问题
I know you can get the first byte by using
int x = number & ((1<<8)-1);
or
int x = number & 0xFF;
But I don\'t know how to get the nth byte of an integer. For example, 1234 is 00000000 00000000 00000100 11010010 as 32bit integer How can I get all of those bytes? first one would be 210, second would be 4 and the last two would be 0.
回答1:
int x = (number >> (8*n)) & 0xff;
where n is 0 for the first byte, 1 for the second byte, etc.
回答2:
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.
回答3:
int nth = (number >> (n * 8)) & 0xFF;
Carry it into the lowest byte and take it in the "familiar" manner.
回答4:
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.
来源：https://stackoverflow.com/questions/7787423/c-get-nth-byte-of-integer