在 Scratch 2.0 中，添加了对自定义堆栈块（过程）的支持。但是有没有办法使用它来“抽象”返回值的逻辑？
例如，我在这里有一个脚本来天真地计算指数：（查看图形表示）
set [base v] to [2]
set [index v] to [3]
... // above is for initializing
set [result v] to (base)
repeat until <(index) = [1]>
set [result v] to ((result) * (base))
change [index v] by (-1)
我如何将此逻辑导出到“自定义报告器”以重用？
这是一个示例（渲染）：
define split [text] by [splitter]
delete (all v) of [output list v]
set [parse v] to [0]
set [cur string v] to []
repeat until ((parse) > (length of (splitter))
if <(letter (parse) of (text)) = (splitter)> then
add (cur string) to [output list v]
set [cur string v] to []
else
set [cur string v] to (join (cur string) (letter (parse) of (text)))
end
end
when GF clicked
split [Hello, world! Do you like this?] by [ ] // That's a space.
// That should output a list: ["Hello,", "world!", "Do", "you", "like", "this?"]
define the answer to life
set [output var v] to (42)
when GF clicked
the answer to life
say (output var)
它展示了如何同时使用列表输出和变量输出。
最简单的方法是创建一个自定义命令块，并将返回值存储在一个变量中。这有一些缺点，例如不允许递归调用，但在大多数情况下都有效。我还建议将块设置为在不刷新屏幕的情况下运行。
像这样简单地定义它，返回值可用result：
define (base) ^ (exp)
set [index v] to (exp) // need a variable, as arguments are immutable
set [result v] to (base)
repeat until <(index) = [1]>
set [result v] to ((result) * (base))
change [index v] by (-1)
然后可以这样调用它：
when gf clicked
(4) ^ (3) // the stack block
say (join [4 ^ 3 = ] (result)) // result is set by the [()^()] block
还有第二种更复杂的方法可以做到这一点。它允许递归块，您可以多次运行一个块。我称它为堆叠方法，因为我使用列表作为堆栈。例如，请参阅我制作的这个项目。
此方法也不会弄乱变量调色板。
define (base) ^ (index) recursive
if <(index) = [1]>
add (base) to [stack v]
else
(base) ^ ((index) - (1)) recursive // adds the previous item to the stack
add ((base) * (item (last v) of [stack v])) to [stack v]
delete ((length of [stack v]) - (1)) of [stack v] // clean up
然后可以使用与我在另一个答案中解释的基本相同的方法访问它：
when gf clicked
(4) ^ (3) recursive // the stack block
say (join [4 ^ 3 = ] (item (last v) of [stack v])) // get the item from the end of the stack
delete [last v] of [stack v] // optional, if you want to clean up