我的cs50滤镜边缘功能无法正常工作，编译正常，但是当我运行check50时，第一个测试（边缘正确过滤了中间像素）我们正确，而其他测试仅靠最后一个值是不正确的，就像这样：
:(边缘正确过滤边缘上的像素 预期为“ 213228255 \ n”，而不是“ 213228140 \ n”
但是，当我仅打印红色，绿色和蓝色的gx和gy以及平方根的值时，所有颜色的值都不匹配。
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;
}
答案 0 :(得分：0)
当您将round(sqrt((copia[i][j].rgbtRed * copia[i][j].rgbtRed) + (copia2[i][j].rgbtRed *copia2[i][j].rgbtRed)));存储到变量image[i][j].rgbtRed（或其任何其他变体）中时，就会出现您描述的问题。这是因为在计算sqrt（gx ^ 2 + gy ^ 2）时，会得到一个大于255的数字。例如，四舍五入后可能会得到整数值395。要将该值存储到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。
这意味着您的if语句无用，因为各个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;
}
要解决此问题，必须先将值设置为上限，然后再将其存储到图像中。一个简单的实现是通过创建一个称为cap的函数，如果输入大于255，则该函数返回255。：
int cap(int rgb)
{
if (rgb > 255)
{
return 255;
}
else
{
return rgb;
}
}
您可以按以下方式使用此功能，这将完全解决您的问题：
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))));
这还将缩短您的代码，使其看起来更整洁，并避免不必要的重复。