@PaigeMiller  I understand that it is easier and much cleaner not to apply formats for the macro variable (As stated in my initial statement). I asked this question, because I am working on a system where previous developers used this method (assign a format to the macro variable and then "un-format"), so I was trying to find any benefits in using it. From what I understood you see none.  ... View more

Thanks,   my assumption is that sas uses . as a delimiter with numeric values. Hence the , in the assignment statement would lead to an error because the symbol is not recognized.  Value = 1.184,40;   My solution would be to prevent this with an input statement providing the right numeric format:  value = input("1.184,40",COMMAX22.2); ... View more

@PaigeMiller  You are right, the value was larger than 1000. In this case I should write something along the lines of : value = input("1.184,40",COMMAX22.2); Thank you, that definitely solves the problem. I however still do not grasp the concept of giving a macro variable a format. I guess it is for the log and makes debugging simpler.  Yet is that all to it?   Best ifb10   ... View more

Thank you @Tom ! Learned a lot today thanks to you. Especially the implied %EVAL is useful to know. ... View more

Thank you @Quentin ! I did not even think of this option, since in other languages the and operator meant that both conditions have to be fulfilled. But it makes sense here, I do it until one of the conditions is reached. Funny that it works the other way around with while, which also makes sense after thinking about it.  ... View more

Hi again @Tom ,   you are right, I do not need the data step for this macro. This was again just to test call symputx. The code should result in the generation of "larva" dependent on both conditions: Eggs should be greater than 0 and will be consumed per larva and Time for the transformation of Egg into Larva is taking 20 min. Hence if I have 6 Eggs but only 1 hour (3600 s) I should obtain 3 Larva with 3 Eggs remaining. I thought this would be a nice example to couple the two conditions using %do %until Loops: %macro Larva ( Eggs = 6 ,time_left = 3600 ,Larva = 0 ); %if &Eggs. > 0 %then %do; %do %until((%eval(&Eggs. < 1)) and (%eval(&time_left. < 1200))); %let time_left = %eval(&time_left. - 1200); %let Larva = %eval(&larva. +1); %let Eggs = %eval(&Eggs. - 1); %put &=Larva. &=Eggs. &=time_left.; %end; %end; %mend Larva; %larva()  But this results in: LARVA=1 EGGS=5 TIME_LEFT=2400 LARVA=2 EGGS=4 TIME_LEFT=1200 LARVA=3 EGGS=3 TIME_LEFT=0 LARVA=4 EGGS=2 TIME_LEFT=-1200 LARVA=5 EGGS=1 TIME_LEFT=-2400 LARVA=6 EGGS=0 TIME_LEFT=-3600 Which is not the intent, since the time factor is being ignored. Hence I tried to execute the two loos sequentially but that lead to the same result.  ... View more

Thank you @Tom for the detailed answer.    The main purpose was to understand the macro logic a bit better.  Especially the macro variable definitions in a local environment using proc sql, call symputx and %let statement.  In this regard your answer was perfectly explaining why call symputx can not work under this setting.  Yet I was wondering, why does the same logic not apply to the %let statement? I tried following code which did the job:        %macro Hatchery (Eggs = 6); data crap.hatchery; %do index= 1 %to 3; %put NOTE: &Eggs.; Larva +1; %let Eggs = %eval(&Eggs. - 1); call symputx("Larva", Larva); %end; run; %mend Hatchery; %Hatchery     Is it because the %let statement is directly processed in the macro processor and the symbol table is immediately updated after each iteration? I guess in contrast to that I get similar results to call symputx with proc sql select into clause as you described. My guess is that Macro processor is not triggered by either call symputx nor select into (proc sql) which are instead resolved as you described only after compilation.  data crap.hatchery; Larva +1; call symputx("Eggs", 5); call symputx("Larva", Larva); Larva +1; call symputx("Eggs", 5); call symputx("Larva", Larva); Larva +1; call symputx("Eggs", 5); call symputx("Larva", Larva); run;   Thanks again, now I have only to find out how I can mark your answer as solution as well.    ... View more

Thank you for the detailed answer!  ... View more

@MarkusWeick, have a look ... View more

Thank you this does the trick. I have been trying to overwrite the original value instead of creating a new one. This is a very simple and effective solution! ... View more