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c++ - 如何编码平面 4 :2:0 (fourcc P010)

转载 作者:行者123 更新时间:2023-12-04 22:57:10 24 4
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我正在尝试重新编码fourcc V210 (这是一个打包的 YUV4:2:2 格式)到 P010 (平面 YUV4:2:0)。我想我已经按照规范实现了它,但是渲染器给出了错误的图像,所以有些东西是关闭的。解码 V210 在 ffmpeg 中有一个不错的示例(定义是从他们的解决方案中修改的),但我找不到 P010 编码器来查看我做错了什么。
(是的,我已经尝试过 ffmpeg 并且它有效,但它太慢了,在 Intel Gen11 i7 上每帧需要大约 30 毫秒)
澄清(在@Frank 的问题之后):正在处理的帧是 4k(3840px 宽),因此没有用于进行 128b 对齐的代码。
这是在英特尔上运行的,因此应用了很少的字节序转换。
Try1 - 全绿色图像:
以下代码

#define V210_READ_PACK_BLOCK(a, b, c) \
do { \
val = *src++; \
a = val & 0x3FF; \
b = (val >> 10) & 0x3FF; \
c = (val >> 20) & 0x3FF; \
} while (0)

#define PIXELS_PER_PACK 6
#define BYTES_PER_PACK (4*4)

void MyClass::FormatVideoFrame(
BYTE* inFrame,
BYTE* outBuffer)
{
const uint32_t pixels = m_height * m_width;

const uint32_t* src = (const uint32_t *)inFrame);

uint16_t* dstY = (uint16_t *)outBuffer;

uint16_t* dstUVStart = (uint16_t*)(outBuffer + ((ptrdiff_t)pixels * sizeof(uint16_t)));
uint16_t* dstUV = dstUVStart;

const uint32_t packsPerLine = m_width / PIXELS_PER_PACK;

for (uint32_t line = 0; line < m_height; line++)
{
for (uint32_t pack = 0; pack < packsPerLine; pack++)
{
uint32_t val;
uint16_t u, y1, y2, v;

if (pack % 2 == 0)
{
V210_READ_PACK_BLOCK(u, y1, v);
*dstUV++ = u;
*dstY++ = y1;
*dstUV++ = v;

V210_READ_PACK_BLOCK(y1, u, y2);
*dstY++ = y1;
*dstUV++ = u;
*dstY++ = y2;

V210_READ_PACK_BLOCK(v, y1, u);
*dstUV++ = v;
*dstY++ = y1;
*dstUV++ = u;

V210_READ_PACK_BLOCK(y1, v, y2);
*dstY++ = y1;
*dstUV++ = v;
*dstY++ = y2;
}
else
{
V210_READ_PACK_BLOCK(u, y1, v);
*dstY++ = y1;

V210_READ_PACK_BLOCK(y1, u, y2);
*dstY++ = y1;
*dstY++ = y2;

V210_READ_PACK_BLOCK(v, y1, u);
*dstY++ = y1;

V210_READ_PACK_BLOCK(y1, v, y2);
*dstY++ = y1;
*dstY++ = y2;
}
}
}

#ifdef _DEBUG

// Fully written Y space
assert(dstY == dstUVStart);

// Fully written UV space
const BYTE* expectedVurrentUVPtr = outBuffer + (ptrdiff_t)GetOutFrameSize();
assert(expectedVurrentUVPtr == (BYTE *)dstUV);

#endif
}

// This is called to determine outBuffer size
LONG MyClass::GetOutFrameSize() const
{
const LONG pixels = m_height * m_width;

return
(pixels * sizeof(uint16_t)) + // Every pixel 1 y
(pixels / 2 / 2 * (2 * sizeof(uint16_t))); // Every 2 pixels and every odd row 2 16-bit numbers
}
导致所有绿色图像。根据 P010 规范,将 10 位放置在 16 位值的高位中,结果证明这是一个丢失的位移位。
尝试 2 - Y 有效,UV 加倍?
将代码更新为正确(或者我认为)将 YUV 值移动到其 16 位空间中的正确位置。
#define V210_READ_PACK_BLOCK(a, b, c) \
do { \
val = *src++; \
a = val & 0x3FF; \
b = (val >> 10) & 0x3FF; \
c = (val >> 20) & 0x3FF; \
} while (0)


