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从串行端口读取时,boost :: asio会阻止std :: cout

[英]boost::asio blocks std::cout when reading from serial port

I'm using boost::asio to read from a serial port. 我正在使用boost::asio从串行端口读取。 I continuously read from the serial port and print it out through std::cout . 我不断从串口读取数据,并通过std::cout其打印出来。 But some strange things happens. 但是有些奇怪的事情发生了。

I'm using TimeoutSerial class from here . 我从这里使用TimeoutSerial类。 My code goes like this: 我的代码是这样的:

#include <iostream>
#include "TimeoutSerial.h"

using namespace std;

int main(){  
  TimeoutSerial serial;

  serial.open("/dev/ttyACM0", 9600 );      
  serial.setTimeout(boost::posix_time::seconds(1));

  char c = '0';
  while(true){
    try{
      serial.read( &c, 1 );
      cout << c;
    }catch( std::runtime_error err ){
      cout << "Error: " << err.what()<< endl;
    }
  }         
  return 0;
}

I get no output and I have no idea why. 我没有输出,我也不知道为什么。 When I change cout << c; 当我改变cout << c; to cout << c << endl; cout << c << endl; I get the output I want but each character is on a new line which is undesirable. 我得到了想要的输出,但是每个字符都在换行符上,这是不希望的。

So can anyone tell me why is this happening? 谁能告诉我为什么会这样?

std::cout is buffered by default, so you need to flush it to display it on your terminal. std::cout默认情况下是缓冲的,因此您需要刷新它才能在终端上显示它。 Using std::endl does this implicitly, as will std::flush() . 使用std::endlstd::flush()都会隐式执行此操作。 Change your loop to 将循环更改为

while(true) {
    try {
      serial.read( &c, 1 );
      cout << c;
      std::flush( std::cout );
    } catch ( std::runtime_error err ) {
      cout << "Error: " << err.what() << endl;
    }
}      

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