Why can't we use uninitialized local variable to access static content of its type?Are static class variables possible in Python?Why is the Java main method static?Why can't I use switch statement on a String?Why can't static methods be abstract in JavaStatic variables in JavaScriptWhy doesn't Java allow overriding of static methods?Can a local variable's memory be accessed outside its scope?Why are static variables considered evil?Why attempt to print uninitialized variable does not always result in an error message

Minimize taxes now that I earn a living wage

If people's daily habits are reliable then why is the stock market so unpredictable?

Is Zack Morris's 'time stop' ability in "Saved By the Bell" a supernatural ability?

Talk about Grandpa's weird talk: Who are these folks?

What are the end bytes of *.docx file format

Do household ovens ventilate heat to the outdoors?

How can I create folders in folders in terminal

Simple way to decompose a lists in a certain way?

Is this adjustment to the Lucky feat underpowered?

Tips for remembering the order of parameters for ln?

Can a business put a specific cancel-by date into a contract?

Is it safe to put a microwave in a walk-in closet?

When would open interest equal trading volume?

How do rulers get rich from war?

Other than good shoes and a stick, what are some ways to preserve your knees on long hikes?

Which block cipher parameters should be kept secret?

Exam design: give maximum score per question or not?

Applications of mathematics in clinical setting

Inquiry answerer

Why do we need to use transistors when building an OR gate?

Is a global DNS record a security risk for phpMyAdmin?

Strength of Female Chimpanzees vs. Male Chimpanzees?

What the did the controller say during my approach to land (audio clip)?

Whence comes increasing usage of "I'll do an X" instead of "I'll have" in ordering food?



Why can't we use uninitialized local variable to access static content of its type?


Are static class variables possible in Python?Why is the Java main method static?Why can't I use switch statement on a String?Why can't static methods be abstract in JavaStatic variables in JavaScriptWhy doesn't Java allow overriding of static methods?Can a local variable's memory be accessed outside its scope?Why are static variables considered evil?Why attempt to print uninitialized variable does not always result in an error message






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








8















class Foo
public static int x = 1;

public static void main(String[] args)
Foo foo;
System.out.println(foo.x); // Error: Variable 'foo' might not have been initialized




While trying to access x via an uninitialized local variable Foo foo; I get the compile error Variable 'foo' might not have been initialized.



It could seem like this error makes sense, but only until we realize that to access a static member the compiler doesn't use the value of a variable, but only its type.



For instance I can initialize foo with value null and this will let us access x without any problems:



Foo foo = null;
System.out.println(foo.x);//Now it works!!!


Such scenario works because compiler realizes that x is static and treats foo.x as it was written Foo.x (at least that is what I thought until now).



So why compiler suddenly insists on foo having a value which it will NOT use at that place?



(Disclaimer: This is not code which would be used in real application, just interesting phenomenon which I couldn't find answer to on Stack Overflow so I decided to ask about it.)










share|improve this question





















  • 2





    I'd say a limitation in the compiler that isn't really worth fixing, given that the code raises a warning anyway.

    – manouti
    8 hours ago






  • 1





    @manouti That is also my guess, but I am still interested in why compiler behaves that way. Which part of specification forces it?

    – Pshemo
    8 hours ago











  • I think the compiler is less concerned with initialization (even though that is the warning) as it is with the null pointer exception you are creating. The reference to the initialization is the compiler trying to give reason to why you may have a potential null pointer exception.

    – portfoliobuilder
    7 hours ago






  • 2





    @portfoliobuilder There is no risk of NPE here since as mentioned in question, in case of accessing static member compiler doesn't use value of variable but its type. We can even write ((Foo)null).x and this will compile and work because compiler will recognize that x is static (unless I misunderstood your comment).

    – Pshemo
    7 hours ago







  • 3





    Accessing a static variable from a non-static context (such as foo.x) should have been a compiler error when Java was first created. Sadly, that ship sailed over 25 years ago and would be a breaking change if they changed it now.

    – Powerlord
    7 hours ago


















8















class Foo
public static int x = 1;

public static void main(String[] args)
Foo foo;
System.out.println(foo.x); // Error: Variable 'foo' might not have been initialized




While trying to access x via an uninitialized local variable Foo foo; I get the compile error Variable 'foo' might not have been initialized.



It could seem like this error makes sense, but only until we realize that to access a static member the compiler doesn't use the value of a variable, but only its type.



