SAS间接宏处理

Question

我在考试试卷上有一个问题，询问SAS日志中的内容

%let test=one;
%let one=two;
%let two=three;
%let three=last;

%put what displays is &&&&&test;

我很惊讶地发现答案是：两个，因为我认为此引用将完全解析为最后。 SAS也同意答案是两个。

任何人都可以解释一下SAS如何得出答案 2 ，因为我读过的所有理论说明都表明宏处理器应该执行以下操作

• scan1＆amp;＆amp;＆amp;＆amp;＆amp; test - ＆gt; ＆amp;＆amp;＆amp;＆amp; test（即＆amp;＆amp;＆amp;＆amp;＆amp;＆amp;并告诉处理器继续从右向左扫描）
• scan2＆amp;＆amp;＆amp;＆amp; test - ＆gt; ＆amp;＆amp;＆amp; one
• scan3＆amp;＆amp;＆amp; one - ＆gt; ＆amp;＆amp; two
• scan4＆amp;＆amp; two - ＆gt; ＆安培;三
• scan5＆amp; three - ＆gt;最后

Answer 1

使用symbolgen选项可以帮助查看日志中发生的事情：

1    options symbolgen;

2    %let test=one;
3    %let one=two;
4    %let two=three;
5    %let three=last;
6
7    %put what displays is &&&&&test;
SYMBOLGEN:  && resolves to &.
SYMBOLGEN:  && resolves to &.
SYMBOLGEN:  Macro variable TEST resolves to one
SYMBOLGEN:  && resolves to &.
SYMBOLGEN:  Macro variable ONE resolves to two
what displays is two

从左到右，使用括号显示代币：

&&&&&test
(&&)(&&)(&test)
(&) (&) (one)
&&one
(&&)(one)
(&)(one)
&one
two