float representation

Projects Programming
1

Hello, do you know how float number are represented in arduino? I know it uses 4 byte but is fixed or mobile comma? how many bits for exponent and how for the base?
I ask this because I need to send a float value on the Xbee in API mode, this is my code for send

unsigned long d = *(unsigned long *)&myValue;   
    payload[1] = d & 0x00FF;  
    payload[2] = (d & 0xFF00) >> 8;
    payload[3] = (d & 0xFF0000) >> 16;
    payload[4] = (d & 0xFF000000) >> 24;

and this the one for rebuild the float once received 4 bytes:

d =  (rx.getData()[4] << 24) | ( (rx.getData()[3] & 0xff) << 16) 
          | ((rx.getData()[2] & 0xff) << 8) | (rx.getData()[1] & 0xff)  ;
        float valRilevato = *(float *)&d;

I always obtain valRilevato=0.0 so I have done some mistakes but without knowing how float are represented it becams difficult for me to debug.
Thanks!

2

Hello, do you know how float number are represented in arduino?

Single precision (even if you specify "double")
IEEE 754

 payload[1] = d & 0x00FF;  
    payload[2] = (d & 0xFF00) >> 8;
    payload[3] = (d & 0xFF0000) >> 16;
    payload[4]

indices normally go from zero - is that deliberate?

3

without knowing how float are represented it becams difficult for me to debug.

You are converting the float into 4 bytes. Use the Serial.print() function to print the 4 bytes to the Serial Monitor. Print the value of the float, too.

On the receiving end, print the 4 bytes received. Are they the same?

If not, then there is a storage/transmission problem. If so, then it is a disassemble/reassemble issue. Knowing where the problem lies is half the battle.

4

Thank you very much I will chek out on IEEE 754. The index starts from 1 because on index 0 i put another information. There are no trasmission problems what I send is the same I recived. For example if my float value is 1.1 value sent (an received) in hex are CD CC 8C 3F if i send 13.26 value sent (an received) are F6 28 54 41 but converting I always obtain 0.0!

5

How is d declared on the receiving end? What do you get if you print d on the sender, in hex (the unsigned long)? What do you get in d on the receiver, in hex?

The conversion from d to valRilevato looks questionable.

6

PaulS:
How is d declared on the receiving end? What do you get if you print d on the sender, in hex (the unsigned long)? What do you get in d on the receiver, in hex?

The conversion from d to valRilevato looks questionable.

You are right! I have done some test and all is fine except the conversion from d to valRilevato!

7 
byte b[4];

void setup()
{
  Serial.begin(9600);
  
  b[0] = 0xCD;
  b[1] = 0xCC;
  b[2] = 0x8C;
  b[3] = 0x3F;
  
  float f1 = *(float *)&b;
  Serial.print("f1 = ");
  Serial.println(f1);
  
  b[0] = 0xF6;
  b[1] = 0x28;
  b[2] = 0x54;
  b[3] = 0x41;
  
  float f2 = *(float *)&b;
  Serial.print("f2 = ");
  Serial.println(f2);
}

void loop()
{
}

produces:
f1 = 1.10
f2 = 13.26

So, you have some other problem.

8

Thanks to AWOL i have understood how float numbers are stored, thanks to PaulS I have found my error: i was copying the MSB for first instead of LSB.
So thank you very much