这失败了:
my @a = ("a", "b", "c", "d", "e");
my %h = map { "prefix-$_" => 1 } @a;
出现此错误:
Not enough arguments for map at foo.pl line 4, near "} @a"
但这有效:
my @a = ("a", "b", "c", "d", "e");
my %h = map { "prefix-" . $_ => 1 } @a;
为什么?
答案 0 :(得分:21)
因为Perl猜测EXPR(例如哈希引用)而不是BLOCK。这应该有用(注意'+'符号):
my @a = ("a", "b", "c", "d", "e");
my %h = map { +"prefix-$_" => 1 } @a;
答案 1 :(得分:13)
我更喜欢把它写成
my %h = map { ("prefix-$_" => 1) } @a;
显示意图,我将返回一个2元素列表。
答案 2 :(得分:11)
来自perldoc -f map
:
"{" starts both hash references and blocks, so "map { ..."
could be either the start of map BLOCK LIST or map EXPR, LIST.
Because perl doesn’t look ahead for the closing "}" it has to
take a guess at which its dealing with based what it finds just
after the "{". Usually it gets it right, but if it doesn’t it
won’t realize something is wrong until it gets to the "}" and
encounters the missing (or unexpected) comma. The syntax error
will be reported close to the "}" but you’ll need to change
something near the "{" such as using a unary "+" to give perl
some help:
%hash = map { "\L$_", 1 } @array # perl guesses EXPR. wrong
%hash = map { +"\L$_", 1 } @array # perl guesses BLOCK. right
%hash = map { ("\L$_", 1) } @array # this also works
%hash = map { lc($_), 1 } @array # as does this.
%hash = map +( lc($_), 1 ), @array # this is EXPR and works!
%hash = map ( lc($_), 1 ), @array # evaluates to (1, @array)
or to force an anon hash constructor use "+{"
@hashes = map +{ lc($_), 1 }, @array # EXPR, so needs , at end
and you get list of anonymous hashes each with only 1 entry.
答案 3 :(得分:6)
另外,另一种方法是做你正在做的事情,初始化哈希,你可以这样做:
my @a = qw( a b c d e );
my %h;
@h{@a} = ();
这将为五个键中的每个键创建undef条目。如果你想给他们所有真正的价值,那么就这样做。
@h{@a} = (1) x @a;
您也可以使用循环显式地执行此操作;
@h{$_} = 1 for @a;
答案 4 :(得分:1)
我认为
map { ; "prefix-$_" => 1 } @a;
更具惯用性,只要指定它是一个语句块而不是散列引用。你只是用空语句踢它。