xcode，带有arduino的c ++串口

Question

i'm making a very simple c++ program which send an angle to arduino through a serial port and then arduino apply that angle to a servo-motor. I know that Unix see serial ports device like a file, in fact this is the c++ code:

#include <iostream>
#include <unistd.h>

using namespace std;

int main()
{
    int angole;
    FILE * arduino;

    do
    {
        arduino = fopen("/dev/tty.usbmodem3a21","w");

        cout<<"\n\give me the angle\n\n";
        cin>>angole;

        fprintf(arduino,"%d",angole);
        sleep(1);

    }while(angole>=0 && angole<=179);

}

and this is arduino's:

#include <Servo.h>

Servo servo;
const int pinServo = 2;
int angle;

void setup()
{
    Serial.begin(9600);
    servo.attach(pinServo);
    servo.write(0);

}

void loop()
{
    if(Serial.available()>0)
    {  
       angle = Serial.read();
      servo.write(angle);
    }
}

i also checked in the arduino app, in tools>serial port>/div/tty.usbmodem3a21 that it was the right port.

The problem is that the program stops at arduino = fopen("/dev/tty.usbmodem3a21","w"); because it doesn't even write the message "give me the angle".

for instance , when i write the wrong port in the open function , it writes the message.

Answer 1

Indeed, " everything in Linux is a file ", but not literally --> the essence is which type of file - in your case you treat the port as plain vanilla file (ie something like txt file) while you need treating it as a device file, so no fopen but :

fd = open("/dev/tty.usbmodem3a21", O_RDWR | O_NOCTTY | O_NDELAY);

The following is a good reference about the file interface of serial ports And this one is even arduino oriented

Answer 2

我得到了这段代码的连接：

arduino = open("/dev/tty.usbmodemfa131", O_RDWR | O_NOCTTY | O_NDELAY);