缩放,调整 bmp 结构,编程 c

scale, resize bmp structure, programming c

我的任务是用给定的因子缩放 bmp 图片:

stream: base64 bmp 图片编码:

Qk1CAAAAAAAAADYAAAAoAAAABAAAAAEAAAABABgAAAAAAAwAAAAjLgAAIy4AAAAAAAAAAAAA/////wAAAP8AAAAA

scale.c

struct bmp_image* scale(const struct bmp_image* image, float factor){
    if(image == NULL || factor <= 0){
        return NULL;
    }

    .........
    int width = bmp->header->width;
    int height = bmp->header->height;
    int widthNew = (int)round((float)width * factor);
    int heightNew = (int)round((float)height * factor);


    for (uint32_t y = 0; y < heightNew; y++){
        for(uint32_t z = 0; z< widthNew; z++){
            bmp->data[(y*widthNew)+z].red=image->data[(int)(floor((float)y/(float)factor)*width+floor((float)z/(float)factor))].red;
            bmp->data[(y*widthNew)+z].green=image->data[(int)(floor((float)y/(float)factor)*width+floor((float)z/(float)factor))].green;
            bmp->data[(y*widthNew)+z].blue=image->data[(int)(floor((float)y/(float)factor)*width+floor((float)z/(float)factor))].blue;
        }
    }

    return bmp;     
}

main.c

FILE *output_p8 = fopen("scale.bmp", "w");
    struct bmp_image* newimage8 = NULL;
    newimage8 = scale(image, 1.025656);  
    free_bmp_image(image);    
    write_bmp(output_p8,newimage8);
    free_bmp_image(newimage8);    
    fclose(output_p8);

数据输出我之后得到的:

ffffff ffffff ff0000 00ff00

我应该有的数据:

ffffff ff0000 00ff00 000000

有什么建议吗?

计算新维度时,进行浮点计算,然后四舍五入为最接近的整数:

int widthNew = round(width * factor);

(您不需要强制转换:factor 是一个 float,整数与 float 相乘的结果也是一个 float ; widthNew 是一个整数,因此 float 结果必须转换为整数。)

因为你四舍五入,你的有效比例因子(我们称之为factor_eff)可能与标称值不同factor:

float factor_eff = (float) widthNew / width;

你的情况:

factor == 1.025656;
width == 4;
widthNew == round(1.025656 * 4) == round(4.102624) == 4;
factor_eff = 4.0 / 4 == 1.0;

但是您使用旧因子进行指数计算:

int zNew = floor(z / factor);

(同样,您不需要转换。)对于您的四个 z 索引,这将产生:

z == 0;    zNew == floor(0 / 1.025656) == floor(0.0)       == 0;   // ok
z == 1;    zNew == floor(1 / 1.025656) ~~ floor(0.9749858) == 0;   // ! should be 1
z == 2;    zNew == floor(2 / 1.025656) ~~ floor(1.9499715) == 1;   // ! should be 2
z == 3;    zNew == floor(3 / 1.025656) ~~ floor(2.9249573) == 2;   // ! should be 3

您可以修复浮点因子,但我认为使用整数算法计算新索引更好(也更快):

int zNew = z * width / widthNew;
int yNew = y * height / heightNew;

确保先做乘法。这种方法应该可以安全地用于高达 46,000 的尺寸。