通用Lisp宏本质上扩展了代码的“语法原语”。

例如，在C语言中，switch/case结构只适用于整型，如果你想将它用于浮点数或字符串，你就只能使用嵌套的if语句和显式比较。你也不可能编写一个C宏来为你做这项工作。

但是，由于lisp宏(本质上)是一个lisp程序，它接受代码片段作为输入，并返回代码来替换宏的“调用”，因此您可以尽可能地扩展您的“原语”库，通常最终会得到一个更可读的程序。

要在C中做同样的事情，您必须编写一个自定义预处理器，它会吃掉您的初始(不完全是C)源代码，并吐出C编译器可以理解的东西。这不是一种错误的方法，但它不一定是最简单的。

我从通用的lisp烹饪书中得到了这个，我认为它解释了为什么lisp宏是有用的。

宏是一段普通的Lisp代码，它对另一段假定的Lisp代码进行操作，将其翻译成(更接近于)可执行的Lisp。这听起来可能有点复杂，所以让我们举一个简单的例子。假设您想要一个版本的setq，将两个变量设置为相同的值。所以如果你写

(setq2 x y (+ z 3))

当z=8时，x和y都被设为11。(我想不出这有什么用，但这只是一个例子。)

It should be obvious that we can't define setq2 as a function. If x=50 and y=-5, this function would receive the values 50, -5, and 11; it would have no knowledge of what variables were supposed to be set. What we really want to say is, When you (the Lisp system) see (setq2 v1 v2 e), treat it as equivalent to (progn (setq v1 e) (setq v2 e)). Actually, this isn't quite right, but it will do for now. A macro allows us to do precisely this, by specifying a program for transforming the input pattern (setq2 v1 v2 e)" into the output pattern (progn ...)."

如果你觉得这很好，你可以继续读下去: http://cl-cookbook.sourceforge.net/macros.html

想想在C或c++中可以用宏和模板做什么。它们是管理重复代码的非常有用的工具，但它们在相当严重的方面受到限制。

有限的宏/模板语法限制了它们的使用。例如，不能编写扩展为类或函数以外内容的模板。宏和模板不容易维护内部数据。 C和c++复杂且不规则的语法使得编写非常通用的宏非常困难。

Lisp和Lisp宏解决了这些问题。

Lisp宏是用Lisp编写的。您拥有Lisp的全部功能来编写宏。 Lisp有一个非常规则的语法。

与任何精通c++的人交谈，问他们花了多长时间来学习模板元编程所需的所有模板。或者是《现代c++设计》等(优秀)书籍中的所有疯狂技巧，尽管语言已经标准化了10年，但这些技巧仍然很难调试，而且(在实践中)无法在真实的编译器之间移植。如果用于元编程的语言与用于编程的语言相同，那么所有这些问题都消失了!

我不确定我能给每个人的(优秀的)帖子添加一些见解，但是……

Lisp宏工作得很好，因为Lisp语法的本质。

Lisp是一种非常规则的语言(把所有东西都想象成一个列表);宏使您能够将数据和代码视为相同的(不需要字符串解析或其他技巧来修改lisp表达式)。将这两个特性结合起来，就有了一种非常简洁的方式来修改代码。

编辑:我想说的是Lisp是同构的，这意味着Lisp程序的数据结构是用Lisp本身编写的。

因此，您最终可以在语言之上创建自己的代码生成器，使用语言本身的所有功能(例如。在Java中，你必须破解字节码编织的方法，尽管一些框架(如AspectJ)允许你使用不同的方法来做到这一点，但它基本上是一种破解)。

在实践中，使用宏可以在lisp的基础上构建自己的迷你语言，而不需要学习其他语言或工具，并使用语言本身的全部功能。

您将在这里找到关于lisp宏的全面辩论。

这篇文章的一个有趣的子集:

In most programming languages, syntax is complex. Macros have to take apart program syntax, analyze it, and reassemble it. They do not have access to the program's parser, so they have to depend on heuristics and best-guesses. Sometimes their cut-rate analysis is wrong, and then they break. But Lisp is different. Lisp macros do have access to the parser, and it is a really simple parser. A Lisp macro is not handed a string, but a preparsed piece of source code in the form of a list, because the source of a Lisp program is not a string; it is a list. And Lisp programs are really good at taking apart lists and putting them back together. They do this reliably, every day. Here is an extended example. Lisp has a macro, called "setf", that performs assignment. The simplest form of setf is (setf x whatever) which sets the value of the symbol "x" to the value of the expression "whatever". Lisp also has lists; you can use the "car" and "cdr" functions to get the first element of a list or the rest of the list, respectively. Now what if you want to replace the first element of a list with a new value? There is a standard function for doing that, and incredibly, its name is even worse than "car". It is "rplaca". But you do not have to remember "rplaca", because you can write (setf (car somelist) whatever) to set the car of somelist. What is really happening here is that "setf" is a macro. At compile time, it examines its arguments, and it sees that the first one has the form (car SOMETHING). It says to itself "Oh, the programmer is trying to set the car of somthing. The function to use for that is 'rplaca'." And it quietly rewrites the code in place to: (rplaca somelist whatever)

While the above all explains what macros are and even have cool examples, I think the key difference between a macro and a normal function is that LISP evaluates all the parameters first before calling the function. With a macro it's the reverse, LISP passes the parameters unevaluated to the macro. For example, if you pass (+ 1 2) to a function, the function will receive the value 3. If you pass this to a macro, it will receive a List( + 1 2). This can be used to do all kinds of incredibly useful stuff.

Adding a new control structure, e.g. loop or the deconstruction of a list Measure the time it takes to execute a function passed in. With a function the parameter would be evaluated before control is passed to the function. With the macro, you can splice your code between the start and stop of your stopwatch. The below has the exact same code in a macro and a function and the output is very different. Note: This is a contrived example and the implementation was chosen so that it is identical to better highlight the difference. (defmacro working-timer (b) (let ( (start (get-universal-time)) (result (eval b))) ;; not splicing here to keep stuff simple ((- (get-universal-time) start)))) (defun my-broken-timer (b) (let ( (start (get-universal-time)) (result (eval b))) ;; doesn't even need eval ((- (get-universal-time) start)))) (working-timer (sleep 10)) => 10 (broken-timer (sleep 10)) => 0