CODEWITHSUNDEEP
cstm32i2c
Mohammad Aldawsari
Mohammad Aldawsari

Reputation: 31

Reading and Storing MSB first

I'm considered a beginner in programming, so sorry if this is a very simple question.

I'm using STM32f411 uC to read from a pressure sensor MS5525DSO using I2C protocol. The programming is in c language through Eclipse.

when I read a data from sensor "PROM" memory, I expect 2 bytes "clocked with the MSB first" as data sheet said in page10.

My code: 

uint8_t Cmd= 0xA6;
  // Transmit a request to read from 0xA6 memory address:
  HAL_I2C_Master_Transmit(&I2C_HandleTypeDef, sensor_address, Cmd, 1 byte, timeout);

uint8_t PROM [2];
  // Read from PROM 0xA6
  HAL_I2C_Master_Receive(&I2C_HandleTypeDef, sensor_address, &PROM[0], 2 bytes, timeout);

Now, I stored the two bytes in an array variable unint8_t PROM [2],

the result is then:

PROM [0]= 61 decimal or (0011 1101 b)

PROM [1]= 66 decimal or (0100 0010 b)

With memcpy function the value then should be = 15,682 or (0011 1101 0100 0010‬ b)

Then, I change my mind and said let me store the data directly in one 16-bit variable so,

unint8_t PROM --->> unint16_t PROM

When I execute that I got this value:

PROM = 16,957 decimal or (0100 0010 0011 1101 b)

It is clear that the second byte (66d) is written first then the (61d) byte after,

So, I'm confused. Is my first method correct or am I missing something? does the second method swaps the bytes? which one is correct?

Upvotes: 2

Views: 1379

Answers (2)

Mohammad Aldawsari
Mohammad Aldawsari

Reputation: 31

Thanks guys for comments and answers.

It is an endianness issue as you said. So, I added this line to the code giving that "PROM" is a 16-bit uint.

           PROM= ((PROM<<8)&0xff00)|((PROM>>8)&0x00ff);

This will do the byte swapping.

Fortunately, my sensor has a temperature measurement. So, I could validate my code and got a correct result "23 C" (my room temp.) using the above byte swapping line code.

However, I'm still interested why the function used in my code did the little-endian ordering, and is there a way to change that setting.

Thank you again for contributions.

Upvotes: 1

ralf htp
ralf htp

Reputation: 9412

you can store two 8-bits into a 16-bit with (x1 << 8) | x2, where x1 is PROM[0] and x2 is PROM[1]

This is in combine multiple chars to an int, << and | are logical operators, << is left-shift, | is or

There are many other bit hacking solutions, cf https://graphics.stanford.edu/~seander/bithacks.html

Upvotes: 2

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