C++ ifstream 从 linux 到 arduino [英] C++ ifstream from linux to arduino

问题描述

原始代码

#include<iostream>
#include<fstream>
using namespace std;

int main()
{
    ofstream arduino_output("/dev/ttyACM0");
    ifstream arduino_input("/dev/ttyACM0");

    int value;
    string txt;

    while(cin >> value){
        arduino_output << value << endl;
        arduino_input >> txt;//I never recieve the "OK" (Which I should get)
        cout << txt;
    }

    arduino_input.close();
    arduino_output.close();
    return(0);
} 

问题来了:

        cin >> value;
        arduino_output << value << endl;
        arduino_input >> txt;//I never recieve the "OK" (Which I should get)
        cout << txt;

但如果我这样做，它会起作用:

but if I do this instead it works:

        cin >> value;
        arduino_output << value << endl;

        for(int i=0;i<10000;++i)
        for(int j=0;j<10000;++j){ //Waste a lot of time
           ++value;
           --value;
        }

        arduino_input >> txt; //I always recieve the "OK"
        cout << txt; //I can see the "OK"

那么如何让我的快速计算机能够读取来自 arduino 的慢输出?(不使用for循环浪费时间)

So how do I make my fast computer able to read the slow output from the arduino? (Without using for-loops to waste time)

这里说一些关于回调的事情 http://www.cplusplus.com/reference/ios/ios_base/register_callback/ 但我永远无法让它工作.它说它支持 3 个事件，但没有一个是:如果输入缓冲区不为空，则调用此函数".

Here it says some things about callback http://www.cplusplus.com/reference/ios/ios_base/register_callback/ but I could never get it to work. It says it supports 3 events, and none of them are: "If input buffer is not empty, call this function".

因为最终的解决方案是在输入缓冲区不为空时使用回调函数.

Because the ultimate solution would be a callback function for whenever the input buffer is not empty.

可接受的解决方案是 arduino 版本Serial.available()"的 C++ 等效版本.

An acceptable solution would be a c++ equivalent version of the arduino version "Serial.available()".

另一个可接受的解决方案是任何迫使我不依赖两个 for 循环的东西.如果这就是您的想法，那么 3 个 for 循环是不可接受的.

Another acceptable solution would be anything that forces me to not rely on two for-loops. 3 for-loops is not acceptable if that's what you're thinking.

显示原始代码
我正在使用 linux(lubuntu)
有人对代码的编写位置感到困惑.奇怪.

Showed the original code
I am using linux(lubuntu)
Someone got confused where the code was written. Strange.

推荐答案

你的问题很奇怪.一般来说，问题在于较慢的一方无法读取较快的一方发送的内容.所以，您这里似乎有一个更根本的问题.

Your problem is weird. In general, the problem is that the slower party can't read what the faster party sends. So, it seems you have a more fundamental problem here.

如果 arduino_output 是串行端口 (UART) 的表示，我建议使用特定于平台的访问方式.在 Windows 上，有 UART 函数，在 Linux 上有 termios(可能在大多数其他类似 POSIX 的系统上也是如此).这将为您提供一种方法来控制通信参数，并获取有关事件(包括奇偶校验/帧错误)的信息和/或通知.

If arduino_output is a representation of a serial port (UART), I suggest to use a platform specific way of accessing it. On Windows, there are UART functions, and on Linux there's termios (probably on most other POSIX-like, too). This will give you a way to control the parameters of communication, and get information and/or notification about events (including parity/framing errors).

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