CODEWITHSUNDEEP
coptimizationendianness
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Reputation: 523

Optimized code for big to little endian conversion

In an interview, I was asked to implement big_to_little_endian() as a macro. I implemented using shift operator. But the interviewer want me to optimize this further. I could not do it. Later I googled & searched but could not find it. Can someone help in understanding how to further optimize this code?

#define be_to_le (((x) >> 24) | (((x) & 0x00FF0000) >> 8) | (((x) & 0x0000FF00) << 8) | ((x) << 24))

Upvotes: 3

Views: 496

Answers (1)

Jason Nordwick
Jason Nordwick

Reputation: 1556

He might have been referring to using a 16-bit op to swap the top two words then using 8-bit ops to swap the bytes in them -- saves a couple instructions, easiest done in a union, though C technically doesn't like it (but many compilers will accept it), and it still compiler dependent since you are hoping the compiler optimizes a couple things out:

union dword {
  unsigned int i;
  union shorts {
    unsigned short s0, s1;
    union bytes {
      unsigned char c0, c1, c2, c3;
    } c;
  } s;
};

union dword in = (union dword)x;
union dword temp = { x.s.s1, x.s.s0 };
union dword out = { temp.s.c.c1, temp.s.c.c0, temp.s.c.c3, temp.s.c.c2 };

Not even valid C, but you get the idea (and I don't think the compiler will even emit what I'm hoping it will).

Or you can save an op, but introduce a data dependency so probably runs slower.

temp = (x << 16) | ( x >> 16)
out = ((0xff00ff00 & temp) >> 8) | (0x00ff00ff & temp) << 8)

Best is just use the compiler intrinsic since it maps to a single bswap instruction.

Upvotes: 3

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