#define P010_WRITE_VALUE(d, v) (*d++ = (v << 6))

#define PIXELS_PER_PACK 6
#define BYTES_PER_PACK (4 * sizeof(uint32_t))

// Snipped constructor here which guarantees that we're processing
// something which does not violate alignment.

void MyClass::FormatVideoFrame(
const BYTE* inBuffer,
BYTE* outBuffer)
{
const uint32_t pixels = m_height * m_width;
const uint32_t aligned_width = ((m_width + 47) / 48) * 48;
const uint32_t stride = aligned_width * 8 / 3;

uint16_t* dstY = (uint16_t *)outBuffer;

uint16_t* dstUVStart = (uint16_t*)(outBuffer + ((ptrdiff_t)pixels * sizeof(uint16_t)));
uint16_t* dstUV = dstUVStart;

const uint32_t packsPerLine = m_width / PIXELS_PER_PACK;

for (uint32_t line = 0; line < m_height; line++)
{
// Lines start at 128 byte alignment
const uint32_t* src = (const uint32_t*)(inBuffer + (ptrdiff_t)(line * stride));

for (uint32_t pack = 0; pack < packsPerLine; pack++)
{
uint32_t val;
uint16_t u, y1, y2, v;

if (pack % 2 == 0)
{
V210_READ_PACK_BLOCK(u, y1, v);
P010_WRITE_VALUE(dstUV, u);
P010_WRITE_VALUE(dstY, y1);
P010_WRITE_VALUE(dstUV, v);

V210_READ_PACK_BLOCK(y1, u, y2);
P010_WRITE_VALUE(dstY, y1);
P010_WRITE_VALUE(dstUV, u);
P010_WRITE_VALUE(dstY, y2);

V210_READ_PACK_BLOCK(v, y1, u);
P010_WRITE_VALUE(dstUV, v);
P010_WRITE_VALUE(dstY, y1);
P010_WRITE_VALUE(dstUV, u);

V210_READ_PACK_BLOCK(y1, v, y2);
P010_WRITE_VALUE(dstY, y1);
P010_WRITE_VALUE(dstUV, v);
P010_WRITE_VALUE(dstY, y2);
}
else
{
V210_READ_PACK_BLOCK(u, y1, v);
P010_WRITE_VALUE(dstY, y1);

V210_READ_PACK_BLOCK(y1, u, y2);
P010_WRITE_VALUE(dstY, y1);
P010_WRITE_VALUE(dstY, y2);

V210_READ_PACK_BLOCK(v, y1, u);
P010_WRITE_VALUE(dstY, y1);

V210_READ_PACK_BLOCK(y1, v, y2);
P010_WRITE_VALUE(dstY, y1);
P010_WRITE_VALUE(dstY, y2);
}
}
}

#ifdef _DEBUG

// Fully written Y space
assert(dstY == dstUVStart);

// Fully written UV space
const BYTE* expectedVurrentUVPtr = outBuffer + (ptrdiff_t)GetOutFrameSize();
assert(expectedVurrentUVPtr == (BYTE *)dstUV);

#endif
}
这导致 Y 是正确的,并且 U 和 V 的行数也是正确的,但不知何故 U 和 V 没有正确重叠。它有两个版本似乎通过中心垂直镜像。将 V 归零时类似但不太明显的东西。所以这两个都以一半的宽度呈现?任何提示表示赞赏:)
修复:
发现了错误,我不是按包而是按 block 翻转 VU
if (pack % 2 == 0)
应该
if (line % 2 == 0)

最佳答案

有 2 个错误。第一个是因为我没有按照规范的要求将 10 位值推到更高位。第二个是因为我不是按奇数行而是按奇数包编写 UV。
把它留在这里是为了迪斯科效果值,也许其他人需要玩这个并走同样的路。我了解到即使在完全未知的领域中,“只需遵循规范”也可以使用 :) 感谢所有看过它的人。

关于c++ - 如何编码平面 4 :2:0 (fourcc P010),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/68433598/

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