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logic behind the shortcut operators  RSS feed

 
Dror Astricher
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hi guys

i know that the shortcut operators are || and && and i know why.
i was wondering about the logic that | and & not are not shortcut operators as well?

(a | b) - when a is true, it doesn't metter what b is and it will always be true.
(a & b) - when a is false, it doesn't metter what b is and it will always be false.

can someone explaine me the logic behind it?
thanks
dror
 
Jeroen Wenting
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they're not shortcut operators but mathematical operators.
 
Jim Yingst
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Well, consider what happens if a nd b are replaced with methods a() and b() (both of which return booleans). If I write

and if a() returns false - well, obviously the value of c will be false, regardless of what b() returns. However - method b() may have a side effect. As a programmer, I may want to ensure that method b() executes, and has its side effect, regardless of the fact that the value of c may not depend on b() at all. Or, maybe I don't want b() to execute unless it really effectus things. Either one of these situations is possible, really, depending on what the programmer is trying to do. So Java offers a choice - if you want short-circuiting, use && and ||; if you don't want short-circuiting, use & and |. In my experience, there's rarely much reason not to use short circuiting - which is why && and || are much more common (in my experience) than & and |. However the latter are still available as an option, for the (relatively rare) cases where this is what the programmer desires.
 
Dror Astricher
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i know that they are not shortcut operators. but waht is the logic behind it?
the logic behind && being a shortcut operator is, that when given (A && B) and A is false, then doesn't metter what will be B and it will always be false. thats why we don't even evaluate B.

in (A & B), if A is false, it's also doesn't metter what B will be.

what is the logic then not to make it also as a shortcut operator and not to save some operations?

thanks
dror
 
Dror Astricher
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thanks Jim, i got the logic
i wrote my second comment before i got your explanation.

wish you all a great day
dror
 
Ravi Shankar R
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I agree with Jim Yingst ....

See Using the mathematical operator what is the ouput of the program

public class ShortCrkOpr {

boolean a(){
return false;
}

boolean b() throws Exception {
throw new Exception();
}
public static void main(String[] args) {
ShortCrkOpr opr = new ShortCrkOpr();
try {
if(opr.a() & opr.b()){
System.out.println("True");
}else{
System.out.println("False");
}

if(opr.a() && opr.b()){
System.out.println("True");
}else{
System.out.println("False");
}

} catch (Exception e) {
System.out.println("Exception caught!!!");
e.printStackTrace();
}
}
}

output....

Exception caught!!!
java.lang.Exception .......

so the latter effect of the method b()has come...


public class ShortCrkOpr {

boolean a(){
return false;
}

boolean b() throws Exception {
throw new Exception();
}
public static void main(String[] args) {
ShortCrkOpr opr = new ShortCrkOpr();
try {
if(opr.a() && opr.b()){
System.out.println("True");
}else{
System.out.println("False");
}
if(opr.a() & opr.b()){
System.out.println("True");
}else{
System.out.println("False");
}

} catch (Exception e) {
System.out.println("Exception caught!!!");
e.printStackTrace();
}
}
}

output is...

False
Exception caught!!!
java.lang.Exception

here once the method a() has returned false it wouldn't move to the next method b() to check....
 
Don't get me started about those stupid light bulbs.
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