next topic »

[on:]

Hey everyone!

I am trying to recieve multiple bytes over serial, store in array, and use atoi to convert to int. However I am getting very different results from atoi

Using atoi like this:
if( Serial.available() >= 3 ) {
  rred[0] = Serial.read();
  rred[1] = Serial.read();
  rred[2] = Serial.read();
  char * thisChar = rred;
  int a = atoi(thisChar);
}

Attaching a pic of terminal, so you can se what it outputs.
gratisupload.dk/download/45675/

As you can see, it works after sending 000 command some times, but only once.

Please help!!

Best regards, Jonas Hald

[Reply #1 on:]

[Reply #2 on:]

Code:

char    buffer[4];
if ( Serial.available() >= 3 )
{
    char*   pbuffer = buffer;
    *pbuffer++ = Serial.read();
    *pbuffer++ = Serial.read();
    *pbuffer++ = Serial.read();
    *pbuffer++ = 0;

    int     n;
    n = atoi(buffer);
}

[Reply #3 on:]

Or, if you're averse to explicit pointer arithmetic and extraneous variables:

Code:

char    buffer[4];
if ( Serial.available() >= 3 )
{
    buffer[0] = Serial.read();
    buffer[1] = Serial.read();
    buffer[2] = Serial.read();
    buffer[3] = '\0';

    int     n;
    n = atoi(buffer);
}

[Reply #4 on:]

Thank you very much!! Both of you!

[Reply #5 on:]

The short answer is you can do:

char * thisChar = rred;

You'd have to use strcpy or something else to copy the array to the other one.

[Reply #6 on:]

The short answer is you can do:

char * thisChar = rred;

He already is doing that - what's your point?

There's nothing wrong with the code in the OP, except there's no "rred[3] ='\0';" before the "char * thisChar = rred;"  (assuming "rred" is declared as having at least four elements.)