Branchless code that maps zero, negative, and positive to 0, 1, 2

Question

Write a branchless function that returns 0, 1, or 2 if the difference between two signed integers is zero, negative, or positive.

Here's a version with branching:

int Compare(int x, int y)
{
    int diff = x - y;
    if (diff == 0)
        return 0;
    else if (diff < 0)
        return 1;
    else
        return 2;
}

Here's a version that may be faster depending on compiler and processor:

int Compare(int x, int y)
{
    int diff = x - y;
    return diff == 0 ? 0 : (diff < 0 ? 1 : 2);
}

Can you come up with a faster one without branches?

SUMMARY

The 10 solutions I benchmarked had similar performance. The actual numbers and winner varied depending on compiler (icc/gcc), compiler options (e.g., -O3, -march=nocona, -fast, -xHost), and machine. Canon's solution performed well in many benchmark runs, but again the performance advantage was slight. I was surprised that in some cases some solutions were slower than the naive solution with branches.

Answer 1

Branchless (at the language level) code that maps negative to -1, zero to 0 and positive to +1 looks as follows

int c = (n > 0) - (n < 0);

if you need a different mapping you can simply use an explicit map to remap it

const int MAP[] = { 1, 0, 2 };
int c = MAP[(n > 0) - (n < 0) + 1];

or, for the requested mapping, use some numerical trick like

int c = 2 * (n > 0) + (n < 0);

(It is obviously very easy to generate any mapping from this as long as 0 is mapped to 0. And the code is quite readable. If 0 is mapped to something else, it becomes more tricky and less readable.)

As an additinal note: comparing two integers by subtracting one from another at C language level is a flawed technique, since it is generally prone to overflow. The beauty of the above methods is that they can immedately be used for "subtractionless" comparisons, like

int c = 2 * (x > y) + (x < y);

Answer 2

int Compare(int x, int y) {
     return (x < y) + (y < x) << 1;
}

Edit: Bitwise only? Guess < and > don't count, then?

int Compare(int x, int y) {
    int diff = x - y;
    return (!!diff) | (!!(diff & 0x80000000) << 1);
}

But there's that pesky -.

Edit: Shift the other way around.

Meh, just to try again:

int Compare(int x, int y) {
    int diff = y - x;
    return (!!diff) << ((diff >> 31) & 1);
}

But I'm guessing there's no standard ASM instruction for !!. Also, the << can be replaced with +, depending on which is faster...

Bit twiddling is fun!

Hmm, I just learned about setnz.

I haven't checked the assembler output (but I did test it a bit this time), and with a bit of luck it could save a whole instruction!:

IN THEORY. MY ASSEMBLER IS RUSTY

subl  %edi, %esi
setnz %eax
sarl  $31, %esi
andl  $1, %esi
sarl  %eax, %esi
mov   %esi, %eax
ret

Rambling is fun.

I need sleep.

Answer 3

Assuming 2s complement, arithmetic right shift, and no overflow in the subtraction:

#define SHIFT (CHARBIT*sizeof(int) - 1)

int Compare(int x, int y)
{
    int diff = x - y;
    return -(diff >> SHIFT) - (((-diff) >> SHIFT) << 1);
}

Answer 4

Two's complement:

#include <limits.h>
#define INT_BITS (CHAR_BITS * sizeof (int))

int Compare(int x, int y) {
    int d = y - x;
    int p = (d + INT_MAX) >> (INT_BITS - 1);
    d = d >> (INT_BITS - 2);
    return (d & 2) + (p & 1);
}

Assuming a sane compiler, this will not invoke the comparison hardware of your system, nor is it using a comparison in the language. To verify: if x == y then d and p will clearly be 0 so the final result will be zero. If (x - y) > 0 then ((x - y) + INT_MAX) will set the high bit of the integer otherwise it will be unset. So p will have its lowest bit set if and only if (x - y) > 0. If (x - y) < 0 then its high bit will be set and d will set its second to lowest bit.

Answer 5

Unsigned Comparison that returns -1,0,1 (cmpu) is one of the cases that is tested for by the GNU SuperOptimizer.

cmpu: compare (unsigned)
int cmpu(unsigned_word v0, unsigned_word v1)
{
    return ( (v0 > v1) ? 1 : ( (v0 < v1) ? -1 : 0) );
}

A SuperOptimizer exhaustively searches the instruction space for the best possible combination of instructions that will implement a given function. It is suggested that compilers automagically replace the functions above by their superoptimized versions (although not all compilers do this). For example, in the PowerPC Compiler Writer's Guide (powerpc-cwg.pdf), the cmpu function is shown as this in Appendix D pg 204:

cmpu: compare (unsigned)
PowerPC SuperOptimized Version
subf  R5,R4,R3
subfc R6,R3,R4
subfe R7,R4,R3
subfe R8,R7,R5

That's pretty good isn't it... just four subtracts (and with carry and/or extended versions). Not to mention it is genuinely branchfree at the machine opcode level. There is probably a PC / Intel X86 equivalent sequence that is similarly short since the GNU Superoptimizer runs for X86 as well as PowerPC.

