Short-circuit evaluation

Short-circuit evaluation is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Assume functions a and b return boolean values, and further, the execution of function b takes considerable resources and is to be minimised.

If we needed to compute:

   x = a() and b()

Then it would be best to not compute the value of b() if the value of a() is computed as False, as the value of x can then only ever be False.

Similarly, if we needed to compute:

   y = a() or b()

Then it would be best to not compute the value of b() if the value of a() is computed as true, as the value of x can then only ever be True.

Some languages will stop further computation of boolean equations as soon as the result is known, so-called short-circuit evaluation of boolean expressions

Task Description
The task is to create two functions named a and b, that take and return the same boolean value. The functions should also print their name whenever they are called. calculate and assign the values of the following equations to a variable in such a way that function b is only called when necessary:

   x = a(i) and b(j)
   y = a(j) or  b(j)

If the language does not have short-circuit evaluation, this might be achieved with nested if statements.

Icon and Unicon

The entire concept of using 'boolean' values for logic control runs counter to the philosophy of Icon. While this task could be written literally, it would be more beneficial to show how an Icon programmer would approach the same problem. Icon already embraces the idea of short circuit evaluation and goes further with the ability of expressions to generate alternate results only if needed. For more information see Failure is an option, Everything Returns a Value Except when it Doesn't, and Goal-Directed Evaluation and Generators. Consequently some small liberties will be taken with this task:

• For false we will use the null value &null and true we will use anything else (a 1 will do).
• Short-circuit evaluation uses success (a result) and failure (a signal that cannot be ignored and no result) so strictly speaking the boolean false will not be returned (only the failure signal).
• Rather than have the tasks print their own name, we will just utilize built-in tracing which will be more informative.

Icon

<lang Icon>procedure main() &trace := -1 # ensures a and b print their names

every (i := &null | 1 ) & ( j := &null | 1) do {

 write("i,j := ",image(i),", ",image(j))
 write("a & b:")
 x := a(i) & b(j)
 write("a | b:")
 y := a(i) | b(j)
 }

end

procedure a(x) #: returns x if x is non-null or fails otherwise return \x end

procedure b(x) return \x end</lang>

Sample output for a single case:

i,j := &null, 1
a & b:
shortcicuit.icn:   10  | a(&null)
shortcicuit.icn:   19  | a failed
a | b:
shortcicuit.icn:   12  | a(&null)
shortcicuit.icn:   19  | a failed
shortcicuit.icn:   12  | b(1)
shortcicuit.icn:   23  | b returned 1

Unicon

The Icon solution works in Unicon.

Java

In Java the boolean operators && and || are short circuit operators. The eager operator counterparts are & and |. <lang java>public class ShortCirc {

   public static void main(String[] args){
       System.out.println("F and F = " + (a(false) && b(false)) + "\n");
       System.out.println("F or F = " + (a(false) || b(false)) + "\n");

       System.out.println("F and T = " + (a(false) && b(true)) + "\n");
       System.out.println("F or T = " + (a(false) || b(true)) + "\n");

       System.out.println("T and F = " + (a(true) && b(false)) + "\n");
       System.out.println("T or F = " + (a(true) || b(false)) + "\n");

       System.out.println("T and T = " + (a(true) && b(true)) + "\n");
       System.out.println("T or T = " + (a(true) || b(true)) + "\n");
   }

   public static boolean a(boolean a){
       System.out.println("a");
       return a;
   }

   public static boolean b(boolean b){
       System.out.println("b");
       return b;
   }

}</lang> Output:

a
F and F = false

a
b
F or F = false

a
F and T = false

a
b
F or T = true

a
b
T and F = false

a
T or F = true

a
b
T and T = true

a
T or T = true

Oz

Oz' andthen and orelse operators are short-circuiting, as indicated by their name. The library functions Bool.and and Bool.or are not short-circuiting, on the other hand.

<lang oz>declare

 fun {A Answer}
    AnswerS = {Value.toVirtualString Answer 1 1}
 in
    {System.showInfo "  % Called function {A "#AnswerS#"} -> "#AnswerS}
    Answer
 end

 fun {B Answer}
    AnswerS = {Value.toVirtualString Answer 1 1}
 in
    {System.showInfo "  % Called function {B "#AnswerS#"} -> "#AnswerS}
    Answer
 end

in

 for I in [false true] do
    for J in [false true] do
       X Y
    in
       {System.showInfo "\nCalculating: X = {A I} andthen {B J}"}
       X = {A I} andthen {B J}
       {System.showInfo "Calculating: Y = {A I} orelse {B J}"}
       Y = {A I} orelse {B J}
    end
 end</lang>

