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Problem Definition:

Given a number x and two positions (from right side) in binary representation of x, write a function that swaps n bits at given two positions and returns the result. It is also given that the two sets of bits do not overlap.

Solution:

We need to swap two sets of bits. XOR can be used in a similar way as it is used to . Following is the algorithm.

1) Move all bits of first set to rightmost side   set1 =  (x >> p1) & ((1U << n) - 1)Here the expression (1U << n) - 1 gives a number that contains last n bits set and other bits as 0. We do & with this expression so that bits other than the last n bits become 0.2) Move all bits of second set to rightmost side   set2 =  (x >> p2) & ((1U << n) - 1)3) XOR the two sets of bits   xor = (set1 ^ set2) 4) Put the xor bits back to their original positions.    xor = (xor << p1) | (xor << p2)5) Finally, XOR the xor with original number so    that the two sets are swapped.   result = x ^ xor

Code:

int swapBits(unsigned int x, unsigned int p1, unsigned int p2, unsigned int n){    /* Move all bits of first set to rightmost side */    unsigned int set1 =  (x >> p1) & ((1U << n) - 1);     /* Moce all bits of second set to rightmost side */    unsigned int set2 =  (x >> p2) & ((1U << n) - 1);     /* XOR the two sets */    unsigned int xor = (set1 ^ set2);     /* Put the xor bits back to their original positions */    xor = (xor << p1) | (xor << p2);     /* XOR the 'xor' with the original number so that the        two sets are swapped */    unsigned int result = x ^ xor;     return result;}

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