为什么我的 CS50 过滤器边缘代码不适用于 check50？
[英]why is my CS50 filter edges code not working with check50?
问题描述
My cs50 filter edges function is not working, it compiles ok but when i run check50 the first test (edges correctly filters middle pixel) us correct while the others are incorrect just by the last value, like this:
我的 cs50 过滤器边缘 function 不工作，它编译正常，但是当我运行 check50 时，第一个测试（边缘正确过滤中间像素）我们是正确的，而其他的只是最后一个值不正确，如下所示：
:( edges correctly filters pixel on edge expected "213 228 255\n", not "213 228 140\n" :( 边缘正确过滤边缘上的像素，预期为“213 228 255\n”，而不是“213 228 140\n”
However, when I print the gx and gy for the red, green and blue alone, and the value of the squareroot, none of the values for the colors match.但是，当我单独打印红色、绿色和蓝色的 gx 和 gy 以及平方根的值时，colors 的值都不匹配。
now, this is my code for edges现在，这是我的边缘代码
void edges(int height, int width, RGBTRIPLE image[height][width])
{
int sr = 0;
int sb = 0;
int sg = 0;
int yr = 0;
int yb = 0;
int yg = 0;
struct RGBTRIPle
{
int rgbtRed;
int rgbtGreen;
int rgbtBlue;
};
struct RGBTRIPle copia[height][width];
struct RGBTRIPLe
{
int rgbtRed;
int rgbtGreen;
int rgbtBlue;
};
struct RGBTRIPLe copia2[height][width];
//Implementing Gx
for (int i = 0; i < height; i++)
{
for (int j = 0; j < width; j++)
{
sr = 0;
sb = 0;
sg = 0;
for (int m = i - 1; m <= i + 1; m++)
{
for (int c = j - 1; c <= j + 1; c++)
{
if (m >= 0 && m < height && c >= 0 && c < width)
{
if (c == j - 1)
{
if (m == i - 1 || m == i + 1)
{
sr += -1 * image[m][c].rgbtRed;
sb += -1 * image[m][c].rgbtBlue;
sg += -1 * image[m][c].rgbtGreen;
}
else
{
sr += -2 * image[m][c].rgbtRed;
sb += -2 * image[m][c].rgbtBlue;
sg += -2 * image[m][c].rgbtGreen;
}
}
if (c == j + 1)
{
if (m == i - 1 || m == i + 1)
{
sr += image[m][c].rgbtRed;
sb += image[m][c].rgbtBlue;
sg += image[m][c].rgbtGreen;
}
else
{
sr += 2 * image[m][c].rgbtRed;
sb += 2 * image[m][c].rgbtBlue;
sg += 2 * image[m][c].rgbtGreen;
}
}
else //c = j
{
sr += 0 * image[m][c].rgbtRed;
sb += 0 * image[m][c].rgbtBlue;
sg += 0 * image[m][c].rgbtGreen;
}
}
}
}
copia[i][j].rgbtRed = sr;
copia[i][j].rgbtGreen = sg;
copia[i][j].rgbtBlue = sb;
}
}
//Implementing Gy
for (int i = 0; i < height; i++)
{
for (int j = 0; j < width; j++)
{
yr = 0;
yb = 0;
yg = 0;
for (int m = i - 1; m <= i + 1; m++)
{
for (int c = j - 1; c <= j + 1; c++)
{
if (m >= 0 && m < height && c >= 0 && c < width)
{
if (m == i - 1)
{
if (c == j - 1 || c == j + 1)
{
yr += -1 * image[m][c].rgbtRed;
yb += -1 * image[m][c].rgbtBlue;
yg += -1 * image[m][c].rgbtGreen;
}
else
{
yr += -2 * image[m][c].rgbtRed;
yb += -2 * image[m][c].rgbtBlue;
yg += -2 * image[m][c].rgbtGreen;
}
}
if (m == i + 1)
{
if (c == j + 1 || c == j - 1)
{
yr += image[m][c].rgbtRed;
yb += image[m][c].rgbtBlue;
yg += image[m][c].rgbtGreen;
}
else
{
yr += 2 * image[m][c].rgbtRed;
yb += 2 * image[m][c].rgbtBlue;
yg += 2 * image[m][c].rgbtGreen;
}
}
else //c = j
{
yr += 0 * image[m][c].rgbtRed;
yb += 0 * image[m][c].rgbtBlue;
yg += 0 * image[m][c].rgbtGreen;
}
}
}
}
copia2[i][j].rgbtRed = yr;
copia2[i][j].rgbtGreen = yg;
copia2[i][j].rgbtBlue = yb;
}
}
//Implementing math operation to calculate resulting color
