C ++将float转换为unsigned char?

时间:2010-10-10 02:50:44

标签: c++ floating-point unsigned-char

我是C ++的新手,并且做了一些谷歌搜索,我认为sprintf可以完成这项工作,但是我在编译时遇到错误,我无法在unsigned charchar。我需要一个unsigned char,因为我要打印到一个图像文件(0-255 RGB)。

unsigned char*** pixels = new unsigned char**[SIZE];
vector<float> pixelColors;

...

sprintf(pixels[i][j][k], "%.4g", pixelColors.at(k));

(pixelColors的大小为3,'k'指的是'for loop'变量)

2 个答案:

答案 0 :(得分:7)

我猜这些花车的范围是0.0 ... 1.0,那么你确实喜欢这个:

float redf = 0.5f;
unsigned char reduc = redf * 255;

变量reduc现在是128。


编辑:完整示例,以网络PPM格式输出图片。

// Usage 
//  program > file.ppm

#include <vector>
#include <iostream>

typedef struct
{   /* colors in range 0..1 anything else is out of gamut */
    float red, green, blue;
} color;

using namespace std;

int main ( int argc, char **argv )
{   
    int width = 10, height = 10;
    vector<color> bitmap; // This should maybe be called floatmap? ;)

    // Make an image in memory as a float vector

    for( int y = 0; y < height; y++ )
    {   
        for( int x = 0; x < width; x++ )
        {
            color temp;
            temp.red = ((float)x / width);
            temp.green = 0;
            temp.blue = ((float)y / height);
            bitmap.push_back(temp);
        }
    }

    // output image as an Netppm pixmap
    cout << "P3" << endl << width << " " << height << endl << 255 << endl;
    for( int y = 0; y < height; y++ )
    {   
        for( int x = 0; x < width; x++ )
        {
            int red, green, blue;
            red = (unsigned char)(bitmap[y*width+x].red * 255);
            green = (unsigned char)(bitmap[y*width+x].green * 255);
            blue = (unsigned char)(bitmap[y*width+x].blue * 255);
            cout << red << " ";
            cout << green << " ";
            cout << blue << " ";
        }
        cout << endl;
    }
    return 0;

}

我希望这会对你有所帮助。您可以阅读Netpbm format on wikipedia


EDIT2:图像输出为明文 结果是这样的:
test image(很小,不是吗?编辑第16到512x512行或其他东西)

实际输出是这样的:

P3
10 10
255
0 0 0 25 0 0 51 0 0 76 0 0 102 0 0 127 0 0 153 0 0 178 0 0 204 0 0 229 0 0 
0 0 25 25 0 25 51 0 25 76 0 25 102 0 25 127 0 25 153 0 25 178 0 25 204 0 25 229 0 25 
0 0 51 25 0 51 51 0 51 76 0 51 102 0 51 127 0 51 153 0 51 178 0 51 204 0 51 229 0 51 
0 0 76 25 0 76 51 0 76 76 0 76 102 0 76 127 0 76 153 0 76 178 0 76 204 0 76 229 0 76 
0 0 102 25 0 102 51 0 102 76 0 102 102 0 102 127 0 102 153 0 102 178 0 102 204 0 102 229 0 102 
0 0 127 25 0 127 51 0 127 76 0 127 102 0 127 127 0 127 153 0 127 178 0 127 204 0 127 229 0 127 
0 0 153 25 0 153 51 0 153 76 0 153 102 0 153 127 0 153 153 0 153 178 0 153 204 0 153 229 0 153 
0 0 178 25 0 178 51 0 178 76 0 178 102 0 178 127 0 178 153 0 178 178 0 178 204 0 178 229 0 178 
0 0 204 25 0 204 51 0 204 76 0 204 102 0 204 127 0 204 153 0 204 178 0 204 204 0 204 229 0 204 
0 0 229 25 0 229 51 0 229 76 0 229 102 0 229 127 0 229 153 0 229 178 0 229 204 0 229 229 0 229 

答案 1 :(得分:2)

不确定,您的具体要求是什么[因为您没有按照Greg的要求粘贴代码片段],以下示例可能会解决此问题:

#include <iostream>
#include <conio.h>
using namespace std;

int main()
{
    float i=1;
    unsigned char c;
    c = static_cast<unsigned char>(i);
    cout << c << endl;
    getch();
    return 0;
}