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I have some issues with understanding Vertx asynchronous model and how Future behave...

In the startup code of my application, I check some conditions, like db access and other credentials, then I start several management verticles (config verticle, http admin verticle, etc) before starting the core of the application. The code has the following structure:

Vertx vertx = null;
Vertx.clusteredVertx(opts, ar -> {
 if(ar.failed()) 
 System.exit(-1);
 
 else 
 vertx = ar.result();
 Future<Void> f1 = asyncStartupFunction_1(...);
 if(f1.failed()) 
 System.exit(-1);
 
 else // f1 succeeded
 Future<Void> f2 = asyncStartupFunction_2(...);
 if(f2.failed()) 
 System.exit(-1);
 
 else // f2 succeeded
 ...
 // f2 succeeded
 // f1 succeeded
 // vertx creation OK

asyncStartupFunction_x do some asynchronous work (query a db or deploy a verticle) and then they return the Future that represents the result of this work.

I would have thought that once I'm in a else block, the corresponding future has succeeded. Is it possible that the work has not finished yet and that the corresponding future has not yet completed ? And thus, because fx.failed() == false I get to the else block while (in my understanding) I shouldn't?

What would be the right pattern ?

vert.x

share|improve this question

asked Mar 24 at 16:21

mszmurlomszmurlo

591615

add a comment | 

0

I have some issues with understanding Vertx asynchronous model and how Future behave...

In the startup code of my application, I check some conditions, like db access and other credentials, then I start several management verticles (config verticle, http admin verticle, etc) before starting the core of the application. The code has the following structure:

Vertx vertx = null;
Vertx.clusteredVertx(opts, ar -> {
 if(ar.failed()) 
 System.exit(-1);
 
 else 
 vertx = ar.result();
 Future<Void> f1 = asyncStartupFunction_1(...);
 if(f1.failed()) 
 System.exit(-1);
 
 else // f1 succeeded
 Future<Void> f2 = asyncStartupFunction_2(...);
 if(f2.failed()) 
 System.exit(-1);
 
 else // f2 succeeded
 ...
 // f2 succeeded
 // f1 succeeded
 // vertx creation OK

asyncStartupFunction_x do some asynchronous work (query a db or deploy a verticle) and then they return the Future that represents the result of this work.

I would have thought that once I'm in a else block, the corresponding future has succeeded. Is it possible that the work has not finished yet and that the corresponding future has not yet completed ? And thus, because fx.failed() == false I get to the else block while (in my understanding) I shouldn't?

What would be the right pattern ?

vert.x

share|improve this question

asked Mar 24 at 16:21

mszmurlomszmurlo

591615

add a comment | 

I have some issues with understanding Vertx asynchronous model and how Future behave...

In the startup code of my application, I check some conditions, like db access and other credentials, then I start several management verticles (config verticle, http admin verticle, etc) before starting the core of the application. The code has the following structure:

Vertx vertx = null;
Vertx.clusteredVertx(opts, ar -> {
 if(ar.failed()) 
 System.exit(-1);
 
 else 
 vertx = ar.result();
 Future<Void> f1 = asyncStartupFunction_1(...);
 if(f1.failed()) 
 System.exit(-1);
 
 else // f1 succeeded
 Future<Void> f2 = asyncStartupFunction_2(...);
 if(f2.failed()) 
 System.exit(-1);
 
 else // f2 succeeded
 ...
 // f2 succeeded
 // f1 succeeded
 // vertx creation OK

asyncStartupFunction_x do some asynchronous work (query a db or deploy a verticle) and then they return the Future that represents the result of this work.

I would have thought that once I'm in a else block, the corresponding future has succeeded. Is it possible that the work has not finished yet and that the corresponding future has not yet completed ? And thus, because fx.failed() == false I get to the else block while (in my understanding) I shouldn't?

What would be the right pattern ?

vert.x

share|improve this question

asked Mar 24 at 16:21

mszmurlomszmurlo

591615

Vertx vertx = null;
Vertx.clusteredVertx(opts, ar -> {
 if(ar.failed()) 
 System.exit(-1);
 
 else 
 vertx = ar.result();
 Future<Void> f1 = asyncStartupFunction_1(...);
 if(f1.failed()) 
 System.exit(-1);
 
 else // f1 succeeded
 Future<Void> f2 = asyncStartupFunction_2(...);
 if(f2.failed()) 
 System.exit(-1);
 
 else // f2 succeeded
 ...
 // f2 succeeded
 // f1 succeeded
 // vertx creation OK

share|improve this question

asked Mar 24 at 16:21

mszmurlomszmurlo

591615

share|improve this question

asked Mar 24 at 16:21

mszmurlomszmurlo

591615

asked Mar 24 at 16:21

mszmurlomszmurlo

591615

asked Mar 24 at 16:21

mszmurlomszmurlo

591615

1 Answer
1

active

oldest

votes

3

asyncStartupFunction_1 returns a Future and there is no guarantee the future is completed at this point.

Consequently f1.failed() has good chances not to give your a proper result. It's not a blocking call, and it may return false simply because the future is not completed yet.

What you need is async coordination. Depending on your needs, you can either execute the async jobs in parallel or sequentially.

share|improve this answer

answered Mar 25 at 8:52

tsegismonttsegismont

3,4651819

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I remember I already read that some time ago but just could remember where when I asked the question... Thanks for your time.
  
  – mszmurlo
  Mar 25 at 10:23
  

  
  
  
  

add a comment | 

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3

asyncStartupFunction_1 returns a Future and there is no guarantee the future is completed at this point.

Consequently f1.failed() has good chances not to give your a proper result. It's not a blocking call, and it may return false simply because the future is not completed yet.

What you need is async coordination. Depending on your needs, you can either execute the async jobs in parallel or sequentially.

share|improve this answer

answered Mar 25 at 8:52

tsegismonttsegismont

3,4651819

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I remember I already read that some time ago but just could remember where when I asked the question... Thanks for your time.
  
  – mszmurlo
  Mar 25 at 10:23
  

  
  
  
  

add a comment | 

3

asyncStartupFunction_1 returns a Future and there is no guarantee the future is completed at this point.

Consequently f1.failed() has good chances not to give your a proper result. It's not a blocking call, and it may return false simply because the future is not completed yet.

What you need is async coordination. Depending on your needs, you can either execute the async jobs in parallel or sequentially.

share|improve this answer

answered Mar 25 at 8:52

tsegismonttsegismont

3,4651819

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I remember I already read that some time ago but just could remember where when I asked the question... Thanks for your time.
  
  – mszmurlo
  Mar 25 at 10:23
  

  
  
  
  

add a comment | 

asyncStartupFunction_1 returns a Future and there is no guarantee the future is completed at this point.

Consequently f1.failed() has good chances not to give your a proper result. It's not a blocking call, and it may return false simply because the future is not completed yet.

What you need is async coordination. Depending on your needs, you can either execute the async jobs in parallel or sequentially.

share|improve this answer

answered Mar 25 at 8:52

tsegismonttsegismont

3,4651819

share|improve this answer

answered Mar 25 at 8:52

tsegismonttsegismont

3,4651819

answered Mar 25 at 8:52

tsegismonttsegismont

3,4651819

answered Mar 25 at 8:52

tsegismonttsegismont

3,4651819