打开串行端口ANSI C.

Question

I have been googling my last couple of hours(days) to find a simple, working example of how one opens a serial port in ANSI C on Windows. I try to open a txt file and send it to the serial port char by char. (Also I would be happy if you could give me a hint how to send char by char with 1 second delay!) This is my code but somehow it won't work :(

FILE *file;
file = fopen("text.txt", "r");
if (file)
{
    while ((c = getc(file)) != EOF)
    {
        FILE *fp;
        fp = open("COM1", O_RDWR | O_NOCTTY | O_NDELAY);
        if (fp == -1) puts("Couldn't open port!");
        else puts("Port opened!");
        fprintf(fp,c);  
    }
    fclose(file);
}
else puts("Couldn't find text.txt!");

Answer 1

The C standard does not know about serial ports. Your question is Windows specific. Perhaps you want to

         fp = open("COM1:", O_RDWR | O_NOCTTY | O_NDELAY);
         if (fp<0) perror("open COM1:");

I'm not sure that open , O_NDELAY exist on Windows (looks like Posix or Linux code!)

Answer 2

I don't think the code you have written will work in Windows, unless maybe you are using some POSIX library that I don't know about. Otherwise, you have to use the Windows API functions used for serial communication.

All documentation and some examples are found at Microsoft's documentation site MSDN .

Windows regards serial ports as files. You open them with CreateFile() and read/write with ReadFile() and WriteFile(). And then there are various functions used for setting up baudrate, handshaking, timeouts etc.

How to do this in detail is quite a big topic, but at least the information on MSDN should get you started.