在C中将yuv420p转换为rgb888

时间:2015-02-19 10:06:23

标签: c algorithm rgb yuv

我正在尝试将平面YUV(yuv420p)转换为C中的RGB24。

我不控制输入,并将YUV数据作为3个单独的缓冲区,一个用于 Y ,一个用于 U ,一个用于 V < /强>

我想从这3个输入缓冲区创建单个输出RGB缓冲区。

我在wiki上遵循了以下架构,但我的代码有问题,因为输出图像颜色错误。

这是我的功能(请原谅......)

static uint8_t* yuv420p_to_rgb2(const uint8_t* y, const uint8_t* u, const uint8_t* v, const size_t width, const size_t height)
{
    const size_t size = width * height;
    uint8_t* rgb = (uint8_t*)calloc((size * 3), sizeof(uint8_t));

    int uv_index = 0, pass = 0;
    int b,g,r;
    uint8_t* ptr = rgb;
    for (size_t i = 0; i < size; i += 6)
    {
        if (pass == 2)
        {
            pass = 0;
            uv_index += 3;
        }
        int y1 = y[i];
        int y2 = y[i+1];
        int y3 = y[i+2];
        int y4 = y[i+3];
        int y5 = y[i+4];
        int y6 = y[i+5];

        int u1 = u[uv_index];
        int u2 = u[uv_index+1];
        int u3 = u[uv_index+2];

        int v1 = v[uv_index];
        int v2 = v[uv_index+1];
        int v3 = v[uv_index+2];
        //1
        r = 1.164 * (y1 - 16) + 1.596 * (v1 - 128);
        g = 1.164 * (y1 - 16) - 0.813 * (v1 - 128) - 0.391 * (u1 - 128);
        b = 1.164 * (y1 - 16) + 2.018 * (u1 - 128);
        *ptr++ = CLAMP(r);
        *ptr++ = CLAMP(g);
        *ptr++ = CLAMP(b);
        //2
        r = 1.164 * (y2 - 16) + 1.596 * (v1 - 128);
        g = 1.164 * (y2 - 16) - 0.813 * (v1 - 128) - 0.391 * (u1 - 128);
        b = 1.164 * (y2 - 16) + 2.018 * (u1 - 128);
        *ptr++ = CLAMP(r);
        *ptr++ = CLAMP(g);
        *ptr++ = CLAMP(b);
        //3
        r = 1.164 * (y3 - 16) + 1.596 * (v2 - 128);
        g = 1.164 * (y3 - 16) - 0.813 * (v2 - 128) - 0.391 * (u2 - 128);
        b = 1.164 * (y3 - 16) + 2.018 * (u2 - 128);
        *ptr++ = CLAMP(r);
        *ptr++ = CLAMP(g);
        *ptr++ = CLAMP(b);
        //4
        r = 1.164 * (y4 - 16) + 1.596 * (v2 - 128);
        g = 1.164 * (y4 - 16) - 0.813 * (v2 - 128) - 0.391 * (u2 - 128);
        b = 1.164 * (y4 - 16) + 2.018 * (u2 - 128);
        *ptr++ = CLAMP(r);
        *ptr++ = CLAMP(g);
        *ptr++ = CLAMP(b);
        //5
        r = 1.164 * (y5 - 16) + 1.596 * (v3 - 128);
        g = 1.164 * (y5 - 16) - 0.813 * (v3 - 128) - 0.391 * (u3 - 128);
        b = 1.164 * (y5 - 16) + 2.018 * (u3 - 128);
        *ptr++ = CLAMP(r);
        *ptr++ = CLAMP(g);
        *ptr++ = CLAMP(b);
        //6
        r = 1.164 * (y6 - 16) + 1.596 * (v3 - 128);
        g = 1.164 * (y6 - 16) - 0.813 * (v3 - 128) - 0.391 * (u3 - 128);
        b = 1.164 * (y6 - 16) + 2.018 * (u3 - 128);
        *ptr++ = CLAMP(r);
        *ptr++ = CLAMP(g);
        *ptr++ = CLAMP(b);

        pass++;
    }

    return rgb;
}

结果(顶部图像是我的结果,底部是它应该是什么): enter image description here

感谢您的帮助。

1 个答案:

答案 0 :(得分:7)

您的U和V值不正确。

像素(ij)的U和V值为:

u[((j / 2) * (width / 2)) + (i / 2)];
v[((j / 2) * (width / 2)) + (i / 2)];

所以,尝试这样的循环:

for (int j = 0; j < height; j++) {
    for (int i = 0; i < width; i++) {
        int yy = y[(j * width) + i];
        int uu = u[((j / 2) * (width / 2)) + (i / 2)];
        int vv = v[((j / 2) * (width / 2)) + (i / 2)];

        r = 1.164 * (yy - 16) + 1.596 * (vv - 128);
        g = 1.164 * (yy - 16) - 0.813 * (vv - 128) - 0.391 * (uu - 128);
        b = 1.164 * (yy - 16) + 2.018 * (uu - 128);
        *ptr++ = CLAMP(r);
        *ptr++ = CLAMP(g);
        *ptr++ = CLAMP(b);
    }
}