CODEWITHSUNDEEP
javascriptnode.jspromise
isaac.hazan
isaac.hazan

Reputation: 3874

Node JS: Not calling promise reject() callback stops the execution of the program

It seems I misunderstand the Promise mechanism. In the following code:

async function CreateSession()
{
      var sessionId = await new Promise((resolve, reject) => {
        request(options, function (error, response, body) {
            if (!error && response.statusCode == 200) {
                log.debug("Session created successfully")
                resolve(body["session-id"]);
            }
            else
            {
                log.error("Failed to create session:");
                log.error(error);
                // reject(error);
            }
        });
    });

    log.debug(`Session Id: ${sessionId}`);
    return sessionId;
}

The reject() is commented on purpose to understand the mechanism. When calling the above CreateSession() function as follows and making sure an error is happenning:

async function MyFunction()
{
    var sessionId = await CreateSession();
    AnotherCode();
}

The entire program stops, the code after await in CreateSession is never reached, AnotherCode() is never called, and even if I put a handler for uncaught exception it doesn't catch any error.

My questions:

Upvotes: 1

Views: 1185

Answers (2)

Telmo Dias
Telmo Dias

Reputation: 4168

For a start you're mixing 2 concepts that achieve the same thing, the creation of a Promise. You are mixing async/await and new Promise.

That makes it actually like you have 2 chained Promises.

Why does the program stops when reject() is not called?

Because you are returning a Promise and that by itself demands that either a resolve or a reject callbacks are called.

If it's an error why it is not called by the handler for uncaught exceptions?

There is no actual error, the only thing that happens is that your if clause is not met.

How do I ignore the an error when a promise fails to run its code?

Again, the Promise does not fail to run it's code, you're failing to understand the concept.

To make it easier either use something like :

function CreateSession(){
   return new Promise((resolve, reject) => {
      if( <yourconditionhere> ){
         resolve(res);
      }else{
         reject(error);
      }
   });
}


CreateSession
.then((res) => {
   // code to execute in case of success (resolve)
   // receiving as argument the argument of the resolve call.
   AnotherCode();
})
.catch( error => {
   // code to execute in case of non-success (reject)
   // receiving as argument the argument of the reject call.
});

or

async function CreateSession(){
   let res = await request(options, function (error, response, body) {
      if( <yourconditionhere> ){
         return ...;
      }else{
         return ...;
      }
   }
   return res;
}

async function MyFunction()
{
    var sessionId = await CreateSession();
    AnotherCode();
}

Upvotes: 1

Jb31
Jb31

Reputation: 1391

Why does the program stops when reject() is not called?

Because of the await. The program waits for the promise to either resolve or reject. It can't detect somehow that the code inside the promise constructor has finished executing (even if it could - what should be the "standard" behavior?) and there is no more code to come.

From the MDN:

An async function can contain an await expression that pauses the execution of the async function to wait for the passed Promise's resolution, then resumes the async function's execution and evaluates as the resolved value.

For some "under the hood" insights, you might want to look at the ES5 code generated by transpilers like Babel (IIRC it uses generators to handle this).

If it's an error why it is not called by the handler for uncaught exceptions?

The program doesn't know that error is an error if it's neither thrown nor passed using reject.

How do I ignore the an error when a promise fails to run its code?

Either resolve (instead of reject) or reject and do the classical try/catch around the await code

Upvotes: 2

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