Output: <lang oz>Calculating: X = {A I} andthen {B J}

 % Called function {A false} -> false

Calculating: Y = {A I} orelse {B J}

 % Called function {A false} -> false
 % Called function {B false} -> false

Calculating: X = {A I} andthen {B J}

 % Called function {A false} -> false

Calculating: Y = {A I} orelse {B J}

 % Called function {A false} -> false
 % Called function {B true} -> true

Calculating: X = {A I} andthen {B J}

 % Called function {A true} -> true
 % Called function {B false} -> false

Calculating: Y = {A I} orelse {B J}

 % Called function {A true} -> true

Calculating: X = {A I} andthen {B J}

 % Called function {A true} -> true
 % Called function {B true} -> true

Calculating: Y = {A I} orelse {B J}

 % Called function {A true} -> true</lang>

PureBasic

Logical And & Or operators will not evaluate their right-hand expression if the outcome can be determined from the value of the left-hand expression. <lang PureBasic>Procedure a(arg)

 PrintN("  # Called function a("+Str(arg)+")")  
 ProcedureReturn arg

EndProcedure

Procedure b(arg)

 PrintN("  # Called function b("+Str(arg)+")")
 ProcedureReturn arg

EndProcedure

OpenConsole() For a=#False To #True

 For b=#False To #True
   PrintN(#CRLF$+"Calculating: x = a("+Str(a)+") And b("+Str(b)+")")
   x= a(a) And b(b)
   PrintN("Calculating: x = a("+Str(a)+") Or b("+Str(b)+")")
   y= a(a) Or b(b) 
 Next

Next Input()</lang>

Calculating: x = a(0) And b(0)
  # Called function a(0)
Calculating: x = a(0) Or b(0)
  # Called function a(0)
  # Called function b(0)

Calculating: x = a(0) And b(1)
  # Called function a(0)
Calculating: x = a(0) Or b(1)
  # Called function a(0)
  # Called function b(1)

Calculating: x = a(1) And b(0)
  # Called function a(1)
  # Called function b(0)
Calculating: x = a(1) Or b(0)
  # Called function a(1)

Calculating: x = a(1) And b(1)
  # Called function a(1)
  # Called function b(1)
Calculating: x = a(1) Or b(1)
  # Called function a(1)

Python

Pythons and and or binary, infix, boolean operators will not evaluate their right-hand expression if the outcome can be determined from the value of the left-hand expression. <lang python>>>> def a(answer): print(" # Called function a(%r) -> %r" % (answer, answer)) return answer

>>> def b(answer): print(" # Called function b(%r) -> %r" % (answer, answer)) return answer

>>> for i in (False, True): for j in (False, True): print ("\nCalculating: x = a(i) and b(j)") x = a(i) and b(j) print ("Calculating: y = a(i) or b(j)") y = a(i) or b(j)

Calculating: x = a(i) and b(j)

 # Called function a(False) -> False

Calculating: y = a(i) or b(j)

 # Called function a(False) -> False
 # Called function b(False) -> False

Calculating: x = a(i) and b(j)

 # Called function a(False) -> False

Calculating: y = a(i) or b(j)

 # Called function a(False) -> False
 # Called function b(True) -> True

Calculating: x = a(i) and b(j)

 # Called function a(True) -> True
 # Called function b(False) -> False

Calculating: y = a(i) or b(j)

 # Called function a(True) -> True

Calculating: x = a(i) and b(j)

 # Called function a(True) -> True
 # Called function b(True) -> True

Calculating: y = a(i) or b(j)

 # Called function a(True) -> True</lang>

Pythons if expression can also be used to the same ends (but probably should not): <lang python>>>> for i in (False, True): for j in (False, True): print ("\nCalculating: x = a(i) and b(j) using x = b(j) if a(i) else False") x = b(j) if a(i) else False print ("Calculating: y = a(i) or b(j) using y = b(j) if not a(i) else True") y = b(j) if not a(i) else True

Calculating: x = a(i) and b(j) using x = b(j) if a(i) else False

 # Called function a(False) -> False

Calculating: y = a(i) or b(j) using y = b(j) if not a(i) else True

 # Called function a(False) -> False
 # Called function b(False) -> False

Calculating: x = a(i) and b(j) using x = b(j) if a(i) else False

 # Called function a(False) -> False

Calculating: y = a(i) or b(j) using y = b(j) if not a(i) else True

 # Called function a(False) -> False
 # Called function b(True) -> True

Calculating: x = a(i) and b(j) using x = b(j) if a(i) else False

 # Called function a(True) -> True
 # Called function b(False) -> False

Calculating: y = a(i) or b(j) using y = b(j) if not a(i) else True

 # Called function a(True) -> True

Calculating: x = a(i) and b(j) using x = b(j) if a(i) else False

 # Called function a(True) -> True
 # Called function b(True) -> True

Calculating: y = a(i) or b(j) using y = b(j) if not a(i) else True

 # Called function a(True) -> True</lang>

Scheme

<lang scheme>>(define (a x)

  (display "a\n")
  x)

>(define (b x)

  (display "b\n")
  x)

>(for-each (lambda (i)

  (for-each (lambda (j)
    (display i) (display " and ") (display j) (newline)
    (and (a i) (b j))
    (display i) (display " or ") (display j) (newline)
    (or (a i) (b j))
   ) '(#t #f))
 ) '(#t #f))

1. t and #t

a b

1. t or #t

a

1. t and #f

a b

1. t or #f

a

1. f and #t

a

1. f or #t

a b

1. f and #f

a

1. f or #f

a b </lang>