[英]Get the first item of the list returned by function
A simple simulation of the problem:问题的简单模拟:
use strict;
use warnings;
sub uniq
{
my %seen;
grep !$seen{$_}++, @_;
}
my @a = (1, 2, 3, 1, 2);
print shift @{uniq(@a)};
Can't use string ("3") as an ARRAY ref while "strict refs" in use
在使用“strict refs”时不能使用字符串(“3”)作为 ARRAY ref
Need to impose a list context on the function call, and then pick the first element from the list.需要在 function 调用上施加一个列表上下文,然后从列表中选取第一个元素。
The print
, or any other subroutine call, already supplies a list context. print
或任何其他子例程调用已经提供了列表上下文。 Then one way to extract an element from a returned list然后一种从返回列表中提取元素的方法
print +( func(@ary) )[0];
This disregards the rest of the list.这忽略了列表中的 rest。
That +
is necessary (try without it), unless we equip print itself with parentheses around all its arguments, that is那个
+
是必要的(不用它试试),除非我们用括号围绕它的所有 arguments 来装备print本身,即
print( (func(@ary))[0] );
One option could be to return an array reference:一种选择是返回一个数组引用:
sub uniq {
my %seen;
[grep !$seen{$_}++, @_];
}
If uniq
returned an array reference, then @{uniq(...)}
would be the correct idiom to get an array (which is a suitable argument for shift
).如果
uniq
返回数组引用,那么@{uniq(...)}
将是获取数组的正确习惯用法(这是shift
的合适参数)。 For a more general list, you can cast the list to an array reference and then dereference it.对于更通用的列表,您可以将列表转换为数组引用,然后取消引用它。
print shift @{ [ uniq(@a) ] };
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