如何旋转图像用C 90度 [英] How to rotate an image 90 degrees in C
问题描述
我一直在试图将一个黑白图像，然后旋转在C 90度，但我相当新的节目，这是我不得不远。
＃包括LT＆;＆stdio.h中GT;
＃包括LT＆;＆string.h中GT;诠释主（）{
FILE *第一;
FILE *第二;
FILE *第三;
INT计数器;
焦炭℃;
INT宽度，高度，pixelmax，平均值;
INT pixelred，pixelgreen，pixelblue，黑[300] [300] [3];
INT I，J，定时器= 0;
INT K，F = 0;
第一FOPEN =（blackbuck.ppm，R）;
第二=的fopen（blacktrout.ppm，W）;
这将跳过code的前几行
为（计数器= 1;计数器3;;计数器++）{
做{
C = GETC（第一）;
}而（C ='\\ n'！）;
}
的fscanf（第一％D，＆安培;宽度和放大器;高度）;
的fscanf（第一％d个，＆安培; pixelmax）;
在程序的这一部分，我把像素为黑色和白色，采取他们的平均，这是比较容易的部分。
为（i = 0，J = 0; I＆LT;宽度;我++，定时器++）{
的fscanf（第一％D％D，＆安培; pixelred，＆安培; pixelgreen，＆安培; pixelblue）;
平均=（pixelred + pixelgreen + pixelblue）/ 3;
黑色的[I] [J] [0] =平均值;
黑色的[I] [j]的[1] =平均值;
黑色的[I] [J] [2] =平均值;
fprintf中（第二，％D％d个\\ N，黑色的[I] [j]的[0]，黑色的[I] [j]的[1]，黑色的[I] [J] [2]）;如果（ⅰ==（宽度-1）及与放大器; J＆下;高度）{
I = 0;
J ++;
}
}
FCLOSE（第一）;
FCLOSE（第二）;
第三=的fopen（blackflip.ppm，W）;
这是我在哪里完全失去了一部分，我不知道如何在我的3D阵列90度旋转像素商店。任何帮助吗？一个很好的解释说明会走很长的路是新手程序员喜欢自己。谢谢！
的（... ???）
}
}
返回0;
的基本逻辑是这样的：
original_x = original_height-1-rotated_y
original_y = rotated_x
rotated_image（rotated_x，rotated_y）= original_image（original_x，original_y）
让我们说，（0,0）是左上角，我们是想向左旋转90度。然后将旋转后的图像的左上角是等同于原始图像的右上角，它是在（original_height-1,0）。
随着你在图像的顶部，增加中，x，则是从沿着右侧的原始图像采集像素，增加Y，所以你的旋转图像的x是像原来图像的y
当你走在rotated_image，增加Y，你正将越来越多的original_image的左边，这就是为什么我们从减去rotated_y original_height-1，以获得original_x坐标。
要注意的另一件事是，旋转后的图像的宽度和高度都从原始图片颠倒。
I've been trying to turn an image black and white, and then rotate it 90 degrees in C but i'm fairly new to programming, this is what i have to far.
#include<stdio.h>
#include<string.h>
int main () {
FILE* first;
FILE* second;
FILE* third;
int counter;
char c;
int width, height, pixelmax, average;
int pixelred, pixelgreen, pixelblue, black[300][300][3];
int i, j, timer=0;
int k, f=0;
first=fopen("blackbuck.ppm","r");
second=fopen("blacktrout.ppm","w");
this skips the first few lines of code
for(counter=1;counter<3;counter++){
do{
c=getc(first);
}while(c != '\n');
}
fscanf(first,"%d%d", &width,&height);
fscanf(first,"%d", &pixelmax);
in this part of the program, i turn the pixels to black and white by taking their average, this is the easy part.
for(i=0, j=0; i<width;i++, timer++){
fscanf(first,"%d%d%d",&pixelred,&pixelgreen,&pixelblue);
average=(pixelred+pixelgreen+pixelblue)/3;
black[i][j][0]=average;
black[i][j][1]=average;
black[i][j][2]=average;
fprintf(second,"%d %d %d\n",black[i][j][0],black[i][j][1],black[i][j][2]);
if (i==(width-1)&& j<height){
i=0;
j++;
}
}
fclose(first);
fclose(second);
third=fopen("blackflip.ppm","w");
This is the part where i am completely lost, i have no idea how to rotate the pixels stores in my 3d array 90 degrees. Any help please? A well explained explanation would go a long way for a newbie programmer like myself. thanks!
for(...???)
}
}
return 0;
The basic logic is like this:
original_x = original_height-1-rotated_y
original_y = rotated_x
rotated_image(rotated_x,rotated_y) = original_image(original_x,original_y)
Let's say that (0,0) is the upper-left corner, and we are wanting to rotate left 90 degrees. Then the upper-left corner of the rotated image is equivalent to the upper-right corner of the original image, which is at (original_height-1,0).
As you go across the top of the image, increasing, x, you are grabbing pixels from the original image along the right side, increasing y, so the x of your rotated image is like the y of the original image.
As you go down in the rotated_image, increasing y, you are moving more to the left of the original_image, which is why we're subtracting rotated_y from original_height-1 to get the original_x coordinate.
Another thing to notice is that the width and height of the rotated image are reversed from the orignal image.
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