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Cannot Be Applied To Java Lang String Java

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Nathaniel Stodard
SCJP, SCJD, SCWCD, SCBCD, SCDJWS, ICAD, ICSD, ICED Joel McNary Bartender Posts: 1840 I like... Thanks a lot for your help. Is This A Good Question/Topic? 0 Back to top MultiQuote Quote + Reply Replies To: Object() in java.lang.Object cannot be applied to (java.l #2 GregBrannon D.I.C Lover Reputation: 2250 Posts: What class of object are you trying to create here? http://electrictricycle.net/cannot-be/cannot-be-applied-to-java-lang-string-java-lang-string.html

While similar questions may be on-topic here, this one was resolved in a manner unlikely to help future readers. Reply With Quote 09-23-2011,01:21 PM #12 JosAH Moderator Join Date Sep 2008 Location Voorschoten, the Netherlands Posts 14,323 Blog Entries7 Rep Power 25 Re: Java Error cannot be applied to (java.lang.String), Does The Amazing Lightspeed Horse work, RAW? entryList[i][j] would be for a single object. http://stackoverflow.com/questions/5410758/java-lang-string-cannot-be-applied-to-java-lang-object

Java Lang String Cannot Be Applied To Java Lang String

What are you trying to do with that statement? Free Video Tutorials for Java and Eclipse Beginners Reply With Quote 09-23-2011,10:49 AM #5 iceyferrara Member Join Date Sep 2011 Posts 10 Rep Power 0 Re: Java Error cannot be applied If you'd like to look at all the classes and the compiler download it Here. Hey, Thanx for your suggestion.

How small could an animal be before it is consciously aware of the effects of quantum mechanics? What now? int anInt = Integer.parseInt(aString); There's also the matter of exceptions to consider. || Java Cannot Be Applied To Int Is there a way to add a string and integer in to a two-dimensional array or something?

It seems easier that way. Operator Cannot Be Applied To Java Lang String I am new to Java can someone point me in the right direction? What did John Templeton mean when he said that the four most dangerous words in investing are: ‘this time it’s different'? Best of luck.

Why didn’t Japan attack the West Coast of the United States during World War II? Object In Object Cannot Be Applied I'm trying to understand where its looking for the constructor. Storage of a material that passes through non-living matter Wait... Sorry, im really not that bad with this stuff but this has been puzzling me and even my professor couldn't help(or didn't want to) idk.

Operator Cannot Be Applied To Java Lang String

You're trying to construct an Object given arguments to a constructor that doesn't exist. http://www.java-forums.org/new-java/48853-java-error-cannot-applied-java-lang-string-phone-book-entry-program.html help with area of shapes! Java Lang String Cannot Be Applied To Java Lang String Join them; it only takes a minute: Sign up operator '+' cannot be applied to java.lang.string [closed] up vote -3 down vote favorite catch (IOException e) { Log.e(TAG, "Error Loading ", Operator + Cannot Be Applied To Java.lang.string Java foo("bar", +3) is valid, for instance. :-) But the unary + cannot be applied to a string. –T.J.

That's what 'super' does when it is used as the first line of a child class' constructor. this contact form What's New? I have it all working now. share|improve this answer edited May 8 '15 at 23:35 answered May 8 '15 at 23:33 sockeqwe 5,789114384 Why does it work in NetBeans? –F4LLCON May 8 '15 at 23:34 Operator Cannot Be Applied To Java.lang.object Int

Still trying everything but nothing, do i need to put something after the method call in the ()? Browse other questions tagged java string user-interface java.lang.class or ask your own question. Here is my new code. http://electrictricycle.net/cannot-be/cannot-be-applied-to-int-java-lang-string.html because that is where the problem is.

Posted By MS-POWER (2 replies) Yesterday, 07:58 PM in New To Java configure SSL sockets? Cannot Be Applied To Java.lang.string Int Originally Posted by iceyferrara Still trying everything but nothing, do i need to put something after the method call in the ()? i try to make them match, and it gives me the same error, i even tried renaming them but nothing.

public class PriorityQueue { public String createString(int item, String str) { String space, priort; priort = Integer.toString(item); space = " "; totl = str + space + priort; return totl; }

And if so, what is it looking for? I think maybe you're trying to actually put those items in an array instead of an Object constructor (in which case you'll get an exception saying you have an index that Probably a better approach. Operator Cannot Be Applied Java Can any one tell me what that error is and how to fix it?

Is there a name for the (anti- ) pattern of passing parameters that will only be used several levels deep in the call chain? Related 2(java.lang.String) cannot be applied to (java.lang.Object)2768Java's +=, -=, *=, /= compound assignment operators-4Operator || cannot be applied to java.lang.string-3operator '+' cannot be applied to java.lang.string2Operator == cannot be applied to here is my code /* Determines type of triangle based on length of sides. */ import javax.swing.JOptionPane; public class Triangles { public static void main(String[] args) { //get the length of Check This Out Hot Network Questions Does sputtering butter mean that water is present?

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This forum is now closed to new posts, but you can browse existing content. I still have the problem of the close parenthesis. If the user enters a String that isn't a valid number, those parseXXX methods will "throw an exception." Have you learned about those yet? [Jess in Action][AskingGoodQuestions] matt jamison Greenhorn Results 1 to 13 of 13 Thread: Java Error cannot be applied to (java.lang.String), phone book entry program.

I tried if (alpha[i].equals(c)) { but then I would get no results like I do in NetBeans, which is converting a String to Morse e.g. What do you expect the + to do there? Reply With Quote 09-23-2011,11:10 AM #6 iceyferrara Member Join Date Sep 2011 Posts 10 Rep Power 0 Re: Java Error cannot be applied to (java.lang.String), phone book entry program. First, you can't meaningfully use "==" to compare Strings; you must use the equals() method instead.

Could I work as a Professor in Europe if I only speak English? QuoteI'm trying to understand where its looking for the constructor. The binary + can be applied to strings, but the unary + cannot. if (sideOne == sideTwo && sideThree) { JOptionPane.showMessageDialog(null, "The Triangle is Equilateral"); break; } else if (sideOne == sideTwo || sideThree) { JOptionPane.showMessageDialog(null, "The Triangle is Isosceles"); //break; } else if

kind regards, Jos The only person who got everything done by Friday was Robinson Crusoe. You can't, unfortunately, write a.equals(!b) to mean "a not equal to b".