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11-24-2015 02:34 PM

**Reference to Resolving and Using &&var and Referencing Macro Variables Indirectly**

Hi mates,

I've beeing looking for an analogy to understand the enviroment of macro call's.

The analogy to understand is tricky, because you need to know what is the first value that the macro will solve to. Although, there are some tips that might help us to understand and solve then quickly.

As we all know, the best way to understand is practicing and testing, but if you are going to take a SAS Advance Programming Certification you may know how the macro works.

The code below represent the exactly what i've beeing looking for:

```
%let one=two;
%let two=three;
%let three=Check;
%put &one;
%put &&one;
%put &&&one;
%put &&&&one;
%put &&&&&one;
%put &&&&&&one;
%put &&&&&&&one;
%put &&&&&&&&one;
%put &&&&&&&&&one;
%put &&&&&&&&&&one;
```

The values printed to the SAS Log:

21 %put &one; two 22 %put &&one; two 23 %put &&&one; three 24 %put &&&&one; two 25 %put &&&&&one; three 26 %put &&&&&&one; three 27 %put &&&&&&&one; Check 28 %put &&&&&&&&one; two 29 %put &&&&&&&&&one; three 30 %put &&&&&&&&&&one;

But how SAS interprets this and reaches the certain values ?

**The Analogy i've made **

**1. **When the quantity of "&" is 1 or 2 or 4 or 8 or 16 or 32 or 64 or ...,

and so on, the result will always be the value of the first macro

**2. **The second macro, the analogy is: take off the last exponential

quantity, it will left 1 or 2 "&", it solves to "&one > &two > three"

**3.** The third macro , the analogy is: Same analogy of the second,

but if rests 3 "&" then the value will always be the value of third

**Table for the analogy**

All these content were tested in a SAS enviroment

If you have any other idea please show me, or if you have another analogy please explain.

The table for the analogy is attached

**Hope this helps,**

**Rodrigo Dartibali Elias**

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Posted in reply to DartibaliRodrigo

11-24-2015 04:05 PM

Rodrigo,

There are only a small number of rules you need to know.

Macro language resolves strings from left to right.

Two ampersands resolve into one.

If any ampersands remain, macro language repeats the resolution process.

So to take just one of your examples:

&&&&&&one gets treated as if it were four pieces: && && && one

Since two ampersands resolve into one, this becomes &&&one

Ampersands remain, so resolution continues, and &&&one gets treated as if it were two pieces: && &one

Two ampersands resolve into one, so this becomes &two

Ampersands remain, so this gets resolved and becomes three

The same rules will explain all of your results.

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Posted in reply to Astounding

11-25-2015 06:10 AM - edited 11-25-2015 06:13 AM

But if you use the quantity of ampersands in (1,2,4,8,16,32,64,...) it will always be the first value, right ?

Please make a test. And you will see it. Always beginning of number 1 and multiply by 2 => 1*2=2*2=4*2=8*2=16=...

Thats what i mean for the first macro.

The other rules works like you said.

But try to make the same analogy i've made for all other variables.

You said &&&&&&one gets treated as if it were four pieces: && && && one = Ok, but the last result number multiplied by 2 was 4 before the quantity of 6 ampersands. You take out 4 ampersands, left 2 ampersands put these in the first case >> &one = && &one

The result of &&&one is three, same as &&&&&&one.

But if only left 1 ampersands or &&&&&one, the result of &one = &two then solves to three again.

I know the analogy is this like you said, but i tried to find another one for a better understanding, could you please check if this is right ?

Thanks

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Posted in reply to DartibaliRodrigo

11-25-2015 09:24 AM

While it's true that 1, 2, 4, 8, 16 ... ampersands are equivalent, that doesn't mean you can just remove 4 ampersands any time you would like. All the existing pieces of &&&&&&one must be evaluated from left to right.

Resolving each piece generates &&&one

Since ampersands remain, macro language must now re-resolve &&&one

Resolving each piece generates &two

Again, ampersands remain, so macro language now re-resolves &two and comes up with three.

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Posted in reply to Astounding

11-25-2015 10:17 AM

Not remove four ampersands, but remove the quantity of the last exponential.

Just like: If the last exponential was 2 then in the next remove 2 => &&one - &&&one => &two > three

If the last exponential is 4 ampersands: remove 4 &&&&&one => &one => &two => three

Your analogy is the right one, but i tried to understand by another way.

Please use my analogy and check out.

That's what i mean

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Posted in reply to DartibaliRodrigo

11-25-2015 11:09 AM

Powers of two will reduce to the same as just one &.

For the intervening numbers that are not powers of 2 it is probably harder to see a pattern.

Try this version:

```
%let level1=level2;
%let level2=level3;
%let level3=level4;
%let level4=level5;
%put 1=&level1 2=&&level1 4=&&&&level1 8=&&&&&&&&level1;
%put 3=&&&level1 5=&&&&&level1 7=&&&&&&level1 9=&&&&&&&&&level1 10=&&&&&&&&&&level1 12=&&&&&&&&&&&&level1 ;
%put 11=&&&&&&&&&&&level1 13=&&&&&&&&&&&&&level1 14=&&&&&&&&&&&&&&level1;
%put 15=&&&&&&&&&&&&&&&level1;
```

Results:

```
1=level2 2=level2 4=level2 8=level2
3=level3 5=level3 7=level3 9=level3 10=level3 12=level3
11=level4 13=level4 14=level4
15=level5
```

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11-25-2015 11:13 AM

Oh thanks Tom, That's easier.

Tks