your code.
// start code sample
#include <stdio.h>
long f_ConvDblToSingleArray ( double* double_array, double single_array[] , int num_rows, int num_columns);
int main(int argc, char* argv[])
{
int q=4,u=1000;
int i, j;
double temp_data[4][1000];
double conv_data[4][1000];
double meas[4000];
// create some test data for demonstration purposes
for (i=0, j=0; i<q; i++, j+= u/q)
{
if (j > u)
j=u-1;
temp_data[i][j] = i+j;
}
for (i=0, j=0; i<q; i++, j+= 5)
{
if (j > u)
j=u-1;
conv_data[i][j] = i*j;
}
// put a breakpoint on the next line to see before and after
f_ConvDblToSingleArray( (double*)(temp_data), meas, q, u );
// put a breakpoint on the next line to see before and after
f_ConvDblToSingleArray( (double*)(conv_data), meas, q, u );
return 0;
}
//Library Function...(This compiles...)
long f_ConvDblToSingleArray ( double* double_array, double single_array[] , int num_rows, int num_columns)
{
long error = 0;
int idx1 = 0;
int idx2 = 0;
int index = 0;
for (idx1=0; idx1 < num_rows; idx1++)
{
for ( idx2=0; idx2 < num_columns; idx2++ )
{
// create the equivalent of double_array[idx1][idx2] using pointer math
single_array[index++] = *(double_array + (idx1*num_columns) + idx2);
}
}
return 0;
}
// end code sample
Finally, I'm surprised that you were disappointed that we didn't give you quick credit for your solution.
You post a dubious solution that could not have run without errors and expect us to congratulate you on your programming expertise for using something "alot of people dont know"? Well there are a lot of smart people in other fields who haven't heard of C, but don't ask us to congratulate you for knowing something they don't.
You have not given us credit for the hours we have spent helping you, but you would like some credit for the few lines of questionable code you posted.
Do you realize that most of the people helping you are volunteering their time and don't work for National Instruments? We have our own jobs and don't have time to just sit and wait for your next question. We would all appreciate a little more courtesy, a little more appreciation, and a little more effort on your part.
