How to properly convert an unsigned char array into an uint32_t

Question

So, I'm trying to convert an array of unsigned chars into an uint32_t, but keep getting different results each time:

unsigned char buffer[] = {0x80, 0x00, 0x00, 0x00};;
uint32_t num = (uint32_t*)&buffer;

Now, I keep getting this warning:

warning: initialization makes integer from pointer without a cast

When I change num to *num i don't get that warning, but that's not actually the real problem (UPDATE: well, those might be related now that I think of it.), because every time I run the code there is different results. Secondly the num, once it's cast properly, should be 128, but If I need to change the endianness of the buffer I could manage to do that myself, I think.

Thanks!

Answer 1

Did you try this ?

num = (uint32_t)buffer[0] << 24 |
      (uint32_t)buffer[1] << 16 |
      (uint32_t)buffer[2] << 8  |
      (uint32_t)buffer[3];

This way you control endianness and whatnot.

It's really not safe to cast a char pointer and interpret it as anything bigger. Some machines expect pointers to integers to be aligned.

Answer 2

cnicutar's answer is the best assuming you want a particular fixed endianness. If you want host endian, try:

uint32_t num;
memcpy(&num, buffer, 4);

or apply ntohl to cnicutar's answer. Any method based on type punning is wrong and dangerous.