SAS 宏取消引用

Question

声明了以下宏变量：

%令x=2；

%让 y=3;

%让x3=4；

%让 y2=5;

%令x2=6；

%让 y3=7;

%让 m=x;

%让n=y；

以下解除引用将评估什么： i.&&&m&y - &&&n&y ii.&&&m&x&&&n&y `

Answer 1

您只需运行它即可找到答案。

1    %let x=2;
2    %let y=3;
3    %let x3=4;
4    %let y2=5;
5    %let x2=6;
6    %let y3=7;
7    %let m=x;
8    %let n=y;
9
10   %put i. &&&m&y - &&&n&y ;
i. 4 - 7
11   %Put ii. &&&m&x&&&n&y ;
ii. 67

当宏处理器看到 && 时，它会将其转换为 & 并提醒自己，它需要再次传递结果以进一步解析宏变量引用。因此，在 &&&m&y 的第一遍中，它将 && 转换为 &，将 &m 转换为 x，&y 转换为 3。然后，当它返回处理结果时，它将 &x3 转换为 4。

您可以打开 SYMBOLGEN 并查看它的运行情况。

12
13   options symbolgen;
14   %put i. &&&m&y - &&&n&y ;
SYMBOLGEN:  && resolves to &.
SYMBOLGEN:  Macro variable M resolves to x
SYMBOLGEN:  Macro variable Y resolves to 3
SYMBOLGEN:  Macro variable X3 resolves to 4
SYMBOLGEN:  && resolves to &.
SYMBOLGEN:  Macro variable N resolves to y
SYMBOLGEN:  Macro variable Y resolves to 3
SYMBOLGEN:  Macro variable Y3 resolves to 7
i. 4 - 7
15   %Put ii. &&&m&x&&&n&y ;
SYMBOLGEN:  && resolves to &.
SYMBOLGEN:  Macro variable M resolves to x
SYMBOLGEN:  Macro variable X resolves to 2
SYMBOLGEN:  && resolves to &.
SYMBOLGEN:  Macro variable N resolves to y
SYMBOLGEN:  Macro variable Y resolves to 3
SYMBOLGEN:  Macro variable X2 resolves to 6
SYMBOLGEN:  Macro variable Y3 resolves to 7
ii. 67