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theponcer
Quartz | Level 8

Here's an example:

 

%let n=1;
%let count1 = subcount1;
%let subcount1 = 100;

data test;
set mor_filelist_ma;
n="&n."; *Resolves to 1 as expected;
count1="&count1."; *resolves to subcount1 as expected;
subcount1="&subcount1."; *resolves to 100 as expected;
test1="&count&n."; *resolves to subcount1 as expected;
test2="&&&count&n."; *resolves to subcount1 - expecting 100;
run;

How do I get test2 to resolve to 100 instead of subcount1? I'm thinking this is easy and I'm just missing something...

1 ACCEPTED SOLUTION

Accepted Solutions
Tom
Super User Tom
Super User

If you are using a data step anyway you can make it easier by using SYMGET() and SYMGETN() functions instead of trying to use macro code to resolve the references.

599   %let n=1;
600   %let count1 = subcount1;
601   %let subcount1 = 100;
602
603   data test;
604     length name1 name2 $32 value 8 ;
605     name1=cats('count',symgetn('n'));
606     name2=symget(name1);
607     value=symgetn(name2);
608     put (_all_) (=);
609   run;

name1=count1 name2=subcount1 value=100

View solution in original post

6 REPLIES 6
Astounding
PROC Star

Easy?  Maybe.

 

%let n=1;
%let count1 = subcount1;
%let subcount1 = 100;

test2="&&&&&&count&n."; 
ballardw
Super User

Indirect referencint (nothing "nested" here) can be problematic in many ways, often involving way more references than you think are needed:

Consider this output:

227  %let n=1;
228  %let count1 =subcount1;
229  %let subcount1 =100;
230
231  %let test2=&&&&&&count&n.;
232
233  %put &test2;
100

2,3,4 and 5 & at the beginning are insufficient, 6 works on my system.

 

Which why you really want to determine if the indirect reference is really needed.

 

Tom
Super User Tom
Super User

Don't do that.  https://www.amazon.com/Doctor-hurts-when-Jokes-Cartoons/dp/150043230X

Since you need 3 to get value of COUNT1 then you will probably need 6 to get the value macro variable named in COUNT1.

529   %let n=1;
530   %let count1 = subcount1;
531   %let subcount1 = 100;
532   %put &=n &=count1 &=subcount1 ;
N=1 COUNT1=subcount1 SUBCOUNT1=100
533   %put &&&count&n;
subcount1
534   %put &&&&&&count&n;
100

But it is just much easier on your brain if you do the steps explicitly instead.

535   %let var1=count&n;
536   %let var2=&&&var1;
537   %let var3=&&&var2;
538   %put &=var1 &=var2 &=var3;
VAR1=count1 VAR2=subcount1 VAR3=100

 

Kurt_Bremser
Super User

How do the six ampersands work?

In the first pass, double ampersands are converted to single ampersands, and single ampersands are resolved:

&& && && count &n

resolves to

& & & count 1

(blanks inserted to make it easier to read)

In the next pass, the same happens:

&& &count1

resolves to

& subcount1

and that is resolved in the third pass to 100.

Tom
Super User Tom
Super User

If you are using a data step anyway you can make it easier by using SYMGET() and SYMGETN() functions instead of trying to use macro code to resolve the references.

599   %let n=1;
600   %let count1 = subcount1;
601   %let subcount1 = 100;
602
603   data test;
604     length name1 name2 $32 value 8 ;
605     name1=cats('count',symgetn('n'));
606     name2=symget(name1);
607     value=symgetn(name2);
608     put (_all_) (=);
609   run;

name1=count1 name2=subcount1 value=100

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