I got what you mean.
Neither a (b) is not a 'good' expression, nor a (b+c).

Syntax
SUBSTRN(string, position <, length>)  

Arguments

string
specifies a character constant, variable, or expression.
------------------------------------------------------------------------------------------
When using ”%let       string=notfunc  (abc);“, I guess SAS thought "&string" is an "expression", not a constant string.

when using "%let string=1+a;", although there's an operator "+", the macro processor would still treat it as a constant string.

Maybe the key is on the right parenthesis ")", but I don't know why.



  

anyway, it is weird.

151  %let     string=%str(%(nofunc);%put &string;
(nofunc
152  %let string2=%qsysfunc(substrn(&string,1,&lenx));
153  %put &string2;
nofu

where is the 1st (?

maybe just simply as it is, substrn() is not a GOOD function under macro environment.
JingJU

小弟不才，尝试解答一下这个问题！

首先，substr函数和substrn函数的作用机制不一样，参见下例：

%let       lenx = 3;
%let       string = 2*3*4*5*6*7;
%let       string1 = %sysfunc(substr(&string.,1,&lenx));
%let       string2 = %sysfunc(substrn(&string.,1,&lenx));

%put &string1.;
%put &string2.;
   由此可以推断，substr函数发挥作用时，是直接将string的值作为字符串处理了，而substrn函数发挥作用时，并没有将string的值直接当做字符串处理。

其次，宏变量的值是文本，而文本跟字符串不完全一样。比如上例substrn函数表达式中先进行了运算，才进一步转化为了字符串。

再次，表达式%let       string3 = %sysfunc(substrn(notfunc  (abc),1,&lenx)); 也会报错，为什么报错呢？我猜测原因是因为'()'的出现，并且因为表达式前面没有宏函数%sysfunc出现，排除了使用函数的可能性，最终让SAS认为'notfunc’ 和 ‘(abc)'之间需要运算，故报缺少操作符的错误。

You could use ksubstr to replace substrn.

Yeah. Thanks. That is the answer i am looking for. using %qksubstr() can solve my problem in the best way.

JingJu

How substrn() works may not be of my best interest, because it is sort of 'strange'. The input string is in a flexible form and thus hard to control. if substrn() cannot meet with all i need, there is no reason to use it in this circumstance.
JingJu

that's ok