Why is this code alerting "1111" instead of "1112"?

Question

This is the code of a challenge:

var a = 10;
var b = 11;
var c;
if(!(a++ >= 10 || b++ >= 11)){
c = a+b;
alert(a + "" + c);
}else{
alert(a + "" + b);
}

Answer 1

a++ and b++ returns current values of a and b and increments each.

in the if statement a is 10 so a++>=10 is true so since || in an if statement returns true if any condition is true the next condition wont execute so b remains 11.

now because of the ! (NOT) in the if statement the final state is false so the else block is execute for which a has been incremented to 11 while b hasn't changed.

so inside the else, a= 11 and b = 11

Answer 2

This is what happen :
a=10
a++ >= 10 // true --> 10 >= 10
we have a '||' so the statement is true
However, there is '!' so the condition is false
We go to the 'else'
a was equal to 10 but in the 'if' statement, there was 'a++' so a=11 and b=11

In a 'if' statement with a '||' operator, if the first condition is true, the second isn't evaluated.

Ask if you don't understand

Answer 3

!a++>=10 is false, so it go to the else statement and b never increments

Answer 4

Thanks a lot to you all!
I thought logical expressions are always evaluated completely.

Answer 5

!(a++ >= 10) || !(b++ >= 11) will do that you want

Answer 6

( !(a++ >= 10 || b++ >= 11) )

1. a++ >= 10 is true so b is not evaluated
2. all becomes false and it's incremented
3. go to else statement

( !(a++ >= 10) || !(b++ >= 11) )

1. a++ >= 10 is true
2. it becomes false and it's incremented. b has to be evaluated
3. b++ >= 11 is true
4. it becomes false and it's incremented
5. go to else statement