sRGB <-> 线性 RGB 转换精度答案

【问题标题】：sRGB <-> linear RGB conversion precisionsRGB <-> 线性 RGB 转换精度
【发布时间】：2020-08-09 07:31:35
【问题描述】：

我正在尝试了解图像/视频处理的基础知识，最近我了解到所有处理都应该应用于图像的线性表示。所以，我为 sRGB -> RGB 和 RGB -> sRGB 写了两个函数：

void sRGB2lRGB(std::vector<unsigned char>& in, std::vector<unsigned char>& out)
{
    std::vector<double> temp(in.begin(), in.end());

    for (int i = 0; i < temp.size(); i++)
    {
        temp[i] /= 255.0;
    }

    for (int i = 0; i < temp.size(); i++)
    {
        if (temp[i] <= 0.04045)
        {
            temp[i] /= 12.92;
        }
        else
        {
            temp[i] = std::pow((temp[i] + 0.055) / 1.055, 2.4);
        }
    }

    for (int i = 0; i < temp.size(); i++)
    {
        out[i] = temp[i] * (int)255 + 0.5;
    }
}

void lRGB2sRGB(std::vector<unsigned char>& in, std::vector<unsigned char>& out)
{
    std::vector<double> temp(in.begin(), in.end());

    for (int i = 0; i < temp.size(); i++)
    {
        temp[i] /= 255.0;
    }

    for (int i = 0; i < temp.size(); i++)
    {
        if (temp[i] <= 0.0031308)
        {
            temp[i] *= 12.92;
        }
        else
        {
            temp[i] = 1.055 * std::pow(temp[i], 1.0 / 2.4) - 0.055;
        }
    }

    for (int i = 0; i < temp.size(); i++)
    {
        out[i] = temp[i] * (int)255 + 0.5;
    }
}

为了测试它，我尝试了这个：

int main()
{
    std::vector<unsigned char> in(255);
    std::vector<unsigned char> out(255);
    std::vector<unsigned char> out2(255);

    for (int i = 0; i < 255; i++)
    {
        in[i] = i;
    }

    sRGB2lRGB(in, out);
    lRGB2sRGB(out, out2);

    for (int i = 0; i < 255; i++)
    {
        if (out2[i] != i)
        {
            std::cout << "was: " << (int)in[i] << ", now: " << (int)out2[i] << '\n';
        }
    }
}

但似乎值越接近 0，结果就越不准确。输出是：

was: 1, now: 0
was: 2, now: 0
was: 3, now: 0
was: 4, now: 0
was: 5, now: 0
was: 6, now: 0
was: 7, now: 13
was: 8, now: 13
was: 9, now: 13
was: 10, now: 13
was: 11, now: 13
was: 12, now: 13
was: 14, now: 13
was: 15, now: 13
was: 16, now: 13
was: 17, now: 13
was: 18, now: 22
was: 19, now: 22
was: 20, now: 22
was: 21, now: 22
was: 23, now: 22
was: 24, now: 22
was: 25, now: 22
was: 26, now: 28
was: 27, now: 28
was: 29, now: 28
was: 30, now: 28
was: 31, now: 28
was: 32, now: 34
was: 33, now: 34
was: 35, now: 34
was: 36, now: 34
was: 37, now: 38
was: 39, now: 38
was: 40, now: 38
was: 41, now: 42
was: 43, now: 42
was: 44, now: 42
was: 45, now: 46
was: 47, now: 46
was: 48, now: 50
was: 49, now: 50
was: 51, now: 50
was: 52, now: 53
was: 54, now: 53
was: 55, now: 56
was: 57, now: 56
was: 58, now: 59
was: 60, now: 61
was: 62, now: 61
was: 63, now: 64
was: 65, now: 64
was: 67, now: 66
was: 68, now: 69
was: 70, now: 71
was: 72, now: 73
was: 74, now: 73
was: 76, now: 75
was: 78, now: 77
was: 80, now: 79
was: 82, now: 83
was: 84, now: 85
was: 87, now: 86
was: 89, now: 88
was: 91, now: 92
was: 94, now: 95
was: 97, now: 96
was: 100, now: 99
was: 103, now: 104
was: 107, now: 106
was: 111, now: 112
was: 116, now: 117
was: 123, now: 124

我哪里错了？

【问题讨论】：

在 sRGB2lRGB 中，对于所有小于 7 的输入值，您对 out[i] 的计算将为 0，因此当您尝试反转计算时，您将得到相同的值。
看起来它是线性空间的一个特征，但是由于黑色附近的信息基本上丢失了，我看不出那个对话是如何恢复的：/

标签： c++ image colors

【解决方案1】：

因为您通过将线性浮点值转换为整数表示来量化它们，所以转换确实是破坏性的。

但是，发生这种情况应该不足为奇：这实际上是使用 sRGB 逆光电传递函数 (EOTF) 对数据进行编码时的意图。目的是在 8 位信号上传输数据，同时保持良好的感知均匀性并降低整体带宽。

我会参考 Charles Poynton 的 Gamma Faq 和 Poynton, C., & Funt, B. (2014)。数字图像表示和显示中的感知均匀性。颜色研究与应用，39（1），6-15。 https://doi.org/10.1002/col.21768

注意"Conversation"这里应该是"conversion"。

【讨论】：