Thread: rotate line

You'll have to describe in a bit more detail what you mean. Do you want to rotate a line once it has already been created? What sort of user interface would you like to have? You could, for example, just let the user drag one of the endpoints of the line around to in effect rotate it.

If you want to do actual rotation, do you want to rotate about one of the endpoints, or the centre of the line, or what?

Describe precisely what it means to rotate a line. Apart from the begin/end coordinates, that description will also need to represent what are you rotating about and a measure of rotation. Once you've worked that logic out, write a function to implement it.

I cannot tell from your description: are you intending on rotating just the line or the bitmap that the line is on?

You can't just "rotate 90 degrees" without rotating about some fixed point. If you have a line then I can think of two ways you might want to rotate it: you might want to rotate the line about (or around or whatever you want to call it) one of its end points, or you might want to rotate the line about the centre of the line (the point on the line that is equidistant from both endpoints).

Rotating 90 degrees clockwise is pretty easy. Suppose your line has width x and height y. Then a line which is rotated 90 degrees just has width -y and height x. You can get this by taking the following 2D rotation formulae and plugging in theta = 90 degrees: Transformation matrix - Wikipedia, the free encyclopedia
That page describes it in terms of matrices, but all you really have to know are the formulae for x' and y'.

If you're actually calling sin and cos from C, then don't pass 90 -- these functions take radians, not degrees. You can convert degrees to radians by using a function like

Code:

double deg2rad(double deg) {
    return deg * PI * 180;
}

where PI can be defined as M_PI if your math.h has it or as the literal

Code:

#define PI 3.14159265358979324

Note that this will rotate the line about the origin (0, 0). If you want to rotate the line about, say, the endpoint (x,y), then translate the endpoints of the line by (-x,-y), perform the rotation, and then translate back again by (x,y).

No, no . . . When I say translate, I mean take your object and move it by that offset. Here's how you probably want to do it.

Code:

    // This code rotates the line by 90 degrees clockwise about the point start
    // We want (x,y) to be the location of the end endpoint, and start to be moved to the origin
    // First, set (x,y) to the end
    x = copy->rect->endX;
    y = copy->rect->endY;

    // Now translate by (-startX, -startY)
    x -= copy->rect->startX;
    y -= copy->rect->startY;

    // Next perform the rotation (you seem to have not copied the formulae correctly, I got this from the Wikipedia page, more or less)
    x = (x * cos(deg90InRad)) - (y * sin(deg90InRad));
    y = (x * sin(deg90InRad)) + (y * cos(deg90InRad);
  
    // Now translate back, by (startX, startY)
    x += copy->rect->startX;
    y += copy->rect->startY;

At the end of this, one of the endpoints of your rotated line will be (startX, startY) -- since you rotated about that point, it doesn't move anywhere. And the other endpoint of the line segment will be (x, y).

[edit] You seem to have gotten it working before I typed this up -- good work. You should take some graphics courses sometime. [/edit]

What's the value of deg90InRad? If you replace it with the hard-coded value (3.14159 / 2), what happens?

There was a problem with my pseudo-code -- I was replacing the value of x before it was used in calculating y in the second line involving sin and cos. Oops. But it actually works if you adapt it a little:

Code:

$ cat rotate.c
#include <stdio.h>
#include <math.h>

static const double deg90InRad = -3.14159265358979324 / 2;

int main(int argc, char *argv[]) {
    double startX = 10.0, startY = 10.0;
    double endX = 15.0, endY = 15.0;
    double x, y;

    // This code rotates the line by 90 degrees clockwise about the point start
    // We want (x,y) to be the location of the end endpoint, and start to be moved to the origin
    // First, set (x,y) to the end
    x = endX;
    y = endY;

    // Now translate by (-startX, -startY)
    x -= startX;
    y -= startY;

    // Next perform the rotation (you seem to have not copied the formulae correctly, I got this from the Wikipedia page, more or less)
    endX = (x * cos(deg90InRad)) - (y * sin(deg90InRad));
    endY = (x * sin(deg90InRad)) + (y * cos(deg90InRad));
  
    // Now translate back, by (startX, startY)
    endX += startX;
    endY += startY;    

    printf("New line is from (%f,%f) to (%f,%f)\n", startX, startY, endX, endY);

    return 0;
}

$ gcc -W -Wall -g rotate.c -o rotate -lm
rotate.c: In function ‘main’:
rotate.c:6: warning: unused parameter ‘argc’
rotate.c:6: warning: unused parameter ‘argv’
$ ./rotate 
New line is from (10.000000,10.000000) to (15.000000,5.000000)
$

Turns out I also had to make the angle negative to get a clockwise rotation.

If you're still having trouble with your code, you might try posting a more substantial portion of it.