John Zwinck
John Zwinck

Reputation: 249283

GCC optimization missed opportunity

I'm compiling this C code:

int mode; // use aa if true, else bb
int aa[2];
int bb[2];

inline int auto0() { return mode ? aa[0] : bb[0]; }
inline int auto1() { return mode ? aa[1] : bb[1]; }

int slow() { return auto1() - auto0(); }
int fast() { return mode ? aa[1] - aa[0] : bb[1] - bb[0]; }

Both slow() and fast() functions are meant to do the same thing, though fast() does it with one branch statement instead of two. I wanted to check if GCC would collapse the two branches into one. I've tried this with GCC 4.4 and 4.7, with various levels of optimization such as -O2, -O3, -Os, and -Ofast. It always gives the same strange results:

slow():

        movl    mode(%rip), %ecx
        testl   %ecx, %ecx
        je      .L10

        movl    aa+4(%rip), %eax
        movl    aa(%rip), %edx
        subl    %edx, %eax
        ret
.L10:
        movl    bb+4(%rip), %eax
        movl    bb(%rip), %edx
        subl    %edx, %eax
        ret

fast():

        movl    mode(%rip), %esi
        testl   %esi, %esi
        jne     .L18

        movl    bb+4(%rip), %eax
        subl    bb(%rip), %eax
        ret
.L18:
        movl    aa+4(%rip), %eax
        subl    aa(%rip), %eax
        ret

Indeed, only one branch is generated in each function. However, slow() seems to be inferior in a surprising way: it uses one extra load in each branch, for aa[0] and bb[0]. The fast() code uses them straight from memory in the subls without loading them into a register first. So slow() uses one extra register and one extra instruction per call.

A simple micro-benchmark shows that calling fast() one billion times takes 0.7 seconds, vs. 1.1 seconds for slow(). I'm using a Xeon E5-2690 at 2.9 GHz.

Why should this be? Can you tweak my source code somehow so that GCC does a better job?

Edit: here are the results with clang 4.2 on Mac OS:

slow():

        movq    _aa@GOTPCREL(%rip), %rax   ; rax = aa (both ints at once)
        movq    _bb@GOTPCREL(%rip), %rcx   ; rcx = bb
        movq    _mode@GOTPCREL(%rip), %rdx ; rdx = mode
        cmpl    $0, (%rdx)                 ; mode == 0 ?
        leaq    4(%rcx), %rdx              ; rdx = bb[1]
        cmovneq %rax, %rcx                 ; if (mode != 0) rcx = aa
        leaq    4(%rax), %rax              ; rax = aa[1]
        cmoveq  %rdx, %rax                 ; if (mode == 0) rax = bb
        movl    (%rax), %eax               ; eax = xx[1]
        subl    (%rcx), %eax               ; eax -= xx[0]

fast():

        movq    _mode@GOTPCREL(%rip), %rax ; rax = mode
        cmpl    $0, (%rax)                 ; mode == 0 ?
        je      LBB1_2                     ; if (mode != 0) {
        movq    _aa@GOTPCREL(%rip), %rcx   ;   rcx = aa
        jmp     LBB1_3                     ; } else {
LBB1_2:                                    ; // (mode == 0)
        movq    _bb@GOTPCREL(%rip), %rcx   ;   rcx = bb
LBB1_3:                                    ; }
        movl    4(%rcx), %eax              ; eax = xx[1]
        subl    (%rcx), %eax               ; eax -= xx[0]

Interesting: clang generates branchless conditionals for slow() but one branch for fast()! On the other hand, slow() does three loads (two of which are speculative, one will be unnecessary) vs. two for fast(). The fast() implementation is more "obvious," and as with GCC it's shorter and uses one less register.

GCC 4.7 on Mac OS generally suffers the same issue as on Linux. Yet it uses the same "load 8 bytes then twice extract 4 bytes" pattern as Clang on Mac OS. That's sort of interesting, but not very relevant, as the original issue of emitting subl with two registers rather than one memory and one register is the same on either platform for GCC.

Upvotes: 24

Views: 1169

Answers (3)

user2284570
user2284570

Reputation: 3060

Have you tried to modify internals compilers parameters (--param name=value in man page). Those are not changed with any optimizations level (with three minor excepts).

Some of them control code reduction/deduplication.

For some optimizations in this section you can read things like « larger values can exponentially increase compilation time » .

Upvotes: 1

chill
chill

Reputation: 16888

The reason is that in the initial intermediate code, emitted for slow(), the memory load and the subtraction are in different basic blocks:

slow ()
{
  int D.1405;
  int mode.3;
  int D.1402;
  int D.1379;

  # BLOCK 2 freq:10000
  mode.3_5 = mode;
  if (mode.3_5 != 0)
    goto <bb 3>;
  else
    goto <bb 4>;

  # BLOCK 3 freq:5000
  D.1402_6 = aa[1];
  D.1405_10 = aa[0];
  goto <bb 5>;

  # BLOCK 4 freq:5000
  D.1402_7 = bb[1];
  D.1405_11 = bb[0];

  # BLOCK 5 freq:10000
  D.1379_3 = D.1402_17 - D.1405_12;
  return D.1379_3;
}

whereas in fast() they are in the same basic block:

fast ()
{
  int D.1377;
  int D.1376;
  int D.1374;
  int D.1373;
  int mode.1;
  int D.1368;

  # BLOCK 2 freq:10000
  mode.1_2 = mode;
  if (mode.1_2 != 0)
    goto <bb 3>;
  else
    goto <bb 4>;

  # BLOCK 3 freq:3900
  D.1373_3 = aa[1];
  D.1374_4 = aa[0];
  D.1368_5 = D.1373_3 - D.1374_4;
  goto <bb 5>;

  # BLOCK 4 freq:6100
  D.1376_6 = bb[1];
  D.1377_7 = bb[0];
  D.1368_8 = D.1376_6 - D.1377_7;

  # BLOCK 5 freq:10000
  return D.1368_1;
}

GCC relies on instruction combining pass to handle cases like this (i.e. apparently not on the peephole optimization pass) and combining works on the scope of a basic block. That's why the subtraction and load are combined in a single insn in fast() and they aren't even considered for combining in slow().

Later, in the basic block reordering pass, the subtraction in slow() is duplicated and moved into the basic blocks, which contain the loads. Now there's opportunity for the combiner to, well, combine the load and the subtraction, but unfortunately, the combiner pass is not run again (and perhaps it cannot be run that late in the compilation process with hard registers already allocated and stuff).

Upvotes: 22

jxh
jxh

Reputation: 70422

I don't have an answer as to why GCC is unable to optimize the code the way you want it to, but I have a way to re-organize your code to achieve similar performance. Instead of organizing your code the way you have done so in slow() or fast(), I would recommend that you define an inline function that returns either aa or bb based on mode without needing a branch:

inline int * xx () { static int *xx[] = { bb, aa }; return xx[!!mode]; }
inline int kwiky(int *xx) { return xx[1] - xx[0]; }
int kwik() { return kwiky(xx()); }

When compiled by GCC 4.7 with -O3:

    movl    mode, %edx
    xorl    %eax, %eax
    testl   %edx, %edx
    setne   %al
    movl    xx.1369(,%eax,4), %edx
    movl    4(%edx), %eax
    subl    (%edx), %eax
    ret

With the definition of xx(), you can redefine auto0() and auto1() like so:

inline int auto0() { return xx()[0]; }
inline int auto1() { return xx()[1]; }

And, from this, you should see that slow() now compiles into code similar or identical to kwik().

Upvotes: 9

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