Doubt with switch in JavaScript

Summing up a little bit of the Sam response, the most appropriate would be the use of a if-else block instead of switch , because then you already return with the value of a condition already tested.

function experiencia(anos) {
  if(anos == 0 || anos == 1){
    return 'Iniciante';
  } else if(anos > 1 && anos <= 3){
    return 'Intermediário';
  } else if(anos >= 4 && anos <= 6){
    return 'Avançado';
  } else if(anos > 7) {
    return 'Jedi';
  } else {
    return 'Impossível';  // caso digite nº negativos
  }
}

var anosEstudo = -10;
console.log(experiencia(anosEstudo));

Note that in your case, if anos is exactly 7, it would not fall into any of case s or default . I think the condition of default should be >= 7 instead of > 7 .

The block switch serves to relate each case to exactly one value. You can not map a case to a range of default values. There is no such thing as putting an expression after case or default just like you were trying to do. This is not how switch works.

Here is a valid example of switch . Note that each case maps exactly one value:

function experiencia(anos) {
    if (anos < 0) return 'Impossível';
    switch (anos) {
        case 0:
        case 1:
            return 'Iniciante';
        case 2:
        case 3:
            return 'Intermediário';
        case 4:
        case 5:
        case 6:
            return 'Avançado';
        default:
            return 'Jedi';
    }
}

for (var i = -2; i <= 10; i++) {
    document.write(i + " anos: " + experiencia(i) + ".<br>");
}

But there is a trick that can be done to force switch to run at intervals by putting expressions in cases and true in switch :

function experiencia(anos) {
    switch (true) {
        case anos >= 0 && anos <= 1:
            return 'Iniciante';
        case anos > 1 && anos <= 3:
            return 'Intermediário';
        case anos >= 4 && anos <= 6:
            return 'Avançado';
        case anos >= 7:
            return 'Jedi';
        default:
            return 'Impossível';
    }
}

for (var i = -2; i <= 10; i++) {
    document.write(i + " anos: " + experiencia(i) + ".<br>");
}

This works because JavaScript is an interpreted language with poor typing. In other languages that also have switch but are compiled, such as C, C ++, Objective-C, C #, and Java, that does not work.

However, if you do something like this, maybe giving up switch and using if s would be easier:

function experiencia(anos) {
    if (anos < 0) return 'Impossível';
    if (anos <= 1) return 'Iniciante';
    if (anos <= 3) return 'Intermediário';
    if (anos <= 6) return 'Avançado';
    return 'Jedi';
}

for (var i = -2; i <= 10; i++) {
    document.write(i + " anos: " + experiencia(i) + ".<br>");
}

Or you could use the ternary operator:

function experiencia(anos) {
    return anos < 0 ? 'Impossível'
            : anos <= 1 ? 'Iniciante'
            : anos <= 3 ? 'Intermediário'
            : anos <= 6 ? 'Avançado'
            : 'Jedi';
}

for (var i = -2; i <= 10; i++) {
    document.write(i + " anos: " + experiencia(i) + ".<br>");
}

In the opinion of many people (my own), the switch es are horrible language constructions that should not even exist. In 99% of the time a switch is used, there is something else that could be used instead that would be much better or at least as good as.

You have sent the switch value 3 to the variable anos to%. This will access the case 3 .

Then return 'Avançado' will be the only answer. The code in case 2 will never be accessed because the value passed to switch was not 2. In addition the "comparisons" made within each case are ignored because they say nothing, they are not within if , which would be the method for comparing values.

And even if you put all the comparisons within a if , only the case 3 will be accessed, and would return undefined because the variable anos ( 3 ) does not would fit in the >= 4 && anos <= 6 condition.

In short, when you send the value 3 to switch , only the case 3 block will be accessed.