c get nth byte of integer

For the (n+1)th byte in whatever order they appear in memory (which is also least- to most- significant on little-endian machines like x86):

int x = ((unsigned char *)(&number))[n]; 

For the (n+1)th byte from least to most significant on big-endian machines:

int x = ((unsigned char *)(&number))[sizeof(int) - 1 - n]; 

For the (n+1)th byte from least to most significant (any endian):

int x = ((unsigned int)number >> (n << 3)) & 0xff; 

Of course, these all assume that n < sizeof(int), and that number is an int.

int nth = (number >> (n * 8)) & 0xFF;

Carry it into the lowest byte and take it in the "familiar" manner.

If you are wanting a byte, wouldn't the better solution be:

byte x = (byte)(number >> (8 * n));

This way, you are returning and dealing with a byte instead of an int, so we are using less memory, and we don't have to do the binary and operation & 0xff just to mask the result down to a byte. I also saw that the person asking the question used an int in their example, but that doesn't make it right.

I know this question was asked a long time ago, but I just ran into this problem, and I think that this is a better solution regardless.

//was trying to do inplace, would have been better if I had swapped higher and lower bytes somehow uint32_t reverseBytes(uint32_t value) { uint32_t temp; size_t size=sizeof(uint32_t); for(int i=0; i<size/2; i++){ //get byte i temp = (value >> (8*i)) & 0xff; //put higher in lower byte value = ((value & (~(0xff << (8*i)))) | (value & ((0xff << (8*(size-i-1)))))>>(8*(size-2*i-1))) ; //move lower byte which was stored in temp to higher byte value=((value & (~(0xff << (8*(size-i-1)))))|(temp << (8*(size-i-1)))); } return value; }