x = (x^y);
return x;
}
/*
* byteXor - compare the nth byte of x and y, if it is same, return 0, if not, return 1
* example: byteXor(0x12345678, 0x87654321, 1) = 1
* byteXor(0x12345678, 0x87344321, 2) = 0
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 20
* Rating: 2
*/
int byteXor(int x, int y, int n) {
//把x和y的第n个字节取出来异或，再转换为逻辑的0和1
n = n<<3;
x = x>>n;
y = y>>n;
x = x&(0xff);
y = y&(0xff);
return !!(x^y);
}
/*
* logicalAnd - x && y
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 20
* Rating: 3
*/
int logicalAnd(int x, int y) {
//把x和y分别转化为逻辑的0和1，再相与
x = (!(!x))&(!(!y));
return x;
}
/*
* logicalOr - x || y
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 20
* Rating: 3
*/
int logicalOr(int x, int y) {
//把x和y分别转化为逻辑的0和1，再相或
x = (!(!x))|(!(!y));
return x;
}
/*
* rotateLeft - Rotate x to the left by n
* Can assume that 0 <= n <= 31
* Examples: rotateLeft(0x87654321,4) = 0x76543218
* Legal ops: ~ & ^ | + << >> !
* Max ops: 25
* Rating: 3
*/
int rotateLeft(int x, int n) {
//先构造低n位为1，高（32-n）位为0的数z，x左移n位后的数加上x右移（32-n）位的数&z即可
int z;
z = ~(((1<<31)>>31)<x = ((x>>(32+(~n+1)))&z)+(x<return x;
}
/*
* parityCheck - returns 1 if x contains an odd number of 1's
* Examples: parityCheck(5) = 0, parityCheck(7) = 1
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 20
* Rating: 4
