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perl - else 比 elsif 慢吗?

转载 作者:行者123 更新时间:2023-12-03 08:14:06 25 4
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为什么这里带有 subeins else 比带有 subzwei elsif 慢?

#!/usr/bin/env perl
use warnings;
use 5.012;
use Benchmark qw(:all);

my $d = 0;
my $c = 2;

sub eins {
if ( $c == 1) {
$d = 1;
}
else {
$d = 2;
}
}

sub zwei {
if ( $c == 1) {
$d = 1;
}
elsif ( $c == 2 ) {
$d = 2;
}
}

sub drei {
$d = 1;
$d = 2 if $c == 2;
}

cmpthese( -5, {
eins => sub{ eins() },
zwei => sub{ zwei() },
drei => sub{ drei() },
} );
        Rate eins drei zwei
eins 4167007/s -- -1% -16%
drei 4207631/s 1% -- -15%
zwei 4972740/s 19% 18% --

Rate eins drei zwei
eins 4074356/s -- -8% -16%
drei 4428649/s 9% -- -9%
zwei 4854964/s 19% 10% --

Rate eins drei zwei
eins 3455697/s -- -6% -19%
drei 3672628/s 6% -- -14%
zwei 4250826/s 23% 16% --

Rate eins drei zwei
eins 2832634/s -- -8% -19%
drei 3088931/s 9% -- -12%
zwei 3503197/s 24% 13% --

Rate eins zwei drei
eins 3053821/s -- -17% -26%
zwei 3701601/s 21% -- -10%
drei 4131128/s 35% 12% --

Rate eins drei zwei
eins 3033041/s -- -2% -12%
drei 3092511/s 2% -- -10%
zwei 3430837/s 13% 11% --
Summary of my perl5 (revision 5 version 16 subversion 0) configuration:

Platform:
osname=linux, osvers=3.1.10-1.9-desktop, archname=x86_64-linux
uname='linux linux1 3.1.10-1.9-desktop #1 smp preempt thu apr 5 18:48:38 utc 2012 (4a97ec8) x86_64 x86_64 x86_64 gnulinux '
config_args='-de'
hint=recommended, useposix=true, d_sigaction=define
useithreads=undef, usemultiplicity=undef
useperlio=define, d_sfio=undef, uselargefiles=define, usesocks=undef
use64bitint=define, use64bitall=define, uselongdouble=undef
usemymalloc=n, bincompat5005=undef
Compiler:
cc='cc', ccflags ='-fno-strict-aliasing -pipe -fstack-protector -I/usr/local/include -D_LARGEFILE_SOURCE -D_FILE_OFFSET_BITS=64',
optimize='-O2',
cppflags='-fno-strict-aliasing -pipe -fstack-protector -I/usr/local/include'
ccversion='', gccversion='4.6.2', gccosandvers=''
intsize=4, longsize=8, ptrsize=8, doublesize=8, byteorder=12345678
d_longlong=define, longlongsize=8, d_longdbl=define, longdblsize=16
ivtype='long', ivsize=8, nvtype='double', nvsize=8, Off_t='off_t', lseeksize=8
alignbytes=8, prototype=define
Linker and Libraries:
ld='cc', ldflags =' -fstack-protector -L/usr/local/lib'
libpth=/usr/local/lib /lib/../lib64 /usr/lib/../lib64 /lib /usr/lib /lib64 /usr/lib64 /usr/local/lib64
libs=-lnsl -lndbm -lgdbm -ldb -ldl -lm -lcrypt -lutil -lc -lgdbm_compat
perllibs=-lnsl -ldl -lm -lcrypt -lutil -lc
libc=/lib/libc-2.14.1.so, so=so, useshrplib=false, libperl=libperl.a
gnulibc_version='2.14.1'
Dynamic Linking:
dlsrc=dl_dlopen.xs, dlext=so, d_dlsymun=undef, ccdlflags='-Wl,-E'
cccdlflags='-fPIC', lddlflags='-shared -O2 -L/usr/local/lib -fstack-protector'


Characteristics of this binary (from libperl):
Compile-time options: HAS_TIMES PERLIO_LAYERS PERL_DONT_CREATE_GVSV
PERL_MALLOC_WRAP PERL_PRESERVE_IVUV USE_64_BIT_ALL
USE_64_BIT_INT USE_LARGE_FILES USE_LOCALE
USE_LOCALE_COLLATE USE_LOCALE_CTYPE
USE_LOCALE_NUMERIC USE_PERLIO USE_PERL_ATOF
Built under linux
Compiled at May 24 2012 20:53:15
%ENV:
PERL_HTML_DISPLAY_COMMAND="/usr/bin/firefox -new-window %s"
@INC:
/usr/local/lib/perl5/site_perl/5.16.0/x86_64-linux
/usr/local/lib/perl5/site_perl/5.16.0
/usr/local/lib/perl5/5.16.0/x86_64-linux
/usr/local/lib/perl5/5.16.0
.

最佳答案

[这是一个答案,但它是不适合评论的有用信息。 ]

首先我们并排看一下编译后的表格,如果$c == 2 ,“zwei”的执行路径是“eins”的纯超集。 (用“*”标记。)

*1  <0> enter                            *1  <0> enter
*2 <;> nextstate(main 4 -e:2) v:{ *2 <;> nextstate(main 4 -e:2) v:{
*3 <#> gvsv[*c] s *3 <#> gvsv[*c] s
*4 <$> const[IV 1] s *4 <$> const[IV 1] s
*5 <2> eq sK/2 *5 <2> eq sK/2
*6 <|> cond_expr(other->7) vK/1 *6 <|> cond_expr(other->7) vK/1
7 <0> enter v 7 <0> enter v
8 <;> nextstate(main 1 -e:3) v:{ 8 <;> nextstate(main 1 -e:3) v:{
9 <$> const[IV 1] s 9 <$> const[IV 1] s
a <#> gvsv[*d] s a <#> gvsv[*d] s
b <2> sassign vKS/2 b <2> sassign vKS/2
c <@> leave vKP c <@> leave vKP
goto d goto d
*e <#> gvsv[*c] s
*f <$> const[IV 2] s
*g <2> eq sK/2
*h <|> and(other->i) vK/1
*e <0> enter v *i <0> enter v
*f <;> nextstate(main 2 -e:6) v:{ *j <;> nextstate(main 2 -e:6) v:{
*g <$> const[IV 2] s *k <$> const[IV 2] s
*h <#> gvsv[*d] s *l <#> gvsv[*d] s
*i <2> sassign vKS/2 *m <2> sassign vKS/2
*j <@> leave vKP *n <@> leave vKP
*d <@> leave[1 ref] vKP/REFC *d <@> leave[1 ref] vKP/REFC

问题是,我可以重现你的结果! (v5.16.0 为 x86_64-linux-thread-multi 构建)
           Rate drei eins zwei
drei 8974033/s -- -3% -19%
eins 9263260/s 3% -- -16%
zwei 11034175/s 23% 19% --

Rate drei eins zwei
drei 8971868/s -- -1% -21%
eins 9031677/s 1% -- -20%
zwei 11333871/s 26% 25% --

这不是一个小的不同(这可能是 CPU 缓存的结果),并且它可以在不同的运行之间重现(因此它不是影响基准的另一个应用程序)。我难住了。

每次迭代,少执行 4 次操作需要 22 ns(1/9031677 秒 - 1/11333871 秒)。我希望它减少大约 100 ns。

关于perl - else 比 elsif 慢吗?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/10977661/

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