compiler says I have to declare my hexidacimal I try to compare

Question

#include <IRremote.h>

int RECV_PIN = 11; // IR Receiver - Arduino Pin Number 11
IRrecv irrecv(RECV_PIN);
decode_results results;

int led = 12;

void setup() {                
  pinMode(led, OUTPUT);
  digitalWrite(led, LOW);
  Serial.begin(9600);
  irrecv.enableIRIn(); // Start the receiver     
}

void loop() {
  if(irrecv.decode(&results)) //this checks to see if a code has been received
{
    if(results.value == FF6897) //if the button press equals the hex value FF6897
    {
        digitalWrite(led, HIGH);   // turn the LED on (HIGH is the voltage level)
        delay(1000);               // wait for a second
        digitalWrite(led, LOW);    // turn the LED off by making the voltage LOW
        delay(1000);               // wait for a second//do something useful here
    }
    irrecv.resume(); //receive the next value
}
  }

The issue is in the second if in the void loop the error I get is

exit status 1 'FF6897' was not declared in this scope

why does he do this? Why doesnt it just accept my IRcode for comparison.

Answer 1

Nothing here that hasn't been covered in comments already but just putting it in the form of an answer to save digging...

How is the compiler supposed to know that a number is hex rather than decimal?

Sure 1234beef is obviously hex but what about 12345678? Is that hex or decimal?

You need some method for the compiler to know.

You also need some way for the compiler to know if something is a number or a variable. Take the code if (x == deadbeef) {...} is deadbeef a hex number or a variable name?

C uses a very simple convention. If it starts with a number it's a number, if it starts with a letter it's a variable. This is why variable names can contain numbers but can't start with them.

If an integer starts with an 0 e.g. 0123 then it's in octal.
If an integer starts with 0x e.g. 0x123 then it's in hex.
If an integer starts with something other than a 0 e.g. 123 then its in decimal.
e.g.

printf("123 = %d\r\n0x123 = %d\r\n0123 = %d\r\n", 123 , 0x123 , 0123);

returns

123 = 123
0x123 = 291
0123 = 83

This convention, especially the 0x prefix for hexadecimal, is very widely used in other programming languages.

Answer 2

By comparing my code to another one, I found out that putting 0x in front of the hex will fix it, I don't have a clue why this is though.