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Variable throws variable might not have been initialized error


Variable throws variable might not have been initialized error

By : Pooja Banerjee
Date : November 25 2020, 01:01 AM
help you fix your problem You are getting the error, because that is how it is supposed to behave.
From JLS - Section 8.3.1.2:
code :


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Making sure that a final variable is initialized when a constructor throws an exception

Making sure that a final variable is initialized when a constructor throws an exception


By : Salma Sabry
Date : March 29 2020, 07:55 AM
seems to work fine I have a final member data:
code :
foo = createFoo();

...


private static Foo createFoo() {
    try {
        return new Foo();
    } catch (Exception e) {
        e.printStackTrace();
        return null;
    }
}
Drools: using the same DSL definition twice with a variable throws a duplicate variable error

Drools: using the same DSL definition twice with a variable throws a duplicate variable error


By : user3066862
Date : March 29 2020, 07:55 AM
To fix this issue You can synthesize the name of a binding variable like any other chunk of text:
code :
[when][]there is a qualification of type {qualification}=
      ${qualification}: {qualification}()
[then] print {qualification}=System.out.println( ${qualification} );
[then] print qualification=System.out.println( $qualification );
How to increment the value of initialized variable in mongodb query and that initialized value variable I am using to fe

How to increment the value of initialized variable in mongodb query and that initialized value variable I am using to fe


By : Shriya
Date : March 29 2020, 07:55 AM
wish of those help How can I increment the value of initialized variable in mongodb? , this is the answer
code :
    var FeedbackObj = {
"content": [{
        "feed": "It was interesting",
        "emotion": "interesting"
    },
    {
        "feed": "It was cool",
        "emotion": "cool"
    },
    {
        "feed": "It was amazing",
        "emotion": "amazing"
    },
    {
        "feed": "It was fantastic",
        "emotion": "fantastic"
    },
    {
        "feed": "It was entertaining",
        "emotion": "entertaining"
    },
    {
        "feed": "It was superb",
        "emotion": "superb"
    }]}

  let i = 0 ;

db.Movies.aggregate([
{$match :{imdb_rating:{$gt:9}}},
{"$group":{_id:{content_title:"$content_title",trailer_id:'$trailer_id'}}},
{"$project":{_id:0,content_title:"$_id.content_title",trailer_id:"$_id.trailer_id"}}

])
forEach(function(x){
    var y = {};
    y.contentTitle = x.content_title;
    y.feedbackEmoji = "";
    y.feedbackText = FeedbackObj.content[i].feed;
    y.emotion = FeedbackObj.content[i].emotion;
    y.sentiment = 1.0;
    y.timestamp = new NumberLong(new Date().getTime());
    y.userID = "vikash.srivastava@mailinator.com"; 
    y.trailerID = x.trailer_id;
    console.log("value of i :" + i +" : " + FeedbackObj.content[i].feed +" : " + FeedbackObj.content[i].emotion+" : " + x.trailer_id);
    i++;
    if(i == 6){
        i = 0;
    }
   db.User_Feedback.save(y);
})
Tensor flow throws an uninitialized error even after the variable is initialized

Tensor flow throws an uninitialized error even after the variable is initialized


By : Keral. Patel.
Date : March 29 2020, 07:55 AM
should help you out You also need to run the init function in the graph before computing cross entropy:
code :
with tf.Session() as sess:
  sess.run(init)
  ce = sess.run([crossentropy], feed_dict={p : inputs, t : targets, lr : 0.01})
Why python throws away error for referencing a global variable if later we define that same as local variable?

Why python throws away error for referencing a global variable if later we define that same as local variable?


By : MarshallCoder
Date : December 27 2020, 03:57 PM
I wish this help you How exactly Python does this is an implementation detail but the key is, "interpreted" doesn't mean it scans every line in order and knows nothing about the next line. If that were the case, every code until a syntax error would be run. Python analyzes the entire code to make sure it is valid code before running it.
The rule is, if a name is assigned in the local scope, referencing it before the assignment is an error, even if that name is defined in an outer scope.
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