Byte swapping in bit wise operations

Question

I have this function called byte swap I am supposed to implement. The idea is that the function takes 3 integers (int x, int y, int z) and the function will swap the y and z bytes of the int x. The restrictions are pretty much limited to bit wise operations (no loops, and no if statements or logical operators such as ==).

I don't believe that I presented this problem adequately so Im going to re attempt

I now understand that

byte 1 is referring to bits 0-7 byte 2 is referring to bits 8-15 byte 3 16-23 byte 4 24-31

My function is supposed to take 3 integer inputs, x, y and z. The y byte and z byte on the x then would have to get switched

int byteSwap(int x, int y, int z)

ex of the working function

byteSwap(0x12345678, 1, 3) = 0x56341278 byteSwap(0xDEADBEEF, 0, 2) = 0xDEEFBEAD

My original code had some huge errors in it, namely the fact that I was considering a byte to be 2 bits instead of 8. The main problem that I'm struggling with is that I do not know how to access the bits inside of the given byte. For example, when I'm given byte 4 and 5, how do I access their respected bits? As far as I can tell I can't find a mathematical relationship between the given byte, and its starting bit. I'm assuming I have to shift and then mask, and save those to variables.Though I cannot even get that far.

It's unclear to me how exactly is it supposed to work. It'd be useful if you posted 1-2 examples of x, y and z meeting your criteria. — kamituel
– kamituel, Commented Jan 28, 2015 at 22:55
Casting on the left side of an = operator is usually a sign you're doing something wrong. — Carl Norum
– Carl Norum, Commented Jan 28, 2015 at 22:59
added edits to show examples on how it is supposed to work. Hopefully everything is more clear now, sorry for the confusion — rcbranham
– rcbranham, Commented Jan 28, 2015 at 23:10
Um, bytes have 8 bits each, not 2. Where did that idea come from? — Lee Daniel Crocker
– Lee Daniel Crocker, Commented Jan 28, 2015 at 23:27

mrk · Accepted Answer · 2015-01-29 00:02:56Z

Extract the ith byte by using ((1ll << ((i + 1) * 8)) - 1) >> (i * 8). Swap using the XOR operator, and put the swapped bytes in their places.

int x, y, z; y = 1, z = 3; x = 0x12345678; int a, b; /* bytes to swap */ a = (x & ((1ll << ((y + 1) * 8)) - 1)) >> (y * 8); b = (x & ((1ll << ((z + 1) * 8)) - 1)) >> (z * 8); /* swap */ a = a ^ b; b = a ^ b; a = a ^ b; /* put zeros in bytes to swap */ x = x & (~((0xff << (y * 8)))); x = x & (~((0xff << (z * 8)))); /* put new bytes in place */ x = x | (a << (y * 8)); x = x | (b << (z * 8));

abligh · Accepted Answer · 2015-01-29 00:04:06Z

When you say the 'the y and z bytes of x' this implies x is an array of bytes, not an integer. If so:

x[z] ^= x[y]; x[y] ^= x[z]; x[z] ^= x[y];

will do the trick, by swapping x[y] and x[z]

After your edit, it appears you want to swap individual bytes of a 32 bit integer:

On a little-endian machine:

int swapbytes (int x, int y, int z) { char *b = (char *)&x; b[z] ^= b[y]; b[y] ^= b[z]; b[z] ^= b[y]; return x; }

On a big-endian machine:

int swapbytes (int x, int y, int z) { char *b = (char *)&x; b[3-z] ^= b[3-y]; b[3-y] ^= b[3-z]; b[3-z] ^= b[3-y]; return x; }

With a strict interpretation of the rules, you don't even need the xor trick:

int swapbytes (int x, int y, int z) { char *b = (char *)&x; char tmp = b[z]; b[z] = b[y]; b[y] = tmp; return x; }

On a big-endian machine:

int swapbytes (int x, int y, int z) { char *b = (char *)&x; char tmp = b[3-z]; b[3-z] = b[3-y]; b[3-y] = tmp; return x; }

If you want to do it using bit shifts (note <<3 multiplies by 8):

int swapbytes (unsigned int x, int y, int z) { unsigned int masky = 0xff << (y<<3); unsigned int maskz = 0xff << (z<<3); unsigned int origy = (x & masky) >> (y<<3); unsigned int origz = (x & maskz) >> (z<<3); return (x & ~masky & ~maskz) | (origz << (y<<3)) | (origy << (z<<3)); }

X is supposed to be an integer. I added some edits to hopefully make it more clear, I apologize for the confusion
... and, dear downvoters, the problem here is precisely what? Yes, it's artificial, but so is the question.

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