Computer graphics: DDA method to generate straight lines

The content of this article is quoted from "Computer Graphics Tutorial"

Overview of Numerical Differentiation

The numerical differentiation method is shown in the figure below:

the code

void DDA_line2(int x1, int y1, int x2, int y2)
{
    
    
	float increx, increy, x, y, length;
	int i;
	if (abs(x2 - x1) > abs(y2 - y1))
		length = abs(x2 - x1);
	else
		length = abs(y2 - y1);
	increx = (x2 - x1) / length;
	increy = (y2 - y1) / length;
	x = x1;
	y = y1;
	for (i = 1; i <= length; i++)
	{
    
    
		putpixel(int(x + 0.5), int(y + 0.5), RED);
		x += increx;
		y += increy;
	}
}

Notice

 The formula for calculating the slope of a straight line: K=(y2-y1)/(x2-x1), as we all know, can only be used for straight lines that are not perpendicular to the X axis, otherwise x2=x1 will cause an error of dividing by zero. So is it considered in the DDA algorithm? The answer is yes, test the DDA algorithm, DDA_line(10,0,10,100) will find that the vertical straight line is successfully generated, and no error is reported. Why is this?
 Generate a straight line (x1, y1)->(x2, y2), set condition 1 as: x1≠x2, and condition 2 as: y1≠y2, then in general, at least one of condition 1 and condition 2 is established. When x1=x2, y1 is not equal to y2, so abs(y2-y1)>abs(x2-x1), so it will choose to calculate the increment Δy in the y direction to cause the transformation of x, that is, k=(x2-x1) /(y2-y1), then y is +1 or -1 each time, while x remains unchanged. Vice versa, a line perpendicular to the x-axis or y-axis can be generated.
 Then when x1=x2, y1=y2, this special case will result in division by zero, but the length in the above code is of float type, dividing by length=0 will not report an error, but the type value will be: -nan (ind) . Since length=0, the loop body will not be executed once, so this point will not be generated.
 improve algorithm:

void DDA_line(int x1, int y1, int x2, int y2)
{
    
    

	float k, length;
	if (abs(x2 - x1) > abs(y2 - y1))
		length = abs(x2 - x1);
	else
		length = abs(y2 - y1);
	float dx = (x2 - x1) / length;
	float dy = (y2 - y1) / length;
	float x = x1, y = y1;
	putpixel(x, y, RED);
	for (int i = 1; i < length  ; i++)
	{
    
    
		x += dx;
		y += dy;
		putpixel(int(x+0.5), int(y+0.5), RED);
	}
}

 The improved algorithm above will always generate a starting point.