Difference between memcpy and copy by assignment

Question

Struct A
{
   uint16_t len;
   uint8_t  cnt;
   uint8_t  unit;
   uint32_t seq; 
};

This struct A is serialized into a char * buf. If I want to deserialize the individual values eg:

uint16_t len = 0;
memcpy(&len, buf, sizeof(len));

or I can just do

uint16_t len = (uint16_t) buf;

Which one is better or are both the same?

Also to deserialize the whole struct, if I just do

A tmp;

memcpy(&tmp, buf, sizeof(A));

Would this work fine or should I be worried about padding etc from the compiler?

Answer 1

When the data is copied into char[] buffer, it may not be properly aligned in memory for access as multi-byte types. Copying the data back into struct restores proper alignment.

If I want to deserialize the individual values eg:

uint16_t len = 0;
memcpy(&len, buf, sizeof(len));

Assuming that you have copied the struct into buf, this is perfectly valid, because the language guarantees that the initial member would be aligned with the beginning of the structure. However, casting buf to uint16_t* is invalid, because the buffer many not be properly aligned in memory to be addressed as uint16_t.

Note that getting elements of the struct other than the initial one require computing proper offset:

uint32_t seq;
memcpy(&seq, buf+offsetof(struct A, seq), sizeof(seq));

Also to deserialize the whole struct, if I just do

A tmp;
memcpy(&tmp, buf, sizeof(A));

Would this work fine or should I be worried about padding etc from the compiler?

This would work fine. Any padding embedded in the struct when you copied it into the buf would come back into tmp, along with the actual data.