Exchange the bytes

I want to exchange the bytes of a number. Example the binary representation of a number is 00000001 00011000 00000100 00001110. I want to reverse it: 00001110 00000100 00011000 00000001.

Can you please help? The current code is below:

void showBits(int n)
{
    int i,k,andMask;
    for(i=15;i>=0;i--)
    {
开发者_Go百科        andMask=1<<i;
        k=n&andMask;
        k==0?(cout<<"0"):(cout<<"1");
    }
}
int reverse(int a)
{
    int b=a<<8;
    int c=a>>8;
    return (b|c);
}
int main()
{
    int a=10;
    showBits(a);
    int b=reverse(a);
    showBits(b);
    cin.get();
} 

---

Something like this should work:

result = ((number & 0xFF) << 24) | ((number & 0xFF00) << 8) | 
   ((number & 0xFF0000) >> 8) | ((number & 0xFF000000) >> 24);

---

int myInt = 0x012345678

__asm {
  mov eax, myInt
  bswap eax
  mov myInt, eax
}

---

If you want to simply reverse a 32-bit number, you can use a bit-shifting technique that isolates each byte region using bit-masks and logical AND, and then swaps those bytes by the appropriate number of shifted bits using the bit-shift operators >> and <<. You can then recombined the bits using logical OR like so:

int temp = 0x12345678;
temp = ((0xFF & temp) << 24) | ((0xFF00 & temp) << 8) | ((0xFF0000 & temp) >> 8) |
       ((0xFF000000 & temp) >> 24));

You'll now end up with a final value in temp of 0x78563412.

Update: Okay, I'm looking over you code, and noting the following:

1. The size of int it seems like you're wanting to work with from the binary digit you have posted is 32-bits ... so your showBits function is not cycling through enough bits to display an entire 32-bit integer. As far as I can see right now, it will only show up to the 16 lower bits. So you'll need to be clear on whether you're working on a platform that defines int as 16 or 32-bits.
2. Your reverse functions is not correct. If you have a 32-bit int like 0x12345678, and you shift it left by 8-bits, you will end up with 0x34567800. Likewise, when you shift it right by 8-bits, you will just end up with 0x00123456. Shifting a number does not rotate the values through their respective positions (i.e., a shift left of 8-bits would not give you 0x34567812). Plus, even if it did, the logical OR of the rotated values would still not be a correct reversal of the values. Instead, you must use the technique described above that uses bit-masks and logical AND to isolate each byte, and then shift those bits the appropriate number of places in order to reverse the bits in a 32-bit word.

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If your original task to to convert from machine-specific byte order to big endian and back, take a look at the hton* and ntoh* functions.

http://minix1.woodhull.com/manpages/man3/hton.3.html