Calculating points of a line

I am writing a function that takes two coordinates, then calculates the points in between and draw the line.
The coordinates represent points in my array, the array is viewed by a print comand that prints it into a grid in console.
The function works,
output on
0,2 -> 3,9 in a 10 by 10 grid would look like


0000000000
0000000000
1000000000
1000000000
0100000000
0100000000
0100000000
0010000000
0010000000
0001000000

I'm wondering if i am doing this properly, or if there would be a better way to do it.
Links to similar functions or the code to them would help, trying to learn how best to do this

my code- i commented as best i could

Code:

void Line(int xs, int ys, int xe, int ye, int vl, int* array)
{
	// first calculate slope using  rise/run
	float oy; // store last y coord		
	bool first = true; //  to handle first time in loop

	if (xe != xs) // can't have a slope of infinity
	{

		float s =  ((float)ye - (float)ys) / ((float)xe - (float)xs); // slope
		float b = ys - ((float) xs * s); // the B in the equation y = ax + b
		
		for( float c = float(xs); c <= (float)xe; c++) // drawing loop
		{
			float y = (c * s) + b;  // calculate the y
			if ( (y + .5) >= ceil(y) ) // means it was  x.5 or greater
			{
				y = ceil(y); //  round up
			}
			else
			{
				y = floor(y); // its less then x.5  so  round down
			}
			if (first) // first time in the loop
			{
				write(c, (int)y, vl, array);
				oy = y;
				first = false;
			}
			else // not first time in loop
			{
				for( int ty = oy ; ty < y; ty++)  // draw from last coord, draw line towards  next y, stop 1 before
				{
					write(c - 1, ty, vl, array);
				}
				oy = y;
			}

			
		}
		// last  point, should be xe, ye
		write(xe, ye, vl, array);


	}
}

Thanks for your time

If you stick with integers, you wouldnt have to worry about rounding the y, and I think you'll get pretty good results.

Code:

if (xe != xs)

if this fails, won't you still want to draw a straight line up and down?

why are you checking for the first iteration of the loop?

Are you trying to implement this
http://www.google.com/search?q=bresenham+line+algorithm
or something of your own?

Something of my own, but i had read up on Bresenham algo about 6 months ago, so you could say it was inspire.

First iteration of loop thing is gone now, it was just a solution to a problem bu i found a better way.
This work, draws the line way i figured it would. But i'll read s'more on Bresenham's algo to see how i can improve mine.
Thanks