Rotate picture?

smeezekitty · 03-19-2010, 12:32 PM

How do i rotate a picture in C without using any libraries unless they can be compiled on Linux, Windows, Dos protected mode and Dos real mode?

MTK358 · 03-19-2010, 12:57 PM

If you don't have to worry about encoding/decoding common image file formats and will only need to rotate images in 90 degree increments, then it might be feasible to do it without image libraries.

smeezekitty · 03-19-2010, 12:59 PM

The image loading code is already handled.
The image is in a two demention array and needs to be rotated in 1 or 22 degree increments from the center-point.

MTK358 · 03-19-2010, 01:04 PM

Assuming that (0, 0) is the center of the image, iterate through every point on the image, rotate the point around (0, 0) using trigonometry, and apply the color at the original point on the old image to the new point on the new image.

MTK358 · 03-19-2010, 01:09 PM

I think this might work:

Code:

void rotatePoint(int *x, int *y, double radians) {
	d = sqrt((*x * *x) + (*y * *y));
	t = atan2(*x, *y);
	t += radians;
	*x = sin(t) * d;
	*y = cos(t) * d;
}

I cannot guarantee the correctness of this, because my trigonometry is a little rusty, so it might not work as expected.

smeezekitty · 03-19-2010, 01:21 PM

I believe the degrees to radian equations is radians = (degrees / 58).
Is that correct?

MTK358 · 03-19-2010, 01:29 PM

360 / (2 * pi) = 57.295...

So I guess 58 is close enough.

smeezekitty · 03-19-2010, 01:41 PM

Uh-oh:

Code:

atan2: DOMAIN error
Floating point error: Domain.
Abnormal program termination

Code:

#include <math.h>
#include <stdio.h>
void rotatePoint(int *x, int *y, double radians) {
 double d = sqrt((*x * *x) + (*y * *y));
 double t = atan2(*x, *y);
 t += radians;
 *x = sin(t) * d;
 *y = cos(t) * d;
}
int main(){
char array[4][4] = {
"0#00",
"0#00",
"0#00",
"0#00"
};
for(int x = 0;x < 4;x++){
for(int y = 0;y < 4;y++){
int jx = x;
int jy = y;
rotatePoint(&jx, &jy, (double)90/(double)58);
array[jx][jy] = array[x][y];
}
}
for((int) x = 0;x < 4;x++){
for(int y = 0;y < 4;y++){
printf("%c", array[x][y]);
}
}
return 0;
}

MTK358 · 03-19-2010, 01:42 PM

Please indent your code.

Most good editors intended for coding retain the current indentation level when you make a new line, so you just press Tab when you start a block and Backspace when you end one. Even with editors that don't do that, IMHO pressing tab after every line doesn't do any harm.

smeezekitty · 03-19-2010, 01:48 PM

The best i can do:

Code:

#include <math.h>
#include <stdio.h>
void rotatePoint(int *x, int *y, double radians) {
	double d = sqrt((*x * *x) + (*y * *y));
	double t = atan2(*x, *y);
	t += radians;
	*x = sin(t) * d;
	*y = cos(t) * d;
}
int main(){
	char array[4][4] = {
		"0#00",
		"0#00",
		"0#00",
		"0#00"
	};
	for(int x = 0;x < 4;x++){
		for(int y = 0;y < 4;y++){
			int jx = x;
			int jy = y;
			rotatePoint(&jx, &jy, (double)90/(double)58);
			array[jx][jy] = array[x][y];
		}
	}
	for((int) x = 0;x < 4;x++){printf("\n");
		for(int y = 0;y < 4;y++){
			printf("%c", array[x][y]);
		}
	}
	return 0;
}

MTK358 · 03-19-2010, 01:54 PM

The indentation is pretty good.

Any way, I don't really understand the error. Maybe try this:

double t = atan2((double) *x, (double) *y);

instead of this:

double t = atan2(*x, *y);

MTK358 · 03-19-2010, 01:57 PM

IMPORTANT:

I just thought that if you take each point in the old image, rotate it, and put it in the new one, it's likely that some pixels will me missed in the new image.

It is probably a much better idea to iterate through the points of the new, empty image, rotate the point backwards, and read that point from the original image.

smeezekitty · 03-19-2010, 02:05 PM

Code:

#include <math.h>
#include <stdio.h>
void rotatePoint(int *x, int *y, double radians) {
	double d = sqrt((*x * *x) + (*y * *y));
	double t = atan2((double)*x, (double)*y);
	t += radians;
	*x = sin(t) * d;
	*y = cos(t) * d;
}
int main(){
	char array[4][4] = {
		"0#00",
		"0#00",
		"0#00",
		"0#00"
	};
	char paw[4][4];
	for(int x = 4;x > 0;x--){
		for(int y = 4;y > 0;y--){
			int jx = x;
			int jy = y;
			rotatePoint(&jx, &jy, -(double)90/(double)58);
			paw[jx][jy] = array[x][y];
		}
	}
	for((int) x = 0;x < 4;x++){
		printf("\n");
		for(int y = 0;y < 4;y++){
			printf("%c", paw[x][y]);
		}
	}
	return 0;

Output:

Code:

*###
 ☼Γ
►☻
! ☻%

MTK358 · 03-19-2010, 02:26 PM

What is that supposed to mean?

smeezekitty · 03-19-2010, 02:48 PM

Desired output is

Code:

0000
####
0000
0000