Can I do this with a Switch case?

[Mario Gimenez]

Hi. I need use a switch case but I don't know if this is correct.

  
Switch (num){
case 1..9 :        system.out,println("Hola"); 
                       break;
case 10..99 :     system.out,println("Hola");
                       break;
case 100..999 : system.out,println("Hola");
                       break;

  

I won't do write each case because I don't know if this ok.

thanks.

[Nicola Garofalo]

switch(num){
  case 1: 
  case 2:
  case 3:
  case 4:
  //... go on until 8
  case 9: System.out.println("Hola");break;
  case 10:
  case 11:
  //.. go on until 98
   case 99: System.out.println("Hola");break;
  //etc..etc..
}

where num must be a type compatible with an int (byte,short,char,int) or, as java6, an enum

case 1..99: does not exist in java 6

Bye

[Rob Spoor]

In this case, an if-else if-else if would seem better:

if (1 <= num && num <= 9)
{
    ...
}
else if (10 <= num && num <= 99)
{
    ...
}
else if (100 <= num && num <= 999)
{
    ...
}

With these ranges there are other tricks using % 10, % 100 and % 1000 as well:

if (num % 10 == 0) // 0 to 9, slightly different here
{
}
else if (num % 100 == 0) // 0-99 but without 0-9 so 10-99
{
}
else if (num % 1000 == 0) // 0-999 but without 0-99 so 100-999
{
}

[marc weber]

Rob Prime wrote:... With these ranges there are other tricks using % 10, % 100 and % 1000 as well:

if (num % 10 == 0) // 0 to 9, slightly different here
{
}
else if (num % 100 == 0) // 0-99 but without 0-9 so 10-99
{
}
else if (num % 1000 == 0) // 0-999 but without 0-99 so 100-999
{
}

Ah, but 100 % 10 is 0. (As with 1000, 10000, etc.)

[marc weber]

Mario Gimenez wrote:Hi. I need use a switch case...

Do you really need to use switch/case?

[marc weber]

switch(new Integer(num).toString().length())

(But we always write clean, readable code, so we would never do this in practice. Right?)

[David Newton]

That's just cruel.

[Mike Simmons]

marc weber wrote:

Rob Prime wrote:... With these ranges there are other tricks using % 10, % 100 and % 1000 as well:

if (num % 10 == 0) // 0 to 9, slightly different here
{
}
else if (num % 100 == 0) // 0-99 but without 0-9 so 10-99
{
}
else if (num % 1000 == 0) // 0-999 but without 0-99 so 100-999
{
}

Ah, but 100 % 10 is 0. (As with 1000, 10000, etc.)

There's something amusing in the fact that a guy named Prime overlooked this. Maybe he's not used to numbers having other factors.

On a related note though, it's worthwhile to pay attention to the ordering of if / else if / else clauses, and avoid needless redundancy. For this example here, I'd rather code it like this:

if (num < 1) {
  // do nothing, or maybe throw an IllegalArgumentException
}
else if (num < 10) {
  System.out.println("A");
}
else if (num < 100) {
  System.out.println("B");
}
else if (num < 1000) {
  System.out.println("C");
}
else {
  // do nothing, or maybe throw an IllegalArgumentException
}

I replaced "Hola" with three different strings A-B-C, since otherwise the whole construct is pointless. Anyway, organizing the statements this way, we don't need to specify (for example) <=99 for one group, and >=100 for the next. We eliminate needless redundancy, and also make it easier to make alterations if we need to change the cutoff from 100 to, say, 256, in a future release. This way we only need to change one number, instead of two. And we're less likely to have problems because we said < instead of <, or because we refactored num to be a double rather than an int.

[Rob Spoor]

marc weber wrote:

Rob Prime wrote:... With these ranges there are other tricks using % 10, % 100 and % 1000 as well:

if (num % 10 == 0) // 0 to 9, slightly different here
{
}
else if (num % 100 == 0) // 0-99 but without 0-9 so 10-99
{
}
else if (num % 1000 == 0) // 0-999 but without 0-99 so 100-999
{
}

Ah, but 100 % 10 is 0. (As with 1000, 10000, etc.)

You are so right of course. I was so busy about trying to prevent nested ifs that this slipped by. "&& num / 10 == 0" and "&& num / 100 == 0" should fix this. Or just stick with the simple range checking.

Mike Simmons wrote:There's something amusing in the fact that a guy named Prime overlooked this.

[adam smith ii]

Rob Prime wrote:In this case, an if-else if-else if would seem better:
With these ranges there are other tricks using % 10, % 100 and % 1000 as well:

if (num % 10 == 0) // 0 to 9, slightly different here
{
}
else if (num % 100 == 0) // 0-99 but without 0-9 so 10-99
{
}
else if (num % 1000 == 0) // 0-999 but without 0-99 so 100-999
{
}

As well as other remarks about 0 remainder, will this code accomplish anything for any number not ending in 0? That is to say, the comparison of (num % (power of 10)) to zero will always be false using the remainder operator.

[Campbell Ritchie]

I think that mistake has already been noticed

It might work better with / than %