函数在返回数组的时候遇到了问题。。。大家都来看看：）

dugu072_cu

可以这样:
shift( @{ [ &func ] } );

dugu072_cu

ft～
有这么麻烦嘛，与什么上下文无关，就一点： list是 value，array 是 variable！
由于 shift 要修改传递的参数，故只能对  variable 进行操作， 而 list 是一个 value，相当于一个常量，当然无法传递给 shift 了……

dugu072_cu

黑色阳光_cu 是对的。编译不通过是因为：
shift(func()); 失败是因为shift的参数只能是array.而不是一个表 ...
toniz 发表于 2010-08-05 15:44

    黑色阳光_cu 的 warn \func(); ，说明了 return 确实是返回 list；但结合你给的代码，反而很难理解了：为什么 return 回来的 list，这里却可以修改？

dugu072_cu

我觉得，一个正常的函数在做return @array;的时候就是返回了一个数组。
不能说func2返回的不是数组，但 ...
py 发表于 2010-08-06 09:19

    perldoc -q "a list and an array"

What is the difference between a list and an array?
  An array has a changeable length. A list does not. An array is something
  you can push or pop, while a list is a set of values. Some people make
  the distinction that a list is a value while an array is a variable.
  Subroutines are passed and return lists, you put things into list
  context, you initialize arrays with lists, and you "foreach()" across a
  list. "@" variables are arrays, anonymous arrays are arrays, arrays in
  scalar context behave like the number of elements in them, subroutines
  access their arguments through the array @_, and "push"/"pop"/"shift"
  only work on arrays.

  As a side note, there's no such thing as a list in scalar context. When
  you say

          $scalar = (2, 5, 7, 9);

  you're using the comma operator in scalar context, so it uses the scalar
  comma operator. There never was a list there at all! This causes the
  last value to be returned: 9.

我是看过这段话，所以一直记得，return 的是 list ……