标量上下文中的列表分配 [英] List Assignment in Scalar Context

问题描述

标量上下文中的列表分配返回右侧的元素数:

A list assignment in scalar context returns the number of elements on the right hand side:

scalar(my ($hello, $there, $world) = (7,8)); #evaluates to 2

为什么它要评估右侧并生成2，而不是评估新定义的列表并返回3?

Why does it evaluate the right hand side and produce 2, instead of the newly defined list being evaluated and returning 3?

在我看来，$hello得到7，$there得到8，而$world得到undef，然后该列表在标量上下文中求值，结果为3，因为这是数字列表中的元素($hello $there $world).对我来说，上下文影响返回所评估表达式的哪一部分似乎很奇怪:

To me, it seems like $hello gets 7, $there gets 8, and $world gets undef, then that list is evaluated in scalar context, which would result in 3, as that is the number of elements in the list ($hello $there $world). It seems weird to me that context affects which part of the evaluated expression is returned:

my $greeting = (($hello, $there, $world) = (7,8)); #2

my @greeting = (($hello, $there, $world) = (7,8));
my $greeting_length = @greeting; #3

推荐答案

该文档记录了 perlop (赋值运算符"部分的最后一句话):

It's documented to count the elements on the right in perlop (the last sentence in the Assignment Operators section):

类似地，列表上下文中的列表赋值产生分配给它的左值的列表，而标量上下文中的列表赋值返回由赋值右侧的表达式产生的元素数.

Similarly, a list assignment in list context produces the list of lvalues assigned to, and a list assignment in scalar context returns the number of elements produced by the expression on the right hand side of the assignment.

之所以这样工作，是因为您可以编写如下内容:

The reason it works like that is so that you can write things like this:

while (my ($key, $value) = each %hash) { ... }

如果它计算赋值左侧的元素数，那将是一个无限循环.

If it counted the number of elements on the left hand side of the assignment, that would be an infinite loop.

考虑一下，左侧的元素数量与右侧的元素数量相同或为常数(当您分配给标量列表时).在第一种情况下，计数哪一侧没有区别，在第二种情况下，计数右手侧更有用.

If you think about it, the number of elements on the left hand side is either the same as on the right hand side or it's a constant (when you're assigning to a list of scalars). In the first case, it makes no difference which side you count, and in the second case, counting the right hand side is more useful.

另一方面，在列表上下文中，赋值运算符返回左侧列表，因为这样做更有用.如果在修改列表元素的上下文中使用它，则要修改刚刚分配给的变量.

On the other hand, in list context the assignment operator returns the left hand list, because that's more useful. If you use it in a context that modifies the list elements, you want to modify the variables that were just assigned to.

回复:您的评论在您的示例中，(7,8)是一个包含两个元素的列表，这就是赋值运算符返回2的原因.当您将一个较短的列表分配给一个较长的标量列表时，分配发生之前，右侧不会用undef填充".相反，从右侧列表中没有与之关联的任何变量都将重置为其默认值.对于标量变量，为undef.对于数组，这是一个空数组.对于哈希，这是一个空哈希.

Re: your comment In your example, (7,8) is a two-element list, which is why the assignment operator returns 2. When you assign a shorter list to a longer list of scalars, the right hand side is not "padded out" with undef before the assignment happens. Instead, any variables that did not have a value associated with them from the right hand list are reset to their default value. For a scalar variable, that's undef. For arrays, that's an empty array. For hashes, that's an empty hash.

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