Problem with typecast from pointers

Projects Programming
1 
uint8_t in_bytes[1024];

int zoffset = 0;

uint16_t CRCget;

CRCget = *((uint16_t *) (in_bytes[5])); // no exception

CRCget = *((uint16_t *) (in_bytes[5 + zoffset])); // raise exception

Also have different reaction on array defenition

  1. uint8_t in_bytes[1024];

  2. uint8_t * in_bytes = new uint8_t[1024];

in_bytes[0] // equally 1th and 2nd variants

uint16_t datasize = *((uint16_t *) &in_bytes[3]); // 1th variant is done, but 2nd variant raised exception
2

in_bytes[5] is a uint8_t. You can't cast that to a pointer.

In fact, you cannot cast any pointer to a different type, this is not allowed by the Standard. If you want to reinterpret the bit pattern of two uint8_ts to a uint16_t, the only valid way is to use memcpy:

uint8_t in_bytes[1024];
uint16_t CRCget;
memcpy(&CRCget, &in_bytes[5], sizeof(CRCget));

A more portable solution that deals with Endianness correctly would be:

uint8_t in_bytes[1024];
uint16_t CRCget = (uint16_t(in_bytes[5]) << 8) | in_bytes[6];

Pieter

3

Thank you!
I'll try.

P.S.
This is not C...

4

alexey_nikitayev:
P.S.
This is not C...

Where did I talk about C? If you're using the Arduino IDE, then it's most likely C++.
Reinterpretation of raw bit patterns is one of the only valid uses of memcpy in C++, at least until C++20's std::bit_cast becomes available on Arduino (most Arduino Cores are still on C++11).

5

All good with memcpy!

uint8_t in_bytes[1024];
...
uint16_t CRCget;
memcpy(&CRCget, &in_bytes[datasize + 5], sizeof(CRCget));
Serial.println("CRC16 -> " + String(CRCget));
uint16_t CRCcalc = GetCrc16Table(in_bytes, datasize + 5);
Serial.println("CRC calculated -> " + String(CRCcalc));

if (CRCcalc == CRCget)
{
  Serial.println("CRC Ok");
  .....

Output from Serial:

All bytes received
CRC16 -> 59441
CRC calculated -> 59441
CRC Ok