Is there a more clever way of writing this piece of code? It does work but I thought there might be a better way of returning a matrix from multiplying the parameter by itself.
public static void Main(string[] args)
{
int[,] m1 = CreateMatrix(5);
}
static int[,] CreateMatrix(int rowscols)
{
int[,] result = new int[rowscols, rowscols];
int res = rowscols * rowscols;
int counter= 0;
for (int i = 0; i < rowscols; ++i)
{
for (int k = 0; k < rowscols; ++k)
{
result[i, k] = (res - counter);
counter += 1;
}
}
return result;
}
1 Answer 1
rowscols
is not a very good name.
Although on one hand it's "accurate",
on another it's kind of awkward.
How about dimension
or dim
instead?
res
is also not a very good name.
Since its value is dim * dim
,
it's the number of cells in the matrix, so cellCount
would be better.
It would be better to move the counter
variable inside the loop,
by declaring it in the outer loop together with i
,
and incrementing it in the innermost loop together with k
.
It's a common practice to name loop variables i
, j
, k
.
You skipped j
and went directly for k
.
It's not a "problem",
but j
would be more natural,
especially because at school (I think) they teach matrices typically with \$m_{ij}\$ notation rather than \$m_{ik}\$
Lastly,
since the result of the function is a matrix,
it will be more meaningful to call the variable matrix
instead of result
.
Suggested implementation
Putting the above suggestions together:
static int[,] CreateMatrix(int dimension)
{
var matrix = new int[dimension, dimension];
var cellCount = dimension * dimension;
for (int i = 0, counter = 0; i < dimension; ++i)
{
for (int j = 0; j < dimension; ++j, ++counter)
{
matrix[i, j] = cellCount - counter;
}
}
return matrix;
}
Special thanks to @Abbas for pointing out to use the var
keyword instead of explicitly declaring the matrix
and cellCount
variables.
-
1\$\begingroup\$ +1, very nice answer. I'd only like to add two things. 1. Use the
var
keyword instead of explicitly declaring your variables.dim
is as meaningful asrowscols
. Ok, a but better but still... I would strongly preferdimension
. \$\endgroup\$Abbas– Abbas2014年10月30日 20:52:58 +00:00Commented Oct 30, 2014 at 20:52 -
1\$\begingroup\$ Thanks @Abbas, I agree with your points and updated my answer accordingly. \$\endgroup\$janos– janos2014年10月30日 22:09:18 +00:00Commented Oct 30, 2014 at 22:09