（４）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 image
        image[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 image
        image[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-1）k +，左边m -，右边m+
    sumRed = 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4，6，9，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 limit
    for (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 boundary
                    if (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 round
            sumRed = 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