IF ... THEN ... followed by AND

[stechafle]

Why does the first IF Statement below not produce an Error?

data test;
x=0; y=1;

if x=0 then z=0
and y=0;

if x=0
and y=0 then w=0;
run;

proc print data=test;
run;

[PGStats]

With SAS the equal sign can be an asignment or an equality operation. z = ... is and assignment because it occurs at the beginning of the statement, but y=0 is an equality operation. SAS represents logical (boolean) values with zeros and ones, 0 = FALSE, 1 = TRUE. The expression

z=0 and y=0

is computed as

z = (0 and (y=0))

z = (0 and (1=0)) /* because y = 1 */

z = (0 and 0) /* because 1=0 is FALSE */

z = 0 /* Because FALSE and FALSE is FALSE */

[ballardw]

---

@stechafle wrote:

Why does the first IF Statement below not produce an Error?

data test;
x=0; y=1;

if x=0 then z=0
and y=0;

if x=0
and y=0 then w=0;
run;

proc print data=test;
run;

---

If the intent of the first if is to assign values to both Z and Y the code would be:

if x= 0 then do;

   z=0;

   y=0;

end;

[Tom]

What error should it produce?  

Note that the compiler cannot read your mind to know whether it is what you intended or not.

[VDD]

think about it,

draw pictures with lines results.

then ask yourself do we give order in that manner?

no we give clear and understandable orders so a computer understands what the hay you are thinking.

[Kurt_Bremser · Accepted solution]

if x=0 then z=0
and y=0;

translates to

if (x = 0) then z = (0 and (y = 0));

(0 and (y = 0)) is a valid numeric expression, and once you look at that, you can see that z will always be set to zero if the condition "x equals zero" is met.

[stechafle]

Thank you Kurt! Your explanation makes perfect sense.