ssr网站怎么做,肥西建设局网站,免费的x网站域名,wordpress 文章自定义排序#xff08;#xff14;#xff09;blur function
简单画图熟悉一下要做什么 可以看到3种情况#xff0c;顶格#xff0c;边界#xff0c;里面如果分开算的话#xff0c;是真的麻烦#xff1b;但是当时还真的没有想到更好的#xff0c;就先写一写#xff08;此处摘取…blur function
简单画图熟悉一下要做什么 可以看到3种情况顶格边界里面如果分开算的话是真的麻烦但是当时还真的没有想到更好的就先写一写此处摘取3处外部都是 for (int i 0; i height; i) { 下边省去不写 for (int j 0; j width; j) { } }
里面
int sumRed 0,sumGreen 0,sumBlue 0;//set 3 conditions by the limit//前面1部分//计算9个格子if (i 0 i height - 1 j 0 j width - 1) {for (int k -1; k 1; k) {for (int m -1; m 1; m) {sumRed copy[i k][j m].rgbtRed;sumGreen copy[i k][j m].rgbtGreen;sumBlue copy[i k][j m].rgbtBlue;}}//after sum it all and give that value to imageimage[i][j].rgbtRed round(sumRed / 9.0);image[i][j].rgbtGreen round(sumGreen / 9.0);image[i][j].rgbtBlue round(sumBlue / 9.0);
}顶格
//1.1 左上角 i j sumRed 0, sumGreen 0, sumBlue 0;if (j 0 i 0) {for (int k 0; k 1; k) {for (int m 0; m 1; m) {sumRed copy[i k][j m].rgbtRed;sumGreen copy[i k][j m].rgbtGreen;sumBlue copy[i k][j m].rgbtBlue;}}//after sum it all and give that value to imageimage[i][j].rgbtRed round(sumRed / 4.0);image[i][j].rgbtGreen round(sumGreen / 4.0);image[i][j].rgbtBlue round(sumBlue / 4.0);}边界 //2.1上边界:移动的方向只有下高度-1k 左边m -右边msumRed 0, sumGreen 0, sumBlue 0;if (i 0) {//计算RGB利用双循环(k,m)for (int k 0; k 1; k) {for (int m -1; m 1; m) {sumRed copy[i k][j m].rgbtRed;sumGreen copy[i k][j m].rgbtGreen;sumBlue copy[i k][j m].rgbtBlue;}}image[i][j].rgbtRed round(sumRed / 6.0);image[i][j].rgbtGreen round(sumGreen / 6.0);image[i][j].rgbtBlue round(sumBlue / 6.0);}简化
然后可以看到很多重复的地方需要开始统一找到不同的地方469然后里层的双循环k,m其实也可以合并只要移动之后的index在范围里面就可以了里层循环的次数加上限定的条件计算count,然后一步步开始删减…
// Blur image
void blur(int height, int width, RGBTRIPLE image[height][width])
{// copy the image[height][width]RGBTRIPLE copy[height][width];for (int i 0; i height; i){for (int j 0; j width; j){copy[i][j] image[i][j];}}int sumRed 0, sumGreen 0, sumBlue 0, count 0;// set 3 conditions by the limitfor (int i 0; i height; i){for (int j 0; j width; j){for (int k -1; k 1; k){for (int m -1; m 1; m){// condition limit boundaryif (i k 0 i k height - 1 j m 0 j m width - 1){sumRed copy[i k][j m].rgbtRed;sumGreen copy[i k][j m].rgbtGreen;sumBlue copy[i k][j m].rgbtBlue;count;}}}// after sum it all and give that value to image//before calculating the results,first convert it to float!image[i][j].rgbtRed round((float) sumRed / count);image[i][j].rgbtGreen round((float) sumGreen / count);image[i][j].rgbtBlue round((float) sumBlue / count);// clear it and for next roundsumRed 0, sumGreen 0, sumBlue 0, count 0;}}
}
问题
这里要在计算之前把int sumRGB值转化成float类型
image[i][j].rgbtRed round((float) sumRed / count);而不是float计算之后的值这样就把小数点之后的都丢了
image[i][j].rgbtRed round((float) sumRed / count);总结
刚开始的时候想不到简洁的方法或者不太理解先慢慢写出来能运行就可以。 后面再去优化一步一步来不需要太大压力
check
:) helpers.c exists
:) filter compiles
:) grayscale correctly filters single pixel with whole number average
:) grayscale correctly filters single pixel without whole number average
:) grayscale leaves alone pixels that are already gray
:) grayscale correctly filters simple 3x3 image
:) grayscale correctly filters more complex 3x3 image
:) grayscale correctly filters 4x4 image
:) sepia correctly filters single pixel
:) sepia correctly filters simple 3x3 image
:) sepia correctly filters more complex 3x3 image
:) sepia correctly filters 4x4 image
:) reflect correctly filters 1x2 image
:) reflect correctly filters 1x3 image
:) reflect correctly filters image that is its own mirror image
:) reflect correctly filters 3x3 image
:) reflect correctly filters 4x4 image
:) blur correctly filters middle pixel
:) blur correctly filters pixel on edge
:) blur correctly filters pixel in corner
:) blur correctly filters 3x3 image
:) blur correctly filters 4x4 image
