Arduino serial.read to int

Question

I am trying to read a serial string which comes through as "R0123" for example then I need the 0123 to be in an int. to send out through another method

Here is my code it is not working the way that I think it should

void loop()
{
  if(Serial.available())
  {
    delay(100);
    if(Serial.read() == 'R')
      {
        int r1 = Serial.read();
        int r2 = Serial.read();
        int r3 = Serial.read();
        int r4 = Serial.read();
        int red = ((int)r1 * 1000) + ((int)r2 * 100) + ((int)r3 * 10) + (int)r4;

        sb.sendColour(red,0,0);

        Serial.print(r1,0);
        Serial.print(r2,0);
        Serial.print(r3,0);
        Serial.print(r4,0);
        Serial.print(red);
      }
  }
}

`

Bladt · Accepted Answer · 2012-08-26 09:46:11Z

7

What about using atoi()? You will still need to trim of the 'R', and you might have to pad it with a NULL character at the end ('\0'). But then it should work as:

int rInteger = atoi(s);

when s is a pointer to a null terminated string.

answered Aug 26, 2012 at 9:46

Bladt

1986 bronze badges

\$\begingroup\$ So if i do Serial.read into a string for the four characters then add \o and run in it atoi it should work \$\endgroup\$

Ashley Hughes
– Ashley Hughes

2012-08-26 09:55:56 +00:00
Commented Aug 26, 2012 at 9:55
\$\begingroup\$ I am getting an error ArduinoShiftyWireless.cpp: In function 'void loop()': ArduinoShiftyWireless:45: error: cannot convert 'String' to 'const char*' for argument '1' to 'int atoi(const char*)' \$\endgroup\$

Ashley Hughes
– Ashley Hughes

2012-08-26 10:04:27 +00:00
Commented Aug 26, 2012 at 10:04
\$\begingroup\$ Like I said, atoi() doesn't take a string, it takes a pointer to a string, or to an array of chars, hence the error :) I am not sure if arduino uses ANSI C(?) Never the less, this should be accurate enough: acm.uiuc.edu/webmonkeys/book/c_guide/2.13.html#atoi \$\endgroup\$

Bladt
– Bladt

2012-08-26 10:13:41 +00:00
Commented Aug 26, 2012 at 10:13
\$\begingroup\$ Frankly, I'm a little rusty on my C, bit something like //code to read the 'R' character char* readString; for(int i=0; i<4; i++) { readString[i]=Serial.Read(); } int a = atoi(readString); I imagine would do the trick. \$\endgroup\$

Bladt
– Bladt

2012-08-26 10:43:24 +00:00
Commented Aug 26, 2012 at 10:43
\$\begingroup\$ Ehh formatting? See it here: pastebin.com/Q4GCLh75 \$\endgroup\$

Bladt
– Bladt

2012-08-26 10:49:07 +00:00
Commented Aug 26, 2012 at 10:49

| Show 1 more comment

Wouter van Ooijen · Accepted Answer · 2012-08-26 12:05:17Z

8

You are confusing integer values and ascii character values. When your

  int r1 = Serial.read();

reads a 0 it will read the ASCII character 0. The integer value of the character 0 is (decimal) 48 (check for instance this ascii table). So instead that line should read

  int r1 = Serial.read() - 48;

or even better

  int r1 = Serial.read() - '0';

answered Aug 26, 2012 at 12:05

Wouter van Ooijen

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stevenvh · Accepted Answer · 2012-08-26 10:17:48Z

5

This reads 4 characters after the "R", places them in a char array and appends a null character. Then converts to integer. The code doesn't check if there are at least 4 chars following the "R", or that they are digits, though.

#define numberOfDigits 4
char theNumberString[numberOfDigits + 1];
int theNumber;

if(Serial.read() == 'R')
  {
  for (int i = 0; i < numberOfDigits; theNumberString[i++] = Serial.Read());
  theNumberString[numberOfDigits] = 0x00;
  theNumber = atoi(theNumberString);
}

answered Aug 26, 2012 at 10:17

stevenvh

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NothinRandom · Accepted Answer · 2013-05-02 04:13:33Z

1

You can try:

int intParse(byte length) //return parsed byte from serial
{    
  char BUFFER[length + 1];
  delay(5); //need for Baud 9600, less delay for faster speed
  for(byte i = 0; i < length; i++)    
    BUFFER[i] = Serial.read();   
  BUFFER[length] = 0; //I had this earlier, but not necessary for Arduino.
  return atoi(BUFFER);
}

For example: R1234 You would write:

void loop()
{
  char input = Serial.read();
  if(input == 'R')
  {
    int output = intParse(4);
  }
}

edited May 2, 2013 at 4:13

answered May 1, 2013 at 16:02

NothinRandom

3491 silver badge6 bronze badges

\$\begingroup\$ You should also set BUFFER[length] = 0;, as C doesn't guarantee this by default. \$\endgroup\$

Ben Voigt
– Ben Voigt

2013-05-02 01:10:58 +00:00
Commented May 2, 2013 at 1:10

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Nick Alexeev · Accepted Answer · 2013-11-12 06:31:15Z

1

more simple code using parseInt()

void loop()
{ 
  if (Serial.available())
  {
   int f = Serial.parseInt();  
   if (f > 0)
   {
     Serial.println(f);
   } 
  } 
}

edited Nov 12, 2013 at 6:31

Nick Alexeev

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answered Nov 12, 2013 at 6:06

ano

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Ritvik Mandyam · Accepted Answer · 2014-08-29 05:19:16Z

-1

Why not just use the "readString()" function? Then you can have your code as:

void setup() {
  Serial.begin(9600);
  Serial.println("Waiting for intstructions...\n");
}

void loop() {
  if(Serial.available()) {
  Serial.println(Serial.readString());
  }
}

answered Aug 29, 2014 at 5:19

Ritvik Mandyam

1

1

\$\begingroup\$ This doesn't really coverting it to an integer, perhaps you could expand the answer a bit to include that as well? \$\endgroup\$

PeterJ
– PeterJ

2014-08-29 05:42:58 +00:00
Commented Aug 29, 2014 at 5:42

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