How a=3 and b=4?

Question

I found an interesting Programing question :

What will be the values of a,b,c,f after executing this programe ?

int i=0,a=0,b=0,c=0,f=0;
while(i<=5){
switch(i++){
    case 1:++a;
    case 2:++b;
    break;
    case 3:
    case 4:++c;a++;b++;
    break;
    default:++f;
}
}

I thought that the values

a=2 , b=2 , c=2 and f=2 but

when i executed this programe i got a = 3, b = 4, c = 2 and f = 2.

I understood how c and f got it values 2 but how come a=3 and b=4.

(As per the syntax ++a and a++ are different as ++a updates the value and then uses it where as a++ uses the value and then updates it )

Can anyone explain how a and b got its values as 3 and 4.

UPDATE:

Hey my doubt is : In i++ the intial value is 0 and not 1. But then how case 4 => a=3

It should be a=2 and should incriment the value if there was any updation of 'a' in case 5 (which is not true)as i haven't given any substitution like a=a++.

Any Help appreciated.

I would step through your code in a debugger to see what each line does. It shows you the value of each variable at each line. — Peter Lawrey, Aug 21 '12 at 12:59
I hope its not his homework or atleast its just extra credit @edze . There are more valuable things to teach like CS Theory then how screwed up the switch statement is. — Adam Gent, Aug 21 '12 at 13:00
@PeterLawrey has the right idea. There is an easy-to-use debugger in Eclipse and all other mainstream IDEs. Stepping through code is always instructive. — slim, Aug 21 '12 at 13:01
a gets incremented when i is 1,3 and 4; b gets incremented when i is 1,2,3,4; ... — milan, Aug 21 '12 at 13:03
Checkout http://docs.oracle.com/javase/tutorial/java/nutsandbolts/switch.html for more details — MadProgrammer, Aug 21 '12 at 13:05
so that means the difference between i++ and ++i is not an issue in this case. — Arjun Sunil Kumar, Aug 21 '12 at 13:12
Hey my doubt is in i++ the intial value is 0 and not 1 but then how code case 4 => a=3 (it should be a=2) — Arjun Sunil Kumar, Aug 21 '12 at 13:22

score 3 · Accepted Answer · answered Aug 21 '12 at 13:01

3

I suggest you to do this exercise with paper and pen. Anyway:

i = 0 ==> f =1;
i = 1 ==> a = 1; b = 1; (ther isn't break after case 1!)
i = 2 ==> b = 2;
i = 3 ==> c = 1; a = 2; b = 3;
i = 4 ==> c = 2; a = 3; b = 4;
i = 5 ==> f = 2;

answered Aug 21 '12 at 13:01

Valerio Emanuele

909
13
27

so that means the difference between i++ and ++i is not an issue in this case.wright! – Arjun Sunil Kumar Aug 21 '12 at 13:10
@R41nB0wM47r1z right not in this case, but it would have been if you where assigning value like so a = ++x or a = x++ ; – MimiEAM Aug 21 '12 at 13:16
If the switch contained ++i then the first iteration would begin from 1 and not from 0. This mean that finally the value of f would been 1 and not 2. – Valerio Emanuele Aug 21 '12 at 13:17
Hey my doubt is : In i++ the intial value is 0 and not 1. But then how case 4 => a=3 (it should be a=2 and should incriment the value if there was any updation of 'a' in case 5 as i haven't give any substitution like a=a++) – Arjun Sunil Kumar Aug 21 '12 at 13:27
Hum no one answered this doubt !In your ans case 5=>a=3(it should be a=2 and should incriment the value if there was any updation of 'a' in case 5 as i haven't give any substitution like a=a++) – Arjun Sunil Kumar Aug 29 '12 at 18:58
What do you mean with 'sobstitution'? If you write a++; is the same as you write a = a + 1; – Valerio Emanuele Aug 30 '12 at 07:31

Vins · Answer 2 · 2012-08-21T13:10:36.667

Hope this will help you...

When i is 0 
    None of the case matched and went to default
    so a=0,b=0,c=0,f=1;

When i is 1 
    Case 1 and 2 will execute as there is no break after 1;
    so a=1,b=1,c=0,f=1;

When i is 2 
    Case 2 will execute
    so a=1,b=2,c=0,f=1; 

When i is 3 
    Case 3 and 4 will execute as there is no break after 3;
    so a=2,b=3,c=1,f=1; 

When i is 4 
    Case 4 will execute
    so a=3,b=4,c=2,f=1; 

When i is 5 
    Default will execute
    so a=3,b=4,c=2,f=2;

score 1 · Answer 3 · answered Aug 21 '12 at 13:00

1

Remember that switch statements support "fall through" - for i=2, only b is incremented, but for i=1 both are incremented.

answered Aug 21 '12 at 13:00

Eyal Schneider

22,166
5
47
78

score 0 · Answer 4 · edited May 23 '17 at 12:12

0

the pre and post increment ( ++a, and a++ ) are best explained in this previous answer

Update As pointed out in the comments bellow there is a fundamental difference in the C++ and Java implementation of those concepts. But here is a simple example in java

        x = 1;
        y = ++x;  Here y = 2 and x =2 because we first increment x and assign it to y

        x = 1;
        y = x++;  But here y = 1 and x = 2 because we first assign x to y and then increment x

in essense , y = x++ is equivalent to those 2 statements
            y =x;
            x = x + 1;

edited May 23 '17 at 12:12

Community

1
1

answered Aug 21 '12 at 13:02

MimiEAM

2,505
1
26
28

1

That's C++. These operators have a more well defined evaluation in Java. – fgb Aug 21 '12 at 13:06
@fgb is there any fundamental difference in both language implementation of the concept ? – MimiEAM Aug 21 '12 at 13:09
@MimiEAM Yes, for example, from your link *"The compiler is free to [...] delay the increment until the end of the expression"* - That is not the case in Java. Also there is not a concept of lvalue in Java. – assylias Aug 21 '12 at 13:17

How a=3 and b=4?

4 Answers4