libyuv API MJPGToI420()的示例用法

Question

我正在尝试使用libyuv API，更具体地说是MJPGToI420()。

我想首先将jpeg图像作为MJPGToI420()的输入，其签名如下：

int MJPGToI420(const uint8_t* sample,
size_t sample_size,
uint8_t* dst_y,
int dst_stride_y,
uint8_t* dst_u,
int dst_stride_u,
uint8_t* dst_v,
int dst_stride_v,
int src_width,
int src_height,
int dst_width,
int dst_height);

然后，我想为dst_y、dst_u和dst_v指针分配空间。然而，我不知道该为它们分配多少空间。我还对步幅应该是什么感到困惑，即参数dst_stride_y、dst_stride_u和dst_stride_v应该是什么

如果有任何正确的指示，我将不胜感激。

EDIT:以下是使用此函数的libyuv源单元测试中的一段代码。然而，测试返回1，这是作为预期行为的功能的失败。该测试也只是对数据使用零，而不是实际的MJPG文件。

TEST_F(LibYUVConvertTest, MJPGToI420) {
const int kOff = 10;
const int kMinJpeg = 64;
const int kImageSize = benchmark_width_ * benchmark_height_ >= kMinJpeg
? benchmark_width_ * benchmark_height_
: kMinJpeg;
const int kSize = kImageSize + kOff;
align_buffer_page_end(orig_pixels, kSize);
align_buffer_page_end(dst_y_opt, benchmark_width_ * benchmark_height_);
align_buffer_page_end(dst_u_opt, SUBSAMPLE(benchmark_width_, 2) *
SUBSAMPLE(benchmark_height_, 2));
align_buffer_page_end(dst_v_opt, SUBSAMPLE(benchmark_width_, 2) *
SUBSAMPLE(benchmark_height_, 2));
// EOI, SOI to make MJPG appear valid.
memset(orig_pixels, 0, kSize);
orig_pixels[0] = 0xff;
orig_pixels[1] = 0xd8;  // SOI.
orig_pixels[kSize - kOff + 0] = 0xff;
orig_pixels[kSize - kOff + 1] = 0xd9;  // EOI.
for (int times = 0; times < benchmark_iterations_; ++times) {
int ret =
MJPGToI420(orig_pixels, kSize, dst_y_opt, benchmark_width_, dst_u_opt,
SUBSAMPLE(benchmark_width_, 2), dst_v_opt,
SUBSAMPLE(benchmark_width_, 2), benchmark_width_,
benchmark_height_, benchmark_width_, benchmark_height_);
// Expect failure because image is not really valid.
EXPECT_EQ(1, ret);
}
free_aligned_buffer_page_end(dst_y_opt);
free_aligned_buffer_page_end(dst_u_opt);
free_aligned_buffer_page_end(dst_v_opt);
free_aligned_buffer_page_end(orig_pixels);
}

EDIT 2:此外，这也是我尝试过的，但是，在yuv查看器(使用缓冲区dst_u_opt和dst_y_opt创建(中甚至看不到结束的yuv文件，这让我相信可能是我把函数搞砸了：

int convertMJPGToI420() {
auto fileSize = filesize(IMG_NAME);
// load image into memory
uint8_t* my_img = (uint8_t*) calloc(fileSize, 1);
std::ifstream fin(IMG_NAME, ios::in | ios::binary);
fin.read(reinterpret_cast<char*>(my_img), fileSize);
// exif data offset 
// This is the size of the exif data
const int kOff = 4096;
// 4k image is being sent in
int benchmark_width_ = 3840;
int benchmark_height_ = 2160;
const int kSize = fileSize;
// align_buffer_page_end is a macro (look at link posted for unit tests above)
// I'm not sure if the size allocation for these is correct
// I have tried to model it based off the example
align_buffer_page_end(orig_pixels, kSize);
align_buffer_page_end(dst_y_opt, benchmark_width_ * benchmark_height_);
align_buffer_page_end(dst_u_opt, SUBSAMPLE(benchmark_width_, 2) * 
SUBSAMPLE(benchmark_height_, 2));
align_buffer_page_end(dst_v_opt, SUBSAMPLE(benchmark_width_, 2) * 
SUBSAMPLE(benchmark_height_, 2));
// EOI, SOI to make MJPG appear valid
memset(orig_pixels, 0, kSize);
orig_pixels[0] = 0xff;
orig_pixels[1] = 0xd8; // SOI
memcpy(orig_pixels + 2, my_img, kSize - kOff - 3);
orig_pixels[kSize - kOff + 0] = 0xff;
orig_pixels[kSize - kOff + 1] = 0xd9; // EOI
// using async as this function might be ansynchronous
std::future<int> ret = std::async(libyuv::MJPGToI420, orig_pixels, kSize, dst_y_opt, benchmark_width_,
dst_u_opt, SUBSAMPLE(benchmark_width_, 2),
dst_v_opt, SUBSAMPLE(benchmark_width_, 2), 
benchmark_width_, benchmark_height_, 
benchmark_width_, benchmark_height_);
ret.wait();
// ret is always one, which means there was a failure
if(ret.get() == 0) {
cout << "return value was zero" << endl;
} else {
cout << "return value was one" << endl;
}
FILE* file = fopen("/data/dst_u_opt", "wb");
fwrite(dst_y_opt, 1, SUBSAMPLE(benchmark_width_, 2) * SUBSAMPLE(benchmark_height_, 2) , file);
fclose(file);
file = fopen("/data/dst_v_opt", "wb");
fwrite(dst_y_opt, 1, SUBSAMPLE(benchmark_width_, 2) *  SUBSAMPLE(benchmark_height_, 2), file);
fclose(file);
free_aligned_buffer_page_end(dst_y_opt);
free_aligned_buffer_page_end(dst_u_opt);
free_aligned_buffer_page_end(dst_v_opt);
free_aligned_buffer_page_end(orig_pixels);
return 0;
}

Answer 1

您需要知道jpeg的宽度和高度。

I420是420个子采样的YUV。Y平面是以字节为单位的宽度*高度。dstrongtride_y值为宽度例如

char* dst_y = malloc(width * height);

U平面和V平面是宽度和高度的一半。要处理奇数尺寸，你应该四舍五入。

dst_stride_u = (width + 1) / 2;
dst_stride_v = (width + 1) / 2;

u和v平面是((宽度+1(/2(*(高度+1(/2个字节。

char* dst_u = malloc(((width + 1) / 2) * ((height + 1) / 2));
char* dst_y = malloc(((width + 1) / 2) * ((height + 1) / 2));

如果你想提交一个问题，包括更好的文档，请在这里发布：https://bugs.chromium.org/p/libyuv/issues/list