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c++ -\r 和\n 或缓冲区刷新问题

转载 作者:太空宇宙 更新时间:2023-11-04 13:40:37 35 4
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我在多个地方读到\n 在使用时不会刷新缓冲区,但是在我的代码中,我将在这个问题的末尾添加,它似乎就是这样做的,或者至少看起来是这样的输出(由于我执行 couts 的方式,可能在后台发生了其他事情?)。

预期输出:

Mining Laser 1 cycle will complete in x seconds...

Mining Laser 2 cycle will complete in x seconds...

Mining Laser 3 cycle will complete in x seconds...

Mining Laser 4 cycle will complete in x seconds...

我在 CLI 中得到的结果:

Mining Laser 1 cycle will complete in x seconds...

Mining Laser 2 cycle will complete in x seconds...

Mining Laser 3 cycle will complete in x seconds...

Mining Laser 4 cycle will complete in x seconds...

Mining Laser 1 cycle will complete in x seconds...

Mining Laser 2 cycle will complete in x seconds...

Mining Laser 3 cycle will complete in x seconds...

Mining Laser 4 cycle will complete in x seconds...

Mining Laser 1 cycle will complete in x seconds...

Mining Laser 2 cycle will complete in x seconds...

Mining Laser 3 cycle will complete in x seconds...

Mining Laser 4 cycle will complete in x seconds...

我希望输出保持原样,就像预期输出示例中的输出一样,并且在我的代码中每次执行时间循环时更新自身。

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();
}

根据 Ben Voigt 的说法,\r 似乎不能按照我尝试使用它的方式使用。除了 Matthew 每次更新关闭命令窗口的建议之外,还有其他建议吗?也许是 Boost 中的某些东西或 c++11 的新东西?

谢谢。

最佳答案

您可以尝试在每次执行后清空控制台像 system("cls");这是一个帖子的链接[ How can I clear console

关于c++ -\r 和\n 或缓冲区刷新问题,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/27772964/

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