无法将函数调用用作s-exp中的第一个参数 [英] Cannot use function call as first argument in s-exp

问题描述

我正在尝试使用函数以通用Lisp返回函数.但是，我遇到了一种奇怪的情况，想解释一下.

I'm trying to use a function to return a function in common lisp. However, I've run into a strange situation I'd like explained.

这就是我想要做的:

(defun makefun(x) ;1
  (lambda (z)
    (+ x z)))

((makefun 1) 2) ;2

这将导致非法的函数调用.但是，以下内容是有效的:

This results in an illegal function call. However, the following are valid:

((lambda (z) (+ 1 z)) 2) ;3

(funcall (makefun 1) 2) ;4

为什么不能像第一个示例那样使用makefun?我希望对2中的调用进行评估，以使其等效于第3行.

Why can I not use makefun as in the first example? I'd expect the call in 2 to be evaluated so it would be equivalent to line 3.

推荐答案

如果已在Stackoverflow上回答了类似的问题.这是重复的.需要找到它.

If answered a similar question already on Stackoverflow. It's a duplicate. Would need to find it.

反正.

(defun makefun(x) ;1
  (lambda (z)
    (+ x z)))

((makefun 1) 2)

Common Lisp不允许在功能位置进行评估.它要求您在此处放置符号或实际的lambda表达式.

Common Lisp does not allow evaluation in the function position. It requires you to put a symbol or an actual lambda expression there.

请记住:Common Lisp有一个单独的 function名称空间. 值命名空间是不同的. MAKEFUN返回一个值，该值在函数名称空间中不可用.

Remember: Common Lisp has a separate function namespace. The value namespace is different. Here MAKEFUN returns a value and this value is not available in the function namespace.

只有两种语法调用方法:

There are only two syntactic ways to call a function:

(foo 1 2)

和

((lambda (a b) (list a b)) 1 2)

请参见 CLHS(以表格形式).这里有一个 Lambda表格.

添加编写((some-function-returning-function) 1 2)的功能会使Common Lisp中的函数调用更加混乱:

Adding the capability to write ((some-function-returning-function) 1 2) would make function calling in Common Lisp more confusing:

• (foo 1 2)将使用函数名称空间
• ((some-function-returning-function) 1 2)将使用值名称空间

• (foo 1 2) would use the function namespace
• ((some-function-returning-function) 1 2) would use the value namespace

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