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问题:
I came across a #define in which they use __builtin_expect.
The documentation says:
Built-in Function: long __builtin_expect (long exp, long c)
You may use __builtin_expect to provide the compiler with branch
prediction information. In general, you should prefer to use actual
profile feedback for this (-fprofile-arcs), as programmers are
notoriously bad at predicting how their programs actually perform.
However, there are applications in which this data is hard to collect.
The return value is the value of exp, which should be an integral
expression. The semantics of the built-in are that it is expected that
exp == c. For example:
if (__builtin_expect (x, 0))
foo ();
would indicate that we do not expect to call foo, since we expect x to be zero.
So why not directly use:
if (x)
foo ();
instead of the complicated syntax with __builtin_expect?
回答1:
Imagine the assembly code that would be generated from:
if (__builtin_expect(x, 0)) {
foo();
...
} else {
bar();
...
}
I guess it should be something like:
cmp $x, 0
jne _foo
_bar:
call bar
...
jmp after_if
_foo:
call foo
...
after_if:
You can see that the instructions are arranged in such an order that the bar case precedes the foo case (as opposed to the C code). This can utilise the CPU pipeline better, since a jump thrashes the already fetched instructions.
Before the jump is executed, the instructions below it (the bar case) are pushed to the pipeline. Since the foo case is unlikely, jumping too is unlikely, hence thrashing the pipeline is unlikely.
回答2:
Let's decompile to see what GCC 4.8 does with it
Blagovest mentioned branch inversion to improve the pipeline, but do current compilers really do it? Let's find out!
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)
puts("a");
return 0;
}
Compile and decompile with GCC 4.8.2 x86_64 Linux:
gcc -c -O3 -std=gnu11 main.c
objdump -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 0a jne 1a <main+0x1a>
10: bf 00 00 00 00 mov $0x0,%edi
11: R_X86_64_32 .rodata.str1.1
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
The instruction order in memory was unchanged: first the puts and then 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 07 je 17 <main+0x17>
10: 31 c0 xor %eax,%eax
12: 48 83 c4 08 add $0x8,%rsp
16: c3 retq
17: bf 00 00 00 00 mov $0x0,%edi
18: R_X86_64_32 .rodata.str1.1
1c: e8 00 00 00 00 callq 21 <main+0x21>
1d: R_X86_64_PC32 puts-0x4
21: eb ed jmp 10 <main+0x10>
The puts was moved to the very end of the function, the retq return!
The new code is basically the same as:
int i = !time(NULL);
if (i)
goto puts;
ret:
return 0;
puts:
puts("a");
goto ret;
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 those 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.
回答3:
The idea of __builtin_expect is to tell the compiler that you'll usually find that the expression evaluates to c, so that the compiler can optimize for that case.
I'd guess that someone thought they were being clever and that they were speeding things up by doing this.
Unfortunately, unless the situation is very well understood (it's likely that they have done no such thing), it may well have made things worse. The documentation even says:
In general, you should prefer to use actual profile feedback for this (-fprofile-arcs), as programmers are notoriously bad at predicting how their programs actually perform. However, there are applications in which this data is hard to collect.
In general, you shouldn't be using __builtin_expect unless:
- You have a very real performance issue
- You've already optimized the algorithms in the system appropriately
- You've got performance data to back up your assertion that a particular case is the most likely
回答4:
Well, as it says in the description, the first version adds a predictive element to the construction, telling the compiler that the x == 0 branch is the more likely one - that is, it's the branch that will be taken more often by your program.
With that in mind, the compiler can optimize the conditional so that it requires the least amount of work when the expected condition holds, at the expense of maybe having to do more work in case of the unexpected condition.
Take a look at how conditionals are implemented during the compilation phase, and also in the resulting assembly, to see how one branch may be less work than the other.
However, I would only expect this optimization to have noticeable effect if the conditional in question is part of a tight inner loop that gets called a lot, since the difference in the resulting code is relatively small. And if you optimize it the wrong way round, you may well decrease your performance.
回答5:
I don't see any of the answers addressing the question that I think you were asking, paraphrased:
Is there a more portable way of hinting branch prediction to the compiler.
The title of your question made me think of doing it this way:
if ( !x ) {} else foo();
If the compiler assumes that 'true' is more likely, it could optimize for not calling foo().
The problem here is just that you don't, in general, know what the compiler will assume -- so any code that uses this kind of technique would need to be carefully measured (and possibly monitored over time if the context changes).
回答6:
I test it on Mac according @Blagovest Buyukliev and @Ciro. The assembles look clear and I add comments;
Commands are
gcc -c -O3 -std=gnu11 testOpt.c; otool -tVI testOpt.o
When I use -O3 ， it looks the same no matter the __builtin_expect(i, 0) exist or not.
testOpt.o:
(__TEXT,__text) section
_main:
0000000000000000 pushq %rbp
0000000000000001 movq %rsp, %rbp // open function stack
0000000000000004 xorl %edi, %edi // set time args 0 (NULL)
0000000000000006 callq _time // call time(NULL)
000000000000000b testq %rax, %rax // check time(NULL) result
000000000000000e je 0x14 // jump 0x14 if testq result = 0， namely jump to puts
0000000000000010 xorl %eax, %eax // return 0 , return appear first
0000000000000012 popq %rbp // return 0
0000000000000013 retq // return 0
0000000000000014 leaq 0x9(%rip), %rdi ## literal pool for: "a" // puts part, afterwards
000000000000001b callq _puts
0000000000000020 xorl %eax, %eax
0000000000000022 popq %rbp
0000000000000023 retq
When compile with -O2 ， it looks different with and without __builtin_expect(i, 0)
First without
testOpt.o:
(__TEXT,__text) section
_main:
0000000000000000 pushq %rbp
0000000000000001 movq %rsp, %rbp
0000000000000004 xorl %edi, %edi
0000000000000006 callq _time
000000000000000b testq %rax, %rax
000000000000000e jne 0x1c // jump to 0x1c if not zero, then return
0000000000000010 leaq 0x9(%rip), %rdi ## literal pool for: "a" // put part appear first , following jne 0x1c
0000000000000017 callq _puts
000000000000001c xorl %eax, %eax // return part appear afterwards
000000000000001e popq %rbp
000000000000001f retq
Now with __builtin_expect(i, 0)
testOpt.o:
(__TEXT,__text) section
_main:
0000000000000000 pushq %rbp
0000000000000001 movq %rsp, %rbp
0000000000000004 xorl %edi, %edi
0000000000000006 callq _time
000000000000000b testq %rax, %rax
000000000000000e je 0x14 // jump to 0x14 if zero then put. otherwise return
0000000000000010 xorl %eax, %eax // return appear first
0000000000000012 popq %rbp
0000000000000013 retq
0000000000000014 leaq 0x7(%rip), %rdi ## literal pool for: "a"
000000000000001b callq _puts
0000000000000020 jmp 0x10
To summarize, __builtin_expect works in the last case.