for (int i = 0; i < height; i++)
{
for (int j = 0; j < width; j++)
{
int r = 0;
int g = 0;
int b = 0;
image[i][j].rgbtRed = (int) round(sqrt((copia[i][j].rgbtRed * copia[i][j].rgbtRed) + (copia2[i][j].rgbtRed *
copia2[i][j].rgbtRed)));
image[i][j].rgbtGreen = (int) round(sqrt((copia[i][j].rgbtGreen * copia[i][j].rgbtGreen) + (copia2[i][j].rgbtGreen *
copia2[i][j].rgbtGreen)));
image[i][j].rgbtBlue = (int) round(sqrt((copia[i][j].rgbtBlue * copia[i][j].rgbtBlue) + (copia2[i][j].rgbtBlue *
copia2[i][j].rgbtBlue)));
r = image[i][j].rgbtRed;
g = image[i][j].rgbtGreen;
b = image[i][j].rgbtBlue;
if (image[i][j].rgbtRed > 255)
{
image[i][j].rgbtRed = 255;
}
if (image[i][j].rgbtGreen > 255)
{
image[i][j].rgbtGreen = 255;
}
if (image[i][j].rgbtBlue > 255)
{
image[i][j].rgbtBlue = 255;
}
}
}
return;
}
1 个解决方案
解决方案1
0 已采纳 2020-06-18 15:38:16
The problem you describe arises when you store round(sqrt((copia[i][j].rgbtRed * copia[i][j].rgbtRed) + (copia2[i][j].rgbtRed *copia2[i][j].rgbtRed)));当您存储round(sqrt((copia[i][j].rgbtRed * copia[i][j].rgbtRed) + (copia2[i][j].rgbtRed *copia2[i][j].rgbtRed))); into the variable image[i][j].rgbtRed (or any other variant thereof).进入变量image[i][j].rgbtRed （或其任何其他变体）。 This is because when calculating sqrt(gx^2 + gy^2) you are getting a number above 255. For example, you may get the integer value 395 after rounding.这是因为在计算 sqrt(gx^2 + gy^2) 时，您得到的数字大于 255。例如，四舍五入后您可能会得到 integer 值 395。 To store that value in to image[i][j].rgbtRed , C will store the value of 395 % 255, or 140, because the image cannot store values greater than 255, by definition.为了将该值存储到image[i][j].rgbtRed中，C 将存储 395 % 255 或 140 的值，因为根据定义，图像不能存储大于 255 的值。
This means that your if statements are useless, because the respective color values will never be greater than 255:这意味着您的 if 语句是无用的，因为相应的颜色值永远不会大于 255：
if (image[i][j].rgbtRed > 255)
{
image[i][j].rgbtRed = 255;
}
if (image[i][j].rgbtGreen > 255)
{
image[i][j].rgbtGreen = 255;
}
if (image[i][j].rgbtBlue > 255)
{
image[i][j].rgbtBlue = 255;
}
To solve this problem you have to cap the value before storing them into the image.要解决这个问题，您必须在将它们存储到图像之前限制值。 A simple implementation of this would be by making a function called cap that returns 255 if an input is above 255.:一个简单的实现是通过创建一个名为 cap 的 function 如果输入高于 255，则返回 255：
int cap(int rgb)
{
if (rgb > 255)
{
return 255;
}
else
{
return rgb;
}
}
You can then use this function in the following manner, which will solve your problem completely:然后您可以通过以下方式使用此 function，这将彻底解决您的问题：
image[i][j].rgbtRed = cap(round(sqrt((copia[i][j].rgbtRed * copia[i][j].rgbtRed) + (copia2[i][j].rgbtRed * copia2[i][j].rgbtRed))));
image[i][j].rgbtGreen = cap(round(sqrt((copia[i][j].rgbtGreen * copia[i][j].rgbtGreen) + (copia2[i][j].rgbtGreen * copia2[i][j].rgbtGreen))));
image[i][j].rgbtBlue = cap(round(sqrt((copia[i][j].rgbtBlue * copia[i][j].rgbtBlue) + (copia2[i][j].rgbtBlue * copia2[i][j].rgbtBlue))));
This will also shorten your code, make it look cleaner, and avoid unnecessary repetition.这也将缩短您的代码，使其看起来更干净，并避免不必要的重复。
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