For instance I can initialize foo with value null and this will let us access x without any problems:



Foo foo = null;
System.out.println(foo.x);//Now it works!!!


Such scenario works because compiler realizes that x is static and treats foo.x as it was written Foo.x (at least that is what I thought until now).



So why compiler suddenly insists on foo having a value which it will NOT use at that place?



(Disclaimer: This is not code which would be used in real application, just interesting phenomenon which I couldn't find answer to on Stack Overflow so I decided to ask about it.)










share|improve this question





















  • 2





    I'd say a limitation in the compiler that isn't really worth fixing, given that the code raises a warning anyway.

    – manouti
    8 hours ago






  • 1





    @manouti That is also my guess, but I am still interested in why compiler behaves that way. Which part of specification forces it?

    – Pshemo
    8 hours ago











  • I think the compiler is less concerned with initialization (even though that is the warning) as it is with the null pointer exception you are creating. The reference to the initialization is the compiler trying to give reason to why you may have a potential null pointer exception.

    – portfoliobuilder
    7 hours ago






  • 2





    @portfoliobuilder There is no risk of NPE here since as mentioned in question, in case of accessing static member compiler doesn't use value of variable but its type. We can even write ((Foo)null).x and this will compile and work because compiler will recognize that x is static (unless I misunderstood your comment).

    – Pshemo
    7 hours ago







  • 3





    Accessing a static variable from a non-static context (such as foo.x) should have been a compiler error when Java was first created. Sadly, that ship sailed over 25 years ago and would be a breaking change if they changed it now.

    – Powerlord
    7 hours ago














8












8








8








class Foo
public static int x = 1;

public static void main(String[] args)
Foo foo;
System.out.println(foo.x); // Error: Variable 'foo' might not have been initialized




While trying to access x via an uninitialized local variable Foo foo; I get the compile error Variable 'foo' might not have been initialized.



It could seem like this error makes sense, but only until we realize that to access a static member the compiler doesn't use the value of a variable, but only its type.



For instance I can initialize foo with value null and this will let us access x without any problems:



Foo foo = null;
System.out.println(foo.x);//Now it works!!!


Such scenario works because compiler realizes that x is static and treats foo.x as it was written Foo.x (at least that is what I thought until now).



So why compiler suddenly insists on foo having a value which it will NOT use at that place?



(Disclaimer: This is not code which would be used in real application, just interesting phenomenon which I couldn't find answer to on Stack Overflow so I decided to ask about it.)










share|improve this question
















class Foo
public static int x = 1;

public static void main(String[] args)
Foo foo;
System.out.println(foo.x); // Error: Variable 'foo' might not have been initialized




While trying to access x via an uninitialized local variable Foo foo; I get the compile error Variable 'foo' might not have been initialized.



It could seem like this error makes sense, but only until we realize that to access a static member the compiler doesn't use the value of a variable, but only its type.



For instance I can initialize foo with value null and this will let us access x without any problems:



Foo foo = null;
System.out.println(foo.x);//Now it works!!!


Such scenario works because compiler realizes that x is static and treats foo.x as it was written Foo.x (at least that is what I thought until now).



So why compiler suddenly insists on foo having a value which it will NOT use at that place?



(Disclaimer: This is not code which would be used in real application, just interesting phenomenon which I couldn't find answer to on Stack Overflow so I decided to ask about it.)







java static initialization local-variables jls






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 5 hours ago









Andrew Tobilko

35.1k10 gold badges54 silver badges105 bronze badges




35.1k10 gold badges54 silver badges105 bronze badges










asked 8 hours ago









PshemoPshemo

99.7k17 gold badges141 silver badges199 bronze badges




99.7k17 gold badges141 silver badges199 bronze badges










  • 2





    I'd say a limitation in the compiler that isn't really worth fixing, given that the code raises a warning anyway.

    – manouti
    8 hours ago






  • 1





    @manouti That is also my guess, but I am still interested in why compiler behaves that way. Which part of specification forces it?

    – Pshemo
    8 hours ago











  • I think the compiler is less concerned with initialization (even though that is the warning) as it is with the null pointer exception you are creating. The reference to the initialization is the compiler trying to give reason to why you may have a potential null pointer exception.

    – portfoliobuilder
    7 hours ago






  • 2





    @portfoliobuilder There is no risk of NPE here since as mentioned in question, in case of accessing static member compiler doesn't use value of variable but its type. We can even write ((Foo)null).x and this will compile and work because compiler will recognize that x is static (unless I misunderstood your comment).

