数组 – “$scalar = @array [n]”的后果是什么？

发布时间：2020-12-16 18:26:36 所属栏目：安全来源：网络整理

导读：use warnings;my @array = (0,1);my $scalar1 = $array[0];my $scalar2 = @array[0];if($scalar1 == $scalar2) { print "scalars are equaln";} 这是运行/usr/bin/perl5.10.1 test.pl时的输出： Scalar value @array[0] better written as $array[0] at tes

use warnings;
my @array = (0,1);
my $scalar1 = $array[0];
my $scalar2 = @array[0];
if($scalar1 == $scalar2) {
    print "scalars are equaln";
}

这是运行/usr/bin/perl5.10.1 test.pl时的输出：

Scalar value @array[0] better written as $array[0] at test.pl line 4.
scalars are equal

我很担心这个警告.

解决方法

您可以在 perldoc perldiag中查找所有警告消息,这解释了后果：

(W syntax) You’ve used an array slice (indicated by @) to select a
single element of an array. Generally
it’s better to ask for a scalar value
(indicated by $). The difference is
that $foo[&bar] always behaves like a
scalar,both when assigning to it and
when evaluating its argument,while
@foo[&bar] behaves like a list when
you assign to it,and provides a list
context to its subscript,which can do
weird things if you’re expecting only
one subscript.

On the other hand,if you were actually hoping to treat the array
element as a list,you need to look
into how references work,because Perl
will not magically convert between
scalars and lists for you. See
perlref.

同样,您可以use diagnostics;获取警告消息的详细说明.

第三种方法是使用splain实用程序.

（编辑：李大同）

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