更改 C 签名以接受字符数组而不是整数数组

Question

首先我必须承认我不是 C 专家，而且在必须进行此类转换时我总是感到困惑。我有下一个函数，它接受前 2 个参数，2 个指向无符号整数数组的指针。如何更改算法以接受 2 个指向 无符号字符数组 的指针，并且当然可以对这 2 个字符指针数组进行操作？ (我的意思是我知道不仅要更改签名，而且我还应该更改算法中的哪些内容？)

这就是我需要的:

void resize(unsigned char *input, unsigned char *output, int sourceWidth, int sourceHeight, int targetWidth, int targetHeight) 

这就是我所拥有的:

void resize(unsigned int *input, unsigned int *output, int sourceWidth, int sourceHeight, int targetWidth, int targetHeight) 
{    
    int a, b, c, d, x, y, index;
    float x_ratio = ((float)(sourceWidth - 1)) / targetWidth;
    float y_ratio = ((float)(sourceHeight - 1)) / targetHeight;
    float x_diff, y_diff, blue, red, green ;
    int offset = 0 ;

    for (int i = 0; i < targetHeight; i++) 
    {
        for (int j = 0; j < targetWidth; j++) 
        {
            x = (int)(x_ratio * j) ;
            y = (int)(y_ratio * i) ;
            x_diff = (x_ratio * j) - x ;
            y_diff = (y_ratio * i) - y ;
            index = (y * sourceWidth + x) ;                
            a = input[index] ;
            b = input[index + 1] ;
            c = input[index + sourceWidth] ;
            d = input[index + sourceWidth + 1] ;

            // blue element
            blue = (a&0xff)*(1-x_diff)*(1-y_diff) + (b&0xff)*(x_diff)*(1-y_diff) +
                   (c&0xff)*(y_diff)*(1-x_diff)   + (d&0xff)*(x_diff*y_diff);

            // green element
            green = ((a>>8)&0xff)*(1-x_diff)*(1-y_diff) + ((b>>8)&0xff)*(x_diff)*(1-y_diff) +
                    ((c>>8)&0xff)*(y_diff)*(1-x_diff)   + ((d>>8)&0xff)*(x_diff*y_diff);

            // red element
            red = ((a>>16)&0xff)*(1-x_diff)*(1-y_diff) + ((b>>16)&0xff)*(x_diff)*(1-y_diff) +
                  ((c>>16)&0xff)*(y_diff)*(1-x_diff)   + ((d>>16)&0xff)*(x_diff*y_diff);

            output [offset++] = 
                    0x000000ff | // alpha
                    ((((int)red)   << 24)&0xff0000) |
                    ((((int)green) << 16)&0xff00) |
                    ((((int)blue)  << 8)&0xff00);
        }
    }
}

Answer 1

原始代码有:

int a, b, c, d, x, y, index;
...

for (int i = 0; i < targetHeight; i++) 
{
    for (int j = 0; j < targetWidth; j++) 
    {
        ...            
        a = input[index] ;
        b = input[index + 1] ;
        c = input[index + sourceWidth] ;
        d = input[index + sourceWidth + 1] ;

        // blue element
        blue = (a&0xff)*(1-x_diff)*(1-y_diff) + (b&0xff)*(x_diff)*(1-y_diff) +
               (c&0xff)*(y_diff)*(1-x_diff)   + (d&0xff)*(x_diff*y_diff);

        // green element
        green = ((a>>8)&0xff)*(1-x_diff)*(1-y_diff) + ((b>>8)&0xff)*(x_diff)*(1-y_diff) +
                ((c>>8)&0xff)*(y_diff)*(1-x_diff)   + ((d>>8)&0xff)*(x_diff*y_diff);

        // red element
        red = ((a>>16)&0xff)*(1-x_diff)*(1-y_diff) + ((b>>16)&0xff)*(x_diff)*(1-y_diff) +
              ((c>>16)&0xff)*(y_diff)*(1-x_diff)   + ((d>>16)&0xff)*(x_diff*y_diff);

green 和 red 的移位操作意味着代码假设来自 input 的每个值都有 3 个字节的数据，这当input的类型是unsigned int *时就可以了。但是，如果将输入类型更改为 unsigned char *，则必须决定是否只有蓝色数据(没有红色或绿色分量)，或者是否有三个连续的字符提供蓝色、红色和绿色分量(或任何其他顺序；通常是 RGB，不是吗？)，或者是否有其他排列。

如果您有 3 个连续的组件，则无需掩码 (& 0xFF)，但您需要更频繁地访问该数组。