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: 2,因为我本来以为这将参考完全解析到最后.SAS也同意答案为两个.

任何人都可以解释一下SAS如何得出答案二,因为我读过的所有理论说明都表明宏处理器应该做到以下几点

• scan1 &&&&& test - > &&&& test(即&&解析为&并告诉处理器继续从右向左扫描)
• scan2 &&&& test - > &&& one
• scan3 &&& one - > && two
• scan4 && two - >&three
• scan5&three - > last

Answer 1

使用symbolgen选项可以帮助查看日志中发生的情况:

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,使用括号显示标记:

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