Is try {} without catch {} possible in JavaScript?

I have a number of functions which either return something or throw an error. In a main function, I call each of these, and would like to return the value returned by each function, or go on to the second function if the first functions throws an error.

So basically what I currently have is:

function testAll() {
    try { return func1(); } catch(e) {}
    try { return func2(); } catch(e) {} // If func1 throws error, try func2
    try { return func3(); } catch(e) {} // If func2 throws error, try func3
}

But actually I'd like to only try to return it (i.e. if it doesn't throw an error). I do not need the catch block. However, code like try {} fails because it is missing an (unused) catch {} block.

I put an example on jsFiddle.

So, is there any way to have those catch blocks removed whilst achieving the same effect?

A try without a catch clause sends its error to the next higher catch, or the window, if there is no catch defined within that try.

If you do not have a catch, a try expression requires a finally clause.

try {
    // whatever;
} finally {
    // always runs
}

It's possible to have an empty catch block, without an error variable, starting with ES2019. This is called optional catch binding and was implemented in V8 v6.6, released in June 2018. The feature has been available since Node 10, Chrome 66, Firefox 58, Opera 53 and Safari 11.1.

The syntax is shown below:

try {
  throw new Error("This won't show anything");
} catch { };

You still need a catch block, but it can be empty and you don't need to pass any variable. If you don't want a catch block at all, you can use the try/finally, but note that it won't swallow errors as an empty catch does.

try {
  throw new Error("This WILL get logged");
} finally {
  console.log("This syntax does not swallow errors");
}

Nope, catch (or finally) is try's friend and always there as part of try/catch.

However, it is perfectly valid to have them empty, like in your example.

In the comments in your example code (If func1 throws error, try func2), it would seem that what you really want to do is call the next function inside of the catch block of the previous.

I wouldn't recommend try-finally without the catch, because if both the try block and finally block throw errors, the error thrown in the finally clause gets bubbled up and the try block's error is ignored, in my own test:

try {
  console.log('about to error, guys!');
  throw new Error('eat me!');
} finally {
  console.log ('finally, who cares');
  throw new Error('finally error');
}

Result:

>     about to error, guys!
>     finally, who cares
>     .../error.js:9
>         throw new Error('finally error');
>         ^
>     
>     Error: finally error

No, it is not possible to have try block without catch (or finally). As a workaround, I believe you might want to define a helper function such as this:

function tryIt(fn, ...args) {
    try {
        return fn(...args);
    } catch {}
}

and use it like:

tryIt(function1, /* args if any */);
tryIt(function2, /* args if any */);

I've decide to look at the problem presented from a different angle.

I've been able to determine a way to to allow closely for the code pattern requested while in part addressing the un-handled error object listed by another commenter.

code can be seen @ http://jsfiddle.net/Abyssoft/RC7Nw/4/

try:catch is placed within a for loop allowing graceful fall through. while being able to iterate through all the functions needed. when explicit error handling is needed additional function array is used. in the even of error and functional array with error handlers element is not a function, error is dumped to console.

Per requirements of stackoverflow here is the code inline [edited to make JSLint compliant (remove leading spaces to confirm), improve readability]

function func1() {"use strict"; throw "I don't return anything"; }
function func2() {"use strict"; return 123; }
function func3() {"use strict"; throw "I don't return anything"; }

// ctr = Code to Run <array>, values = values <array>, 
// eh = error code can be blank.
// ctr and params should match 1 <-> 1
// Data validation not done here simple POC
function testAll(ctr, values, eh) {
    "use strict";
    var cb; // cb = code block counter
    for (cb in ctr) {
        if (ctr.hasOwnProperty(cb)) {
            try {
                return ctr[cb](values[cb]);
            } catch (e) {
                if (typeof eh[cb] === "function") {
                    eh[cb](e);
                } else {
                    //error intentionally/accidentially ignored
                    console.log(e);
                }
            }
        }
    }
    return false;
}

window.alert(testAll([func1, func2, func3], [], []));

​

If you only want functions 2 and 3 to fire if an error occurs why are you not putting them in the catch block?

function testAll() {
  try {
    return func1();
  } catch(e) {
    try {
      return func2();
    } catch(e) {
      try {
        return func3();
      } catch(e) {
        // LOG EVERYTHING FAILED
      }
    }
  }
}

...is there any way to have those catch blocks removed whilst achieving the same effect? As it would seem, no; Javascript requires a try block be followed by either a catch or a finally block.

Having said that, there is a way to use those catch blocks to achieve the effect you want.

// If func1 throws error, try func2 The if throws error condition, is what the catch block is for.

Why remove them when their use is exactly what you are after?

try { return func1(); }
catch {
   // if func1 throws error
   try { return func2(); } 
   catch {
      // if func2 throws error
      try { return func3(); } 
      catch {
         // if func3 throws error
      }
   }
}

I completely understand why you might not need a catch block, and would find it cleaner to be able to omit it entirely. But I don't think this is one of those situations.

They go together in every language that I know that has them (JavaScript, Java, C#, C++). Don't do it.

try & catch are like 2 side of one coin. so not possible without try.

Since ES2019 you can easily use try {} without catch {}:

try {
  parseResult = JSON.parse(potentiallyMalformedJSON);
} catch (unused) {}

For more info please reffer to Michael Ficcara's proposal

No. You have to keep them.

This actually makes sense since errors shouldn't be silently ignored at all.