调用函数在Lua中作为参数传递

Question

我有这个代码

Option = { }


function Option.nothing( )
  local self = { isNone = true, isSome = false }

  function self:orElse( alt )
    return alt
  end

  function self:map( f )
    return Option.nothing( )
  end

  function self:exec( f )
  end

  function self:maybe( alt, f )
    return alt
  end

  return self
end



function Option.just( val )
  local self = { isNone = false, isSome = true }
  local value = val

  function self:orElse( alt )
    return value
  end

  function self:map( f )
    return Option.just( f(value) )
  end

  function self:exec( f )
    f( value )
  end

  function self:maybe( alt, f )
    return f(value)
  end

  return self
end



function printOpt( opt )
  local str = opt.maybe( "Nothing", function(s) return "Just " .. s end )
  print( str )
end


x = Option.nothing( )
y = Option.just( 4 )

printOpt(x)
printOpt(y)

但是我一直在'尝试在这里调用本地'f'(零值)':

  function self:maybe( alt, f )
    return f(value)
  end

看来我在调用作为参数传递的函数时遇到了麻烦.

Answer 1

你将函数声明为self:maybe(),但是你将其称为opt.maybe().你应该把它称为opt:maybe().

声明它self:maybe(alt, f)等同于声明它self.maybe(self, alt, f).因此,如果你用它来调用它.需要3个参数.你传递2,所以self最终为"Nothing",并alt最终成为函数对象.

然而,通过调用它,因为opt:maybe("Nothing", f)这相当于说opt.maybe(opt, "Nothing", f)提供了所需的3个args.