How do the likely/unlikely macros in the Linux kernel work and what is their benefit? How do the likely/unlikely macros in the Linux kernel work and what is their benefit? linux linux

How do the likely/unlikely macros in the Linux kernel work and what is their benefit?


They are hint to the compiler to emit instructions that will cause branch prediction to favour the "likely" side of a jump instruction. This can be a big win, if the prediction is correct it means that the jump instruction is basically free and will take zero cycles. On the other hand if the prediction is wrong, then it means the processor pipeline needs to be flushed and it can cost several cycles. So long as the prediction is correct most of the time, this will tend to be good for performance.

Like all such performance optimisations you should only do it after extensive profiling to ensure the code really is in a bottleneck, and probably given the micro nature, that it is being run in a tight loop. Generally the Linux developers are pretty experienced so I would imagine they would have done that. They don't really care too much about portability as they only target gcc, and they have a very close idea of the assembly they want it to generate.


Let's decompile to see what GCC 4.8 does with it

Without __builtin_expect

#include "stdio.h"#include "time.h"int main() {    /* Use time to prevent it from being optimized away. */    int i = !time(NULL);    if (i)        printf("%d\n", i);    puts("a");    return 0;}

Compile and decompile with GCC 4.8.2 x86_64 Linux:

gcc -c -O3 -std=gnu11 main.cobjdump -dr main.o

Output:

0000000000000000 <main>:   0:       48 83 ec 08             sub    $0x8,%rsp   4:       31 ff                   xor    %edi,%edi   6:       e8 00 00 00 00          callq  b <main+0xb>                    7: R_X86_64_PC32        time-0x4   b:       48 85 c0                test   %rax,%rax   e:       75 14                   jne    24 <main+0x24>  10:       ba 01 00 00 00          mov    $0x1,%edx  15:       be 00 00 00 00          mov    $0x0,%esi                    16: R_X86_64_32 .rodata.str1.1  1a:       bf 01 00 00 00          mov    $0x1,%edi  1f:       e8 00 00 00 00          callq  24 <main+0x24>                    20: R_X86_64_PC32       __printf_chk-0x4  24:       bf 00 00 00 00          mov    $0x0,%edi                    25: R_X86_64_32 .rodata.str1.1+0x4  29:       e8 00 00 00 00          callq  2e <main+0x2e>                    2a: R_X86_64_PC32       puts-0x4  2e:       31 c0                   xor    %eax,%eax  30:       48 83 c4 08             add    $0x8,%rsp  34:       c3                      retq

The instruction order in memory was unchanged: first the printf and then puts and the retq return.

With __builtin_expect

Now replace if (i) with:

if (__builtin_expect(i, 0))

and we get:

0000000000000000 <main>:   0:       48 83 ec 08             sub    $0x8,%rsp   4:       31 ff                   xor    %edi,%edi   6:       e8 00 00 00 00          callq  b <main+0xb>                    7: R_X86_64_PC32        time-0x4   b:       48 85 c0                test   %rax,%rax   e:       74 11                   je     21 <main+0x21>  10:       bf 00 00 00 00          mov    $0x0,%edi                    11: R_X86_64_32 .rodata.str1.1+0x4  15:       e8 00 00 00 00          callq  1a <main+0x1a>                    16: R_X86_64_PC32       puts-0x4  1a:       31 c0                   xor    %eax,%eax  1c:       48 83 c4 08             add    $0x8,%rsp  20:       c3                      retq  21:       ba 01 00 00 00          mov    $0x1,%edx  26:       be 00 00 00 00          mov    $0x0,%esi                    27: R_X86_64_32 .rodata.str1.1  2b:       bf 01 00 00 00          mov    $0x1,%edi  30:       e8 00 00 00 00          callq  35 <main+0x35>                    31: R_X86_64_PC32       __printf_chk-0x4  35:       eb d9                   jmp    10 <main+0x10>

The printf (compiled to __printf_chk) was moved to the very end of the function, after puts and the return to improve branch prediction as mentioned by other answers.

So it is basically the same as:

int main() {    int i = !time(NULL);    if (i)        goto printf;puts:    puts("a");    return 0;printf:    printf("%d\n", i);    goto puts;}

This optimization was not done with -O0.

But good luck on writing an example that runs faster with __builtin_expect than without, CPUs are really smart these days. My naive attempts are here.

C++20 [[likely]] and [[unlikely]]

C++20 has standardized those C++ built-ins: How to use C++20's likely/unlikely attribute in if-else statement They will likely (a pun!) do the same thing.


These are macros that give hints to the compiler about which way a branch may go. The macros expand to GCC specific extensions, if they're available.

GCC uses these to to optimize for branch prediction. For example, if you have something like the following

if (unlikely(x)) {  dosomething();}return x;

Then it can restructure this code to be something more like:

if (!x) {  return x;}dosomething();return x;

The benefit of this is that when the processor takes a branch the first time, there is significant overhead, because it may have been speculatively loading and executing code further ahead. When it determines it will take the branch, then it has to invalidate that, and start at the branch target.

Most modern processors now have some sort of branch prediction, but that only assists when you've been through the branch before, and the branch is still in the branch prediction cache.

There are a number of other strategies that the compiler and processor can use in these scenarios. You can find more details on how branch predictors work at Wikipedia: http://en.wikipedia.org/wiki/Branch_predictor