Perl

Question

根据 Perldoc(https://perldoc.perl.org/perlop#Comma-Operator):

Comma Operator

...

In list context, it's just the list argument separator, and inserts both its arguments into the list. These arguments are also evaluated from left to right.

在下面的代码中，我认为 $n 应该是 1 但它变成了 2。我错过了什么？

my $v = 1;
my $n = ( $v, $v = 2 )[0];
# I thought left "$v" would be evaluated (and return 1) first, then right "$v = 2" would.
say "n = $n";  # but output was 2
say "v = $v";  # 2

或

my $v = 1;
my @l = ( $v, $v = 2 );
say Dumper(\@l);   # [ 2, 2 ], not [ 1, 2 ]

即使我更改了顺序：

my $v = 1;
# my $n = ( $v, $v = 2 )[0];
my $n = ( $v = 2, $v )[1];
say "n = $n";  # still 2
say "v = $v";  # 2

Answer 1

它确实是从左到右计算的。

$ perl -MO=Concise,-exec -e'my $v = 1; my $n = ( $v, $v = 2 )[0];'
1  <0> enter v
2  <;> nextstate(main 1 -e:1) v:{
3  <$> const[IV 1] s                   \
4  <0> padsv[$v:1,3] sRM*/LVINTRO       > my $v = 1;
5  <2> sassign vKS/2                   /
6  <;> nextstate(main 2 -e:1) v:{
7  <0> pushmark s                                  \              \
8  <$> const[IV 0] s                               |              |
9  <0> pushmark s                                  |              |
a  <0> padsv[$v:1,3] s                    $v       > ( ... )[0]   |
b  <$> const[IV 2] s                   \           |              > my $n = ...
c  <0> padsv[$v:1,3] sRM*               > $v = 2   |              |
d  <2> sassign sKS/2                   /           |              |
e  <2> lslice sK/2                                /               |
f  <0> padsv[$n:2,3] sRM*/LVINTRO                                 |
g  <2> sassign vKS/2                                             /
h  <@> leave[1 ref] vKP/REFC
-e syntax OK

要注意的是，与 $v 关联的标量放在堆栈上，而不是整数。你可以在这里看到发生了什么：

use 5.014;
use warnings;

my @stack;

sub sassign {
   my $rhs = pop(@stack);
   my $lhs = pop(@stack);
   $$rhs = $$lhs;
   push @stack, $rhs;
}
                       # Stack     $v   $n
                       # -------   --   --
push @stack, ;       # 1
push @stack, \my $v;   # 1,$v      !d
sassign();             # $v        1
@stack = ();           #           1

push @stack, $v;      # $v        1

push @stack, ;       # $v,2      1
push @stack, $v;      # $v,2,$v   1
sassign();             # $v,$v     2

pop(@stack);           # $v        2          # Net result of slice.
push @stack, \my $n;   # $v,$n     2    !d
sassign();             # $n        2    2
@stack = ();           #           2    2

say $n;  # 2

如果您更喜欢别名而不是引用：

use 5.014;
use warnings;
use experimental qw( refaliasing declared_refs );

my @stack;

sub sassign {
   my $rhs = \pop(@stack);
   my $lhs = \pop(@stack);
   $rhs = $lhs;
   $stack[@stack] = $rhs;
}
                            # Stack     $v   $n
                            # -------   --   --
$stack[@stack] = ;       # 1
$stack[@stack] = \my $v;   # 1,$v      !d
sassign();                  # $v        1
@stack = ();                #           1

$stack[@stack] = $v;      # $v        1

$stack[@stack] = ;       # $v,2      1
$stack[@stack] = $v;      # $v,2,$v   1
sassign();                  # $v,$v     2

pop(@stack);                # $v        2          # Net result of slice.
$stack[@stack] = \my $n;   # $v,$n     2    !d
sassign();                  # $n        2    2
@stack = ();                #           2    2

say $n;  # 2

Perl - 列表元素的评估顺序，在列表内分配

Perl - evaluation order of list elements, with assignment inside the list

evaluation