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Read float and long from external EEPROM chip

I am using an Arduino to read information from an EEPROM chip over I2C with the code I found here: https://playground.arduino.cc/Code/I2CEEPROM

It works great for int and char values but I also need to read short, long, and float values.

They are stored in this order: 1510543923
is stored as: 01011010 00001001 00010010 00110011

How can I read multiple bytes into a single variable of these types?

I tried the following but it gives me a 16 bit number:

i2c_eeprom_read_buffer(0x50, 25, (byte *)bytes, 4);
long j = (bytes[3] << 0) + (bytes[2] << 8) + (bytes[1] << 16) + (bytes[0] << 24);

Answer*

> 1510543923 is stored as: 01011010 00001001 00010010 00110011

This is known as “big endian”, or “MSB first”, because the most
significant byte (MSB, here 01011010) comes first.

I recommend against this order. If you can, swap all the bytes and store
the value LSB first, i.e. with the least significant byte first, and the
most significant byte last. This will make your life easier since it's
the order used internally in the Arduino. If you can go little endian,
then you will be able to read the data straight into the required
variable, with no conversion needed, as in:

 i2c_eeprom_read_buffer(0x50, 25, (byte *) &j, sizeof j);

and it will work identically with any data type.

If you cannot choose the byte order (maybe you cannot control what is
inside the EEPROM), then you will have to reverse all the bytes. You can
do this either with bit shifts or by explicitly moving bytes.

> `bytes[1] << 16`

This will not work. Per the rules of the C++ language, `bytes[1]` is
implicitly promoted to `int` type before the shift, then it is shifted
as an `int`. But on your Arduino an `int` is 16-bits long, thus by
shifting it 16 positions you are dropping off _all_ of the bits.
Actually, this is even worse: you are invoking what is known as
_undefined behavior_, meaning that it is considered nonsense and the
compiler is free to interpret however it wants.

To do this properly, you first have to _explicitly_ cast `bytes[i]` to
`unsigned long`. It has to be unsigned because changing the sign bit on
a `long` by bit shifting is also undefined behavior. Thus, the proper
way of reconstructing the number via bit shifts is:

 long j = ((unsigned long) bytes[3] << 0)
 | ((unsigned long) bytes[2] << 8)
 | ((unsigned long) bytes[1] << 16)
 | ((unsigned long) bytes[0] << 24);

This is portable (it should also work on big-endian architectures) but
only works with integral types, not with floats.

For reconstructing a float, my preferred option would be to use a
`union`, in order to access the bytes of the variable explicitly. Then
you just have to copy the bytes from the buffer to the `union` in
reverse order:

 union { float f; byte b[4]; } data;
 data.b[0] = bytes[3];
 data.b[1] = bytes[2];
 data.b[2] = bytes[1];
 data.b[3] = bytes[0];
 float x = data.f;

One nice thing about this approach is that it works with any data type.

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In reading the float, I correctly get 00110101-00111000-10101000-11001110 from the EEPROM but running it through the union snipit you show, gives 0.0. It should be 6.8790985E-7.

Alphy13
– Alphy13

2017年12月07日 03:19:08 +00:00
Commented Dec 7, 2017 at 3:19
@Alphy13: I am pretty sure you got the correct number. However, Serial.println(6.8790985e-7); prints "0.00".

Edgar Bonet
– Edgar Bonet

2017年12月07日 08:28:50 +00:00
Commented Dec 7, 2017 at 8:28
Thanks for pointing that out. In case anyone has a similar issue: Serial.println(x, 10); gave the precision i needed to see a result.

Alphy13
– Alphy13

2017年12月07日 22:22:27 +00:00
Commented Dec 7, 2017 at 22:22

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