Question

我首先必须承认我不是C专家，而且在进行此类转换时我总是感到困惑。我有下一个函数，它接收前2个参数，2个指针指向无符号整数的数组。如何更改算法以接受指向 unsigned chars 数组的2个指针，当然还要对这两个char指针数组进行操作？（我的意思是我知道不仅要改变签名，还要在算法中改变什么呢？）

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

由于您的代码需要访问具有4字节数据类型的像素，因此最好在函数开头执行转换：

void resize(unsigned char *input_char, unsigned char *output_char, int sourceWidth, int sourceHeight, int targetWidth, int targetHeight) 
{    
    unsigned int* input = reinterpret_cast<unsigned int*>(input_char);        
    unsigned int* output = reinterpret_cast<unsigned int*>(output_char);
    int a, b, c, d, x, y, index;
    //...
}

注意：值int sourceWidth，int sourceHeight，int targetWidth，int targetHeight必须引用数组的整数大小，否则您的代码将无法正常工作或提升错误。

这里的代码如果你不能使用C ++样式：

void resize(unsigned char *input_char, unsigned char *output_char, int sourceWidth, int sourceHeight, int targetWidth, int targetHeight) 
{    
    unsigned int* input = (unsigned int*)(input_char);        
    unsigned int* output = (unsigned int*)(output_char);
    int a, b, c, d, x, y, index;
    //...
}

Answer 2

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的转换操作意味着代码假设input的每个值中有3个字节的数据，当{{1}的类型时这是正常的是input。但是，如果将输入类型更改为unsigned int *，则必须确定是否只有蓝色数据（没有红色或绿色组件），或者是否有三个连续字符提供蓝色，红色和绿色组件（或任何其他排序;它通常是RGB，不是吗？），或者是否有其他排列。

如果您有3个连续的组件，则可以不使用屏蔽（unsigned char *），但您需要更频繁地访问该阵列。

将C签名更改为接受而不是整数数组，即字符数组

2 个答案: