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Only record task id in BGSet #7

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Only record task id in BGSet #7

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13 changes: 7 additions & 6 deletions include/thread_pool.h
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Original file line number Diff line number Diff line change
Expand Up @@ -12,6 +12,7 @@
#include <queue>
#include <sstream>
#include <vector>
#include <set>
#include <boost/function.hpp>
#include "mutex.h"
#include "timer.h"
Expand Down Expand Up @@ -101,7 +102,7 @@ class ThreadPool {
int64_t exe_time = now_time + delay * 1000;
BGItem bg_item(++last_task_id_, exe_time, task);
time_queue_.push(bg_item);
latest_[bg_item.id] = bg_item;
latest_.insert(bg_item.id);
work_cv_.Signal();
return bg_item.id;
}
Expand All @@ -118,7 +119,7 @@ class ThreadPool {
{
MutexLock lock(&mutex_);
if (running_task_id_ != task_id) {
BGMap::iterator it = latest_.find(task_id);
BGSet::iterator it = latest_.find(task_id);
if (it == latest_.end()) {
if (is_running != NULL) {
*is_running = false;
Expand Down Expand Up @@ -194,8 +195,8 @@ class ThreadPool {
int64_t wait_time = (bg_item.exe_time - now_time) / 1000; // in ms
if (wait_time <= 0) {
time_queue_.pop();
BGMap::iterator it = latest_.find(bg_item.id);
if (it != latest_.end() && it->second.exe_time == bg_item.exe_time) {
BGSet::iterator it = latest_.find(bg_item.id);
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@bluebore bluebore Nov 9, 2015

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这块逻辑删了是不是会导致任务执行两次？

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@yvxiang yvxiang Nov 9, 2015

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没太想懂，比如说？

if (it != latest_.end()) {
schedule_cost_sum_ += now_time - bg_item.exe_time;
schedule_count_++;
task = bg_item.task;
Expand Down Expand Up @@ -251,7 +252,7 @@ class ThreadPool {
: id(id_t), exe_time(exe_time_t), task(task_t) {}
};
typedef std::priority_queue<BGItem> BGQueue;
typedef std::map<int64_t, BGItem> BGMap;
typedef std::set<int64_t> BGSet;

int32_t threads_num_;
std::deque<BGItem> queue_;
Expand All @@ -262,7 +263,7 @@ class ThreadPool {
std::vector<pthread_t> tids_;

BGQueue time_queue_;
BGMap latest_;
BGSet latest_;
int64_t last_task_id_;
int64_t running_task_id_;

Expand Down