我已经在多个地方读过，但是在使用时没有刷新缓冲区，但是在我的代码中，我将在这个问题的最后添加，它似乎只是这样做，或者至少看起来是这样的输出（由于我如何执行我的couts，可能还有其他事情在后台进行？）。
预期产出：
采矿激光1循环将在x秒内完成......
采矿激光2循环将在x秒内完成......
采矿激光3循环将在x秒内完成......
采矿激光4循环将在x秒内完成......
我在CLI中获得的内容：
采矿激光1循环将在x秒内完成......
采矿激光2循环将在x秒内完成......
采矿激光3循环将在x秒内完成......
采矿激光4循环将在x秒内完成......
采矿激光1循环将在x秒内完成......
采矿激光2循环将在x秒内完成......
采矿激光3循环将在x秒内完成......
采矿激光4循环将在x秒内完成......
采矿激光1循环将在x秒内完成......
采矿激光2循环将在x秒内完成......
采矿激光3循环将在x秒内完成......
采矿激光4循环将在x秒内完成......
我希望输出做的是保持原样，就像预期的输出示例中那样，并且只是在我的代码执行的每个时间循环中更新自己。
这是我的代码：
#include <iostream>
#include <Windows.h>
#include <string>
#include <vector>
#include <random>
#include <thread>
#include <future>
using namespace std; //Tacky, but good enough fo a poc D:
class mLaser
{
public:
mLaser(int clen, float mamt, int time_left)
{
mlCLen = clen;
mlMAmt = mamt;
mCTime_left = time_left;
}
int getCLen()
{
return mlCLen;
}
float getMAmt()
{
return mlMAmt;
}
void setMCOld(int old)
{
mCTime_old = old;
}
void mCycle()
{
int mCTime_new = GetTickCount(); //Get current tick count for comparison to mCOld_time
if (mCTime_old != ((mCTime_new + 500) / 1000)) //Do calculations to see if time has passed since mCTime_old was set
{
//If it has then update mCTime_old and remove one second from mCTime_left.
mCTime_old = ((mCTime_new + 500) / 1000);
mCTime_left -= 1000;
}
cur_time = mCTime_left;
}
int getCTime()
{
return cur_time;
}
int getCTLeft()
{
return mCTime_left;
}
private:
int mlCLen; //Time of a complete mining cycle
float mlMAmt; //Amoung of ore produced by one mining cycle (not used yet)
int cur_time; //The current time remaining in the current mining cycle; will be removing this as it is just a copy of mCTime_left that I was going to use for another possiblity to make this code work
int mCTime_left; //The current time remaining in the current mining cycle
int mCTime_old; //The last time that mCycle was called
};
void sMCycle(mLaser& ml, int i1, thread& _thread); //Start a mining cycle thread
//Some global defines
random_device rd;
mt19937 gen(rd());
uniform_int_distribution<> laser(1, 3); //A random range for the number of mlaser entities to use
uniform_int_distribution<> cLRand(30, 90); //A random time range in seconds of mining cycle lengths
uniform_real_distribution<float> mARand(34.0f, 154.3f); //A random float range of the amount of ore produced by one mining cycle (not used yet)
int main()
{
//Init some variables for later use
vector<mLaser> mlasers; //Vector to hold mlaser objects
vector<thread> mthreads; //Vector to hold threads
vector<shared_future<int>> futr; //Vector to hold shared_futures (not used yet, might not be used if I can get the code working like this)
int lasers; //Number of lasers to create
int cycle_time; //Mining cycle time
int active_miners = 0; //Number of active mining cycle threads (one for each laser)
float mining_amount; //Amount of ore produced by one mining cycle (not used yet)
lasers = laser(gen); //Get a random number
active_miners = lasers; //Set this to that random number for the while loop later on
//Create the mlaser objects and push them into the mlasers vector
for (int i = 0; i < lasers; i++)
{
int clength = cLRand(gen);
mlasers.push_back(mLaser(clength, mARand(gen), (clength * 1000)));
//Also push thread obects into mthreads for each laser object
mthreads.push_back(thread());
}
//Setup data for mining cycles
for (int i = 0; i < mlasers.size(); i++)
{
int mCTime_start = GetTickCount(); //Get cycle start time
mlasers.at(i).setMCOld(((mCTime_start + 500) / 1000));
}
//Print initial display for mining cycles
for (int i = 0; i < mlasers.size(); i++)
{
cout << "Mining Laser " << i+1 << " cycle will complete in " << (mlasers.at(i).getCTLeft() + 500) / 1000 << " seconds...\n";
}
while (active_miners > 0)
{
for (int i = 0; i < mlasers.size(); i++)
{
//futr.push_back(async(launch::async, [mlasers, i, &mthreads]{return sMCycle(mlasers.at(i), i + 1, mthreads.at(i)); }));
async(launch::async, [&mlasers, i, &mthreads]{return sMCycle(mlasers.at(i), i + 1, mthreads.at(i)); }); //Launch a thread for the current mlaser object
//mthreads.at(i) = thread(bind(&mLaser::mCycle, ref(mlasers.at(i)), mlasers.at(i).getCLen(), mlasers.at(i).getMAmt()));
}
//Output information from loops
cout << " \r" << flush; //Return cursor to start of line and flush the buffer for the next info
for (int i = 0; i < mlasers.size(); i++)
{
if ((mlasers.at(i).getCTLeft() != 0) //If mining cycle is not completed
{
cout << "Mining Laser " << i + 1 << " cycle will complete in " << (mlasers.at(i).getCTLeft() + 500) / 1000 << " seconds...\n";
}
else //If it is completed
{
cout << "Mining Laser " << i + 1 << " has completed its mining cycle!\n";
active_miners -= 1;
}
}
}
/*for (int i = 0; i < mthreads.size(); i++)
{
mthreads.at(i).join();
}*/
//string temp = futr.get();
//float out = strtof(temp.c_str(),NULL);
//cout << out << endl;
system("Pause");
return 0;
}
void sMCycle(mLaser& ml, int i1,thread& _thread)
{
//Start thread
_thread = thread(bind(&mLaser::mCycle, ref(ml)));
//Join the thread
_thread.join();
}
Per Ben Voigt，似乎\ r \ n不能以我尝试使用它的方式使用。除了Matthew建议每次更新时都关闭命令窗口之外，有没有人有任何其他建议？也许在Boost中有什么东西或c ++ 11的新东西？
感谢。