我在 1.txt,2.txt ... 100.txt 等文本文件中编写了一个具有不同值(100次)的C结构
我在Linux上使用Intel TBB。我创造了:
InputFilter从文件读取结构,并将其传递给TransformFilter。
TrasnformFilter更新结构值,并将其传递给OutputFilter。
OutputFilter将新结构写入光盘。
基本上,它是用于结构的简单读写应用程序。
class InputFilter: public tbb::filter {
public:
InputFilter( int );
~InputFilter();
private:
int total_streams;
int count;
struct video_process_object input_obj;
void* operator()( void* );
};
InputFilter::InputFilter( int x )
: filter( serial_in_order ) {
total_streams = x;
count = 1;
}
InputFilter::~InputFilter() {
total_streams = 0;
}
void* InputFilter::operator()( void* ) {
char path[50] = { };
sprintf( path, "input//%d.txt", count );
printf( "Path : %s\n", path );
FILE *fp;
fp = fopen( path, "r" );
if( fp == NULL || count > total_streams ) {
fclose( fp );
printf( "\n*******Cannot find more data.Terminating********\n\n\n" );
return NULL;
}
fscanf( fp, "%d", &input_obj.video_id );
fscanf( fp, "%s", &input_obj.storage_url );
fscanf( fp, "%s", &input_obj.storage_type );
fscanf( fp, "%d", &input_obj.face_detect );
fscanf( fp, "%d", &input_obj.face_recognise );
fscanf( fp, "%d", &input_obj.scene_recognise );
fscanf( fp, "%d", &input_obj.activity_recognise );
fscanf( fp, "%d", &input_obj.speech_recognise );
fclose( fp );
count++;
return &input_obj;
}
class TransformFilter: public tbb::filter {
public:
TransformFilter();
~TransformFilter();
private:
struct video_process_object input_transform;
void* operator()( void* );
};
TransformFilter::TransformFilter()
: filter( serial_in_order ) {
}
TransformFilter::~TransformFilter() {
}
void* TransformFilter::operator()( void *item ) {
input_transform = *static_cast<struct video_process_object*>( item );
input_transform.video_id += 1000;
strcat( input_transform.storage_url, " nabeel" );
strcat( input_transform.storage_type, " N" );
input_transform.face_detect += 1000;
input_transform.face_recognise += 1000;
return &input_transform;
}
class OutputFilter: public tbb::filter {
public:
OutputFilter();
~OutputFilter();
private:
struct video_process_object output_obj;
void* operator()( void* );
};
OutputFilter::OutputFilter()
: filter( serial_in_order ) {
int status = mkdir( "output", S_IRWXU | S_IRWXG | S_IRWXO );
if( status == -1 )
printf( "\nOutput directory already exists\n\n" );
}
OutputFilter::~OutputFilter() {
}
void* OutputFilter::operator()( void *item ) {
output_obj = *static_cast<struct video_process_object*>( item );
FILE *fp;
char path[50] = { };
sprintf( path, "output//%d.txt", output_obj.video_id - 1000 );
printf( "Output Path : %s\t\t %d\n\n", path, output_obj.video_id );
if( (fp = fopen( path, "w" )) == NULL ) {
fprintf( stderr, "Cannot open output file.\n" );
return NULL;
}
fprintf( fp, "%d\n", output_obj.video_id );
fprintf( fp, "%s\n", output_obj.storage_url );
fprintf( fp, "%s\n", output_obj.storage_type );
fprintf( fp, "%d\n", output_obj.face_detect );
fprintf( fp, "%d\n", output_obj.face_recognise );
fprintf( fp, "%d\n", output_obj.scene_recognise );
fprintf( fp, "%d\n", output_obj.activity_recognise );
fprintf( fp, "%d\n", output_obj.speech_recognise );
fclose( fp );
return NULL;
}
int main() {
tbb::pipeline pipeline;
InputFilter input_filter( 100 );
pipeline.add_filter( input_filter );
TransformFilter transform_filter;
pipeline.add_filter( transform_filter );
OutputFilter output_filter;
pipeline.add_filter( output_filter );
tbb::tick_count t0 = tbb::tick_count::now();
tbb::task_scheduler_init init_parallel;
pipeline.run( 1 );
tbb::tick_count t1 = tbb::tick_count::now();
return 0;
}
少量文件(例如5或10),一切正常。当我读取大量文件(例如50或100)时,问题开始。问题是:
我如何确保不会发生这种情况?
最佳答案
有一个数据争用,因为video_process_objects放置在过滤器结构内,并通过过滤器之间的引用(当然是并行运行)传递的。因此,当InputFilter
开始处理下一个 token 将新数据读入其video_process_object时,第一个 token 刚刚开始通过TransformFilter
中的相同地址读取数据时,您将遇到以下情况:
Token 1 || Token 2
input_filter.operator() ||
transform_filter.operator() || input_filter.operator()
...
要解决此问题,请动态分配数据,例如:
struct video_process_object *input_obj_ptr = new video_process_object;
fscanf( fp, "%d", &input_obj_ptr->video_id );
...
return input_obj_ptr;
并在最后一个过滤器中取消分配它,因为无论如何它的返回值都会被忽略。
this old presentation中的幻灯片49-50描绘了类似的代码。
最后,让我挑战您对tbb::pipelene和serial_in_order过滤器类型的选择。 TBB Reference说:
由于处理和文件是独立的,我认为没有理由将这个额外的“按顺序”限制。为了结构更好的代码,还要考虑另一个引号:
关于c++ - TBB管道输出不正确,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/23356308/