Perl中@_的含义是什么? [英] What is the meaning of @_ in Perl?

问题描述

@_ 在 Perl 中是什么意思?

What is the meaning of @_ in Perl?

推荐答案

perldoc perlvar 是检查任何特殊命名的 Perl 变量信息的第一个地方.

perldoc perlvar is the first place to check for any special-named Perl variable info.

引用:

@_:在子例程中，数组 @_ 包含传递给该子例程的参数.

@_: Within a subroutine the array @_ contains the parameters passed to that subroutine.

可以在从 perlvar 链接的 perldoc perlsub(Perl 子例程) 中找到更多详细信息:

More details can be found in perldoc perlsub (Perl subroutines) linked from the perlvar:

传入的任何参数都显示在数组 @_ .

Any arguments passed in show up in the array @_ .

因此，如果你调用一个带有两个参数的函数，那些将存储在 $_[0] 和 $_[1] 中.

Therefore, if you called a function with two arguments, those would be stored in $_[0] and $_[1].

数组@_ 是一个本地数组，但它的元素是实际标量参数的别名.特别地，如果一个元素 $_[0] 被更新，相应的参数被更新(或如果不是，则会发生错误可更新).

The array @_ is a local array, but its elements are aliases for the actual scalar parameters. In particular, if an element $_[0] is updated, the corresponding argument is updated (or an error occurs if it is not updatable).

如果一个参数是一个数组或不存在的散列元素当函数被调用时，元素仅在(和如果)时创建它被修改或对它的引用是采取.(一些早期版本的 Perl无论是否创建元素元素被分配给.)分配给整个数组@_ 删除了别名，并且不更新任何参数.

If an argument is an array or hash element which did not exist when the function was called, that element is created only when (and if) it is modified or a reference to it is taken. (Some earlier versions of Perl created the element whether or not the element was assigned to.) Assigning to the whole array @_ removes that aliasing, and does not update any arguments.

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