Convert Bytes to Int / uint in C

There's no standard function to do it for you in C. You'll have to assemble the bytes back into your 16- and 32-bit integers yourself. Be careful about endianness!

Here's a simple little-endian example:

extern uint8_t *bytes;uint32_t myInt1 = bytes[0] + (bytes[1] << 8) + (bytes[2] << 16) + (bytes[3] << 24);

For a big-endian system, it's just the opposite order:

uint32_t myInt1 = (bytes[0] << 24) + (bytes[1] << 16) + (bytes[2] << 8) + bytes[3];

You might be able to get away with:

uint32_t myInt1 = *(uint32_t *)bytes;

If you're careful about alignment issues.

Yes there is. Assume your bytes are in:

uint8_t bytes[N] = { /* whatever */ };

We know that, a 16 bit integer is just two 8 bit integers concatenated, i.e. one has a multiple of 256 or alternatively is shifted by 8:

uint16_t sixteen[N/2];for (i = 0; i < N; i += 2)    sixteen[i/2] = bytes[i] | (uint16_t)bytes[i+1] << 8;             // assuming you have read your bytes little-endian

Similarly for 32 bits:

uint32_t thirty_two[N/4];for (i = 0; i < N; i += 4)    thirty_two[i/4] = bytes[i] | (uint32_t)bytes[i+1] << 8        | (uint32_t)bytes[i+2] << 16 | (uint32_t)bytes[i+3] << 24;             // same assumption

If the bytes are read big-endian, of course you reverse the order:

bytes[i+1] | (uint16_t)bytes[i] << 8

and

bytes[i+3] | (uint32_t)bytes[i+2] << 8    | (uint32_t)bytes[i+1] << 16 | (uint32_t)bytes[i] << 24

Note that there's a difference between the endian-ness in the stored integer and the endian-ness of the running architecture. The endian-ness referred to in this answer is of the stored integer, i.e., the contents of bytes. The solutions are independent of the endian-ness of the running architecture since endian-ness is taken care of when shifting.

In case of little-endian, can't you just use memcpy?

memcpy((char*)&myint1, aesData.inputData[startindex], length);