在 Perl 中，使用@a[$i] 访问数组元素与使用 $a[$i] 有什么区别? [英] In Perl, what is the difference between accessing an array element using @a[$i] as opposed to using $a[$i]?

问题描述

我遵循的基本语法教程没有说明这一点:

Basic syntax tutorials I followed do not make this clear:

使用前一个或后一个下标符号访问数组之间是否有任何实际/哲学/上下文相关/棘手的区别?

Is there any practical/philosophical/context-dependent/tricky difference between accessing an array using the former or latter subscript notation?

$ perl -le 'my @a = qw(io tu egli); print $a[1], @a[1]'

两种情况下的输出似乎相同.

The output seems to be the same in both cases.

推荐答案

$a[...]   # array element

返回由索引表达式标识的一个元素，并且

returns the one element identified by the index expression, and

@a[...]   # array slice

返回索引表达式标识的所有元素.

returns all the elements identified by the index expression.

因此，

• 当您打算访问单个元素以将此信息传达给读者时，您应该使用 $a[EXPR].事实上，如果您不这样做，您可能会收到警告.
• 当您打算访问许多元素或可变数量的元素时，您应该使用 @a[LIST].

• You should use $a[EXPR] when you mean to access a single element in order to convey this information to the reader. In fact, you can get a warning if you don't.
• You should use @a[LIST] when you mean to access many elements or a variable number of elements.

但这并不是故事的结局.您询问了实际和棘手(微妙?)的差异，但还没有提到:数组元素的索引表达式是在标量上下文中计算的，而数组切片的索引表达式是在列表上下文中计算的.

But that's not the end of the story. You asked for practical and tricky (subtle?) differences, and there's one noone mentioned yet: The index expression for an array element is evaluated in scalar context, while the index expression for an array slice is evaluated in list context.

sub f { return @_; }

$a[ f(4,5,6) ]     # Same as $a[3]
@a[ f(4,5,6) ]     # Same as $a[4],$a[5],$a[6]

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