Why when I throw a new Error in a try/catch, the rest of my script stops executing?

Question

I have created a basic es6 class. I'm using a try/catch block throughout the class. However, in my catch block, when I write:

    try {
      // Do something
    } catch (error) {
        throw new Error('No string to perform lower method on');
    }

The error is thrown, however, the rest of my script stops running. However, if I replace the throw new Error with console.error() - my script continues just fine.

Here is my class in full:

class StringFormatter {
  constructor(props) {

    try {

        // Check if the value passed to props.string is valid
        if (
            props.string === null ||
            props.string === undefined ||
            props.string.length <= 0 ||
            typeof props.string !== "string"
            ) {
                throw new Error(`can't find string - value provided: "${props.string}"`)
            } 

    } catch (error) {
        console.error(error);
    }

  }

  get stringVal() {
    console.log(this._string);
    return this._string;
  }

  set stringVal(val) {
    this._string = val;
  }

  lower() {
    // try to run lower() method, if not, throw error
    try {
       this.stringVal = this._string.toLowerCase();
       return this;
    } catch (error) {
        throw new Error('No string to perform lower method on');
    }
  }

  replaceSpaces(val) {
    // try to run replaceSpaces() method, if not, throw error
   try {
    this.stringVal = this._string.replace(/\s+/g, val);
    return this;
   } catch (error) {
    throw new Error('No string to perform replaceSpaces method on');
   }
  }
}

// initiate StringFormatter class and provide an empty string for error testing 
let myString = new StringFormatter({
    string: ''
});

// perform methods on new instance of class
myString.lower().replaceSpaces("*").stringVal;

// execute alert() - this should always run, even if the class object fails to execute methods.
alert("continues with the rest of script");

I'm trying to set it up, so that if the myString instance fails to run the methods, the rest of my script will continue to run. However, right now, that isn't the case. The alert() never runs.

Answer 1 · 2020-07-18T14:59:48Z

July 18, 2020 2:59pm

A "throw" that is not caught will always cause the program to end. And only code that is inside a "try" block can be caught.

So the "throw" shown here inside the "catch" block is not in a "try", does not get caught, and instead gets handled by the system which ends the program. It's almost as if there were no "try/catch" at all except that this code changes the original event for a different new one.

If your intention is for the code to continue running, then issuing a message instead of performing a "throw" is the appropriate thing to do.

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Kristian Woods

Kristian Woods

Why when I throw a new Error in a try/catch, the rest of my script stops executing?

1 Answer

Steven Parker

Steven Parker