unable to find numeric literal operator

Using Arduino Programming Questions
#1

im using a cottonwood RFID reader and im trying to run this code in an arduino mega

char command_antenna={0x18,0x03,0xFF};//const
char command_inventory={0x43,0x04,0x01,0xcd};//const
int incomingByte;

void setup()
{
Serial.begin(9600);
Serial1.begin(9600);
Serial1.write(command_antenna);
}

void loop()
{
delay(10000);

Serial.println("Tag List");
Serial1.write(command_inventory,8);
delay (1000);

while (Serial1.available())
{

incomingByte = Serial1.read();
if(incomingByte == 6Serial.println(""));
//Serial.print(incomingByte);
Serial.print(incomingByte,HEX);
Serial.print("-");

}

Serial.println("");

}

and i get this message

exit status 1
unable to find numeric literal operator 'operator"" Serial.println'

Please help

#2

Looks like a minor earthquake happened here:

   if (incomingByte == 6Serial.println(""));
#3

do u have any suggestion?

#4

bennykf:
do u have any suggestion?

You tell us what that line is supposed to do

#5

bennykf:
do u have any suggestion?

I'd try one of these )

You can have that one, it's a spare I carry around.

You probably also need a pair of these '

I'm afraid I'll have to charge for that one.

#6

it supouse to give me the 16 bits epc

#7

Are you looking at the same thing we are?

Do you really expect that "incomingByte" is ever going to be equal to "6Serial.println("")"?

Steve

#8

Do you really expect 6Serial.println("") to ever be anything other than a syntax error?

What on earth are you trying to do with that line?!

#9

DrAzzy:
Do you really expect 6Serial.println("") to ever be anything other than a syntax error?

Yeah, you need a space there to make it work. :wink:

struct Foo {
 void print(const char *);
};

Foo operator ""Serial(const char*) {
 Foo a;
 return a;
}

void hoo()
{
 6Serial .print("");
}

However, you will get a warning:

goop.cc:5:14: warning: literal operator suffixes not preceded by '_' are reserved for future standardization
 Foo operator ""Serial(const char*) {
              ^

http://en.cppreference.com/w/cpp/language/user_literal

#10

  if (incomingByte == 6Serial.println(""));looks suspiciously like an attempt to print a blank line if incomingByte equals 6 but reply #5 seems to indicate not.

#11