Quickest way to change endianness

Question

What is the quickest way to reverse the endianness of a 16 bit and 32 bit integer. I usually do something like (this coding was done in Visual Studio in C++):

union bytes4
{
    __int32 value;
    char ch[4];
};

union bytes2
{
    __int16 value;
    char ch[2];
};

__int16 changeEndianness16(__int16 val)
{
    bytes2 temp;
    temp.value=val;

    char x= temp.ch[0];
    temp.ch[0]=temp.ch[1];
    temp.ch[1]=x;
    return temp.value;
}

__int32 changeEndianness32(__int32 val)
{
    bytes4 temp;
    temp.value=val;
    char x;

    x= temp.ch[0];
    temp.ch[0]=temp.ch[1];
    temp.ch[1]=x;

    x= temp.ch[2];
    temp.ch[2]=temp.ch[3];
    temp.ch[3]=x;
    return temp.value;
}

Is there any faster way to do the same, in which I don't have to do so many calculations?

Answer 1

Why aren't you using the built-in swab function, which is likely optimized better than your code?

Beyond that, the usual bit-shift operations should be fast to begin with, and are so widely used they may be recognized by the optimizer and replaced by even better code.

---

Because other answers have serious bugs, I'll post a better implementation:

int16_t changeEndianness16(int16_t val)
{
    return (val << 8) |          // left-shift always fills with zeros
          ((val >> 8) & 0x00ff); // right-shift sign-extends, so force to zero
}

None of the compilers I tested generate rolw for this code, I think a slightly longer sequence (in terms of instruction count) is actually faster. Benchmarks would be interesting.

For 32-bit, there are a few possible orders for the operations:

//version 1
int32_t changeEndianness32(int32_t val)
{
    return (val << 24) |
          ((val <<  8) & 0x00ff0000) |
          ((val >>  8) & 0x0000ff00) |
          ((val >> 24) & 0x000000ff);
}

//version 2, one less OR, but has data dependencies
int32_t changeEndianness32(int32_t val)
{
    int32_t tmp = (val << 16) |
                 ((val >> 16) & 0x00ffff);
    return ((tmp >> 8) & 0x00ff00ff) | ((tmp & 0x00ff00ff) << 8);
}

Answer 2

At least in Visual C++, you can use _byteswap_ulong() and friends: http://msdn.microsoft.com/en-us/library/a3140177.aspx

These functions are treated as intrinsics by the VC++ compiler, and will result in generated code that takes advantage of hardware support when available. With VC++ 10.0 SP1, I see the following generated code for x86:

return _byteswap_ulong(val);

mov     eax, DWORD PTR _val$[esp-4]
bswap   eax
ret     0

return _byteswap_ushort(val);

mov     ax, WORD PTR _val$[esp-4]
mov     ch, al
mov     cl, ah
mov     ax, cx
ret     0

Answer 3

Who says it does too many calculations?

out = changeEndianness16(in);

gcc 4.6.0

movzwl  -4(%rsp), %edx
movl    %edx, %eax
movsbl  %dh, %ecx
movb    %cl, %al
movb    %dl, %ah
movw    %ax, -2(%rsp)

clang++ 2.9

movw    -2(%rsp), %ax
rolw    $8, %ax
movw    %ax, -4(%rsp)

Intel C/C++ 11.1

movzwl    4(%rsp), %ecx
rolw      $8, %cx
xorl      %eax, %eax
movw      %cx, 6(%rsp)

What does your compiler produce?

Answer 4

I used the following code for the 16bit version swap function:

_int16 changeEndianness16(__int16 val)
{
    return ((val & 0x00ff) << 8) | ((val & 0xff00) >> 8);
}    

With g++ (Ubuntu/Linaro 4.4.4-14ubuntu5) 4.4.5 the above code when compiled with g++ -O3 -S -fomit-frame-pointer test.cpp results in the following (non-inlined) assembler code:

movzwl  4(%esp), %eax
rolw    $8, %ax
ret

The next code is equivalent but g++ is not as good at optimizing it.

__int16 changeEndianness16_2(__int16 val)
{
    return ((val & 0xff) << 8) | (val >> 8);
}

Compiling it gives more asm code:

movzwl  4(%esp), %edx
movl    %edx, %eax
sarl    $8, %eax
sall    $8, %edx
orl     %edx, %eax
ret