return statement does not work in javascript try block

Question

return statement normally terminate function execution. But I have been realized that return statement does not work like what I expected? Where is the problem?

function SimpleSymbols() {
    ["a", "b", "c", "d"].forEach(function (char, index) {
        try {
            console.log("alert");
            return "false";  //this line does not work?
        } catch (err) {

        }
    });
}

SimpleSymbols();

It returns from the `forEach` iteration callback, is that not what you wanted? — Bergi, Jan 27 '19 at 12:50

Mamun · Accepted Answer · 2019-01-27T12:58:29.627

forEach() does not return. You can return from outside of the loop once the loop is completed:

function SimpleSymbols() { 
  var r;
  ["a", "b", "c", "d"].forEach(function (char, index) {
    try {
        console.log("alert");
        r = "false";  //this line does not work?
    } catch (err) {
    }
  });
  return r
}

console.log(SimpleSymbols());

Instead you can use normal for loop:

function SimpleSymbols() { 
  var arr = ["a", "b", "c", "d"]
  for(var i=0; i<arr.length; i++){
    try {
        console.log("alert");
        return "false";  //this line does not work?
    } catch (err) {
    }
  }
}

console.log(SimpleSymbols());

Thank you. You are totally right. I couldn't used to write callback functions yet. — mustafa kemal tuna, Jan 27 '19 at 16:01

return statement does not work in javascript try block

1 Answers1