varags函数返回vargas lambdas lua

如果我想编写一个接受varargs并返回一个接受vargas的函数的函数,我会遇到歧义…例如

function bind(func, ...) return function(...)  func(..., ...) end end

首先,你错过了关闭绑定功能的结束.

如果您有歧义,只需使用不同的名称解决它们.

function bind(func, ...)
  return function(...)  func(..., ...) end
end

如果我们测试你的代码：bind(print,“a”,“b”,“c”)(1,2,3)

你会得到输出：

1 1 2 3

如果在函数参数列表中有…或任何其他名称,则该变量将是该函数范围内的本地变量.它优先于优越范围内具有相同名称的任何其他变量.所以…在你的匿名函数中与函数绑定无关.

要解决此问题,您可以简单地执行类似的操作

function bind(func, ...)
  local a = table.pack(...)
  return function(...)  func(table.unpack(a, 1, a.n), ...) end
end

现在调用bind(print,“a”,“b”,“c”)(1,2,3)将输出：

a 1 2 3

要了解b和c发生了什么,请阅读本节：https://www.lua.org/manual/5.3/manual.html#3.4.11

(当然还有Lua手册的其余部分)

When a function is called, the list of arguments is adjusted to the
length of the list of parameters, unless the function is a vararg
function, which is indicated by three dots (‘…’) at the end of its
parameter list. A vararg function does not adjust its argument list;
instead, it collects all extra arguments and supplies them to the
function through a vararg expression, which is also written as three
dots. The value of this expression is a list of all actual extra
arguments, similar to a function with multiple results. If a vararg
expression is used inside another expression or in the middle of a
list of expressions, then its return list is adjusted to one element.
If the expression is used as the last element of a list of
expressions, then no adjustment is made (unless that last expression
is enclosed in parentheses).

所以像func(…,…)这样的东西永远不会起作用,即使……是两个不同的列表.

为避免这种情况,您必须连接两个参数列表.

function bind(func, ...)
  local args1 = table.pack(...)
  return function(...)
      local args2 = table.pack(...)

      for i = 1, args2.n do
          args1[args1.n+i] = args2[i]
      end
      args1.n = args1.n + args2.n

      func(table.unpack(args1, 1, args1.n))
  end
end

bind(print,“a”,nil,“c”)(1,nil,3)

这最终给了我们想要的输出：

a nil c 1 nil 3

但我相信你可以想出一个更好的方法来实现你的目标,而不是连接各种varargs.