作为函数参数传递时，如何停止评估Lisp格式? [英] How to stop evaluating lisp form when passed as function parameter?

问题描述

我正在学习Lisp.现在，我正在尝试创建一个函数，该函数采用一些有效的Lisp形式作为参数，并返回一个在调用时执行Lisp形式的函数.例如:

I am learning Lisp. Now I am trying to create a function that takes some valid Lisp form as argument and returns a function that executes the Lisp forms when called. For example:

(defun fn (name action)
  (setf (symbol-function name)
        #'(lambda () action)))

当我通过时说(+ 4 5 6)时，该函数将使用特定名称创建，并在调用时返回总和.

When I am passing say (+ 4 5 6) the function is getting created with specific name and when called returning the sum.

(fn 'add (+ 4 5 6))
(add) ==> 15

但是，如果我调用(fn 'error (assert (= 2 3))，则会抛出错误(= 2 3) must evaluate to a non-NIL value.，并且不会创建名称为error的函数.

But if I invoke (fn 'error (assert (= 2 3)) it is throwing error (= 2 3) must evaluate to a non-NIL value. and the function with name error is not created.

当作为函数参数传递时，如何停止对此assert的求值?

How can I stop this evaluation of assert when passed as function parameter?

推荐答案

您无法编写此函数；它必须是宏运算符.如果fn是函数，则调用:

You cannot write this function; it has to be a macro operator. If fn is a function, then the call:

(fn 'add (+ 4 5 6))

计算参数(+ 4 5 6)，将其减小为值15.该函数接收15，而不接收表达式.我们可以通过引用以下代码修复"此问题:

evaluates the argument (+ 4 5 6), reducing it to the value 15. The function receives 15, and not the expression. We can "fix" this by quoting the code:

(fn 'add '(+ 4 5 6))

，但是这样的问题是代码不会与词汇环境交互.例如，这将不起作用，因为x在fn内部不可见:

but then we have the problem that the code doesn't interact with the lexical environment. For instance, this won't work, because x is not visible inside fn:

(let ((x 40)) (fn 'add '(+ x 2)))

要创建一个在适当的环境中求值(+ x 2)的函数，我们必须在相同的词法范围内正确使用lambda运算符:

To create a function which evaluates (+ x 2) in the proper environment, we must the lambda operator right in that same lexical scope:

(let ((x 40)) (lambda () (+ x 2)))

您的fn运算符可以写为生成lambda(没有任何名称)的语法糖:

Your fn operator can be written as a syntactic sugar that generates the lambda (without any name):

(defmacro fn (expr) `(lambda () ,expr))

现在我们可以写:

(let ((x 40)) (fn (+ x 2))) ;; returns a function which returns 42

要做命名的事情:

(defmacro fn (name expr) `(setf (symbol-function ',name) (lambda () ,expr)))

但是，这是一个很糟糕的主意；我们在函数中引入了讨厌的全局副作用.更好的"fn"可能是在某些形式上为函数引入词法绑定的函数.也就是说，它可以像这样使用:

However, this is a quite a poor idea; we're introducing a nasty global side effect into a function. A better "named fn" might be one which introduces a lexical binding for a function over some forms. That is, it can be used like this:

(fn (foo (+ x 2)) (foo))
             ;;  ^^^^^^  foo is a lexical function in this scope
             ;;          denoting the function (lambda () (+ x 2))

可以这样做:

(defmacro fn ((name expr) &rest forms)
   `(flet ((,name () ,expr)) ,@forms)))

或者，如果您希望将名称作为变量绑定而不是函数绑定，则用法为(fn (foo (+ x 2)) (funcall foo)):

Or, if you want the name as a variable binding rather than a function binding, so that the usage is (fn (foo (+ x 2)) (funcall foo)):

(defmacro fn ((name expr) &rest forms)
  `(let ((,name (lambda () ,expr))) ,@forms))

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