Note that Unsigned Comparison (cmpu) can be turned into Signed Comparison (cmps) on a 32-bit compare by adding 0x80000000 to both Signed inputs before passing it to cmpu.

cmps: compare (signed)
int cmps(signed_word v0, signed_word v1)
{
    signed_word offset=0x80000000;
    return ( (unsigned_word) (v0 + signed_word),
        (unsigned_word) (v1 + signed_word) );
}

This is just one option though... the SuperOptimizer may find a cmps that is shorter and does not have to add offsets and call cmpu.

To get the version that you requested that returns your values of {1,0,2} rather than {-1,0,1} use the following code which takes advantage of the SuperOptimized cmps function.

int Compare(int x, int y)
{
    static const int retvals[]={1,0,2};
    return (retvals[cmps(x,y)+1]);
}

Answer 6

I'm siding with Tordek's original answer:

int compare(int x, int y) {
    return (x < y) + 2*(y < x);
}

Compiling with gcc -O3 -march=pentium4 results in branch-free code that uses conditional instructions setg and setl (see this explanation of x86 instructions).

push   %ebp
mov    %esp,%ebp
mov    %eax,%ecx
xor    %eax,%eax
cmp    %edx,%ecx
setg   %al
add    %eax,%eax
cmp    %edx,%ecx
setl   %dl
movzbl %dl,%edx
add    %edx,%eax
pop    %ebp
ret 

Answer 7

Good god, this has haunted me.

Whatever, I think I squeezed out a last drop of performance:

int compare(int a, int b) {
    return (a != b) << (a > b);
}

Although, compiling with -O3 in GCC will give (bear with me I'm doing it from memory)

xorl  %eax, %eax
cmpl  %esi, %edi
setne %al
cmpl  %esi, %edi
setgt %dl
sall  %dl, %eax
ret

But the second comparison seems (according to a tiny bit of testing; I suck at ASM) to be redundant, leaving the small and beautiful

xorl  %eax, %eax
cmpl  %esi, %edi
setne %al
setgt %dl
sall  %dl, %eax
ret

(Sall may totally not be an ASM instruction, but I don't remember exactly)

So... if you don't mind running your benchmark once more, I'd love to hear the results (mine gave a 3% improvement, but it may be wrong).

Answer 8

Combining Stephen Canon and Tordek's answers:

int Compare(int x, int y)
{
    int diff = x - y; 
    return -(diff >> 31) + (2 & (-diff >> 30));
} 

Yields: (g++ -O3)

subl     %esi,%edi 
movl     %edi,%eax
sarl     $31,%edi
negl     %eax
sarl     $30,%eax
andl     $2,%eax
subl     %edi,%eax 
ret 

Tight! However, Paul Hsieh's version has even fewer instructions:

subl     %esi,%edi
leal     0x7fffffff(%rdi),%eax
sarl     $30,%edi
andl     $2,%edi
shrl     $31,%eax
leal     (%rdi,%rax,1),%eax
ret

Answer 9

int Compare(int x, int y)
{
    int diff = x - y;

    int absdiff = 0x7fffffff & diff; // diff with sign bit 0
    int absdiff_not_zero = (int) (0 != udiff);

    return
        (absdiff_not_zero << 1)      // 2 iff abs(diff) > 0
        -
        ((0x80000000 & diff) >> 31); // 1 iff diff < 0
}

Answer 10

The basic C answer is :

int v;           // find the absolute value of v
unsigned int r;  // the result goes here 
int const mask = v >> sizeof(int) * CHAR_BIT - 1;

r = (v + mask) ^ mask;

Also :

r = (v ^ mask) - mask;

Value of sizeof(int) is often 4 and CHAR_BIT is often 8.

Answer 11

For 32 signed integers (like in Java), try:

return 2 - ((((x >> 30) & 2) + (((x-1) >> 30) & 2))) >> 1;

where (x >> 30) & 2 returns 2 for negative numbers and 0 otherwise.

x would be the difference of the two input integers