Serializing Int values to char* buffer in ANSI C

Question

I tried to serialize a structure field (int) to a char* buffer but I think I am doing things wrong.

This is what I am using to copy this field.

memcpy(payload + offset, &packet->payload.offset, sizeof(long int));

packet->payload.offset is a long int and contains the value '5' (hardcoded), note that:

If I print: printf("offset: %ld \n", packet->payload.offset); I get: offset: 5

But if I print: printf("offset: %ld \n",&packet->payload.offset), I get: offset: 167120959

If in memcopy I remove the '&' to the second argument to write '5' to the buffer it causes Segmentation Fault, but if I don't, it copies 167120959 (which I supose it is it's address) to the buffer instead of the actual value.

Thank you very much for your help! I tried to be as descriptive as possible, hope you can help me out.

score 2 · Answer 1 · answered May 20 '12 at 18:03

But if I print: printf("offset: %ld \n",&packet->payload.offset), I get: offset: 167120959

That's because you're printing an address, not a value. Your "serialization" method is OK (endianness aside) because memcpy expects addresses.

I remove the '&' to the second argument to write '5' to the buffer [...] but if I don't, it copies 167120959 (which I supose it is it's address) to the buffer instead of the actual value.

Wrong. It copies sizeof(long int) bytes from that address.

That's what I expected, but when I de-serialize the stream, and print packet->payload.offset I get 167120959 when I was suposed to get "5" — Guillermo Gruschka, May 20 '12 at 19:01

score 0 · Answer 2 · answered May 20 '12 at 18:07

0

But if I print: printf("offset: %ld \n",&packet->payload.offset), I get: offset: 167120959

That's wrong, it prints the address of packet->payload.offset instead of packet->payload.offset itself.

Memcpy has to be supplied with the address of the variable, else it can't modify it (there's no byref parameter passing in C). So you pass memcpy() the address of the variale, but printf() only needs to read it, so you can (and should) pass the variable directly.

answered May 20 '12 at 18:07

got your point, but how am I suposed to do to write the variable content to the char* ? – Guillermo Gruschka May 20 '12 at 18:59
char *buf = (wherever it points to); *(int *)buf = packet->payload.offset; – May 20 '12 at 19:07
Cant I use memcpy() to do so ? – Guillermo Gruschka May 20 '12 at 19:23
you can, but then that's the same (good) method that you used-- then why are you asking? – May 20 '12 at 20:00
thanks for answering :) Because when I use memcopy as detailed in my question, memcopy writes the address into the char* instead of the variable content. – Guillermo Gruschka May 20 '12 at 22:47

Serializing Int values to char* buffer in ANSI C

2 Answers2