Serial.print strange problem

Forum 2005-2010 (read only) Software Troubleshooting
1

Hello guys,
i have recently run into this problem, and i don't know why it happens. If i use this code, everything works fine:

if(Serial.available()){
    for(i=0;i<3;i++){
      incoming=Serial.read();
      buffer[i]=incoming;
      Serial.println(buffer[i]);
    }
    Serial.println(buffer[0]& 0xf);
    Serial.println(buffer[1]& 0xf);
    Serial.println(buffer[2]& 0xf);
    
  }

and, if i send "123", output is like:
49
50
51
1
2
3

but....if i take off this line Serial.println(buffer[i]); (which is found in the for routine in the above code, it stops working and i get data like:
1
15
15
2
3
15

(data sent is still "123"). I have no idea why this happens! Btw, 0xf is 15 in decimal.

Thank you :=)

2

Because the "ln" bit of the print statement tells the computer to output the ASCII codes that put it on a new line, that is a carriage return and a line feed. (CR & LF)

3

I still don't get why that doesn't also happen in the first code though...
and why there are two, and not one, 15s after the "1" is "outputted".

4

Serial.read() returns -1 (0xffff) if there is no character available. You are probably seeing the masked-off 0xf from the -1.

Suppose Serial.available() returns 1 and there really is only one character available. Then the code reads three bytes. One will be the value, and two will be (int) -1, thus printed as 15.

-br

http://entropymouse.com

5

but...why doesn't that happen when the first Serial.println is there?

6

The Serial.println takes time, so the buffer can fill up with the next bytes you need to read. Correct would be, as stated before a few times:

if(Serial.available() [glow]>= 3[/glow]){

Korman