    – Pshemo
    7 hours ago







  • 3





    Accessing a static variable from a non-static context (such as foo.x) should have been a compiler error when Java was first created. Sadly, that ship sailed over 25 years ago and would be a breaking change if they changed it now.

    – Powerlord
    7 hours ago













  • 2





    I'd say a limitation in the compiler that isn't really worth fixing, given that the code raises a warning anyway.

    – manouti
    8 hours ago






  • 1





    @manouti That is also my guess, but I am still interested in why compiler behaves that way. Which part of specification forces it?

    – Pshemo
    8 hours ago











  • I think the compiler is less concerned with initialization (even though that is the warning) as it is with the null pointer exception you are creating. The reference to the initialization is the compiler trying to give reason to why you may have a potential null pointer exception.

    – portfoliobuilder
    7 hours ago






  • 2





    @portfoliobuilder There is no risk of NPE here since as mentioned in question, in case of accessing static member compiler doesn't use value of variable but its type. We can even write ((Foo)null).x and this will compile and work because compiler will recognize that x is static (unless I misunderstood your comment).

    – Pshemo
    7 hours ago







  • 3





    Accessing a static variable from a non-static context (such as foo.x) should have been a compiler error when Java was first created. Sadly, that ship sailed over 25 years ago and would be a breaking change if they changed it now.

    – Powerlord
    7 hours ago








2




2





I'd say a limitation in the compiler that isn't really worth fixing, given that the code raises a warning anyway.

– manouti
8 hours ago





I'd say a limitation in the compiler that isn't really worth fixing, given that the code raises a warning anyway.

– manouti
8 hours ago




1




1





@manouti That is also my guess, but I am still interested in why compiler behaves that way. Which part of specification forces it?

– Pshemo
8 hours ago





@manouti That is also my guess, but I am still interested in why compiler behaves that way. Which part of specification forces it?

– Pshemo
8 hours ago













I think the compiler is less concerned with initialization (even though that is the warning) as it is with the null pointer exception you are creating. The reference to the initialization is the compiler trying to give reason to why you may have a potential null pointer exception.

– portfoliobuilder
7 hours ago





I think the compiler is less concerned with initialization (even though that is the warning) as it is with the null pointer exception you are creating. The reference to the initialization is the compiler trying to give reason to why you may have a potential null pointer exception.

– portfoliobuilder
7 hours ago




2




2





@portfoliobuilder There is no risk of NPE here since as mentioned in question, in case of accessing static member compiler doesn't use value of variable but its type. We can even write ((Foo)null).x and this will compile and work because compiler will recognize that x is static (unless I misunderstood your comment).

– Pshemo
7 hours ago






@portfoliobuilder There is no risk of NPE here since as mentioned in question, in case of accessing static member compiler doesn't use value of variable but its type. We can even write ((Foo)null).x and this will compile and work because compiler will recognize that x is static (unless I misunderstood your comment).

– Pshemo
7 hours ago





3




3





Accessing a static variable from a non-static context (such as foo.x) should have been a compiler error when Java was first created. Sadly, that ship sailed over 25 years ago and would be a breaking change if they changed it now.

– Powerlord
7 hours ago






Accessing a static variable from a non-static context (such as foo.x) should have been a compiler error when Java was first created. Sadly, that ship sailed over 25 years ago and would be a breaking change if they changed it now.

– Powerlord
7 hours ago













3 Answers
3






active

oldest

votes


















5
















§15.11. Field Access Expressions:




If the field is static:




The Primary expression is evaluated, and the result is discarded. If evaluation of the Primary expression completes abruptly, the field access expression completes abruptly for the same reason.





Where earlier it states that field access is identified by Primary.Identifier.



This shows that even though it seems to not use the Primary, it is still evaluated and the result is then discarded which is why it will need to be initialized. This can make a difference when the evaluation halts the access as stated in the quote.






share|improve this answer
































    4

















    Chapter 16. Definite Assignment



    Each local variable (§14.4) and every blank final field (§4.12.4, §8.3.1.2) must have a definitely assigned value when any access of its value occurs.




    It doesn't really matter what you try to access via a local variable. The rule is that it should be definitely assigned before that.



    To evaluate a field access expression foo.x, the primary part of it (foo) must be evaluated first. It means that access to foo will occur, which will result in a compile-time error.




    For every access of a local variable or blank final field x, x must be definitely assigned before the access, or a compile-time error occurs.







    share|improve this answer



























    • I think correct answer is combination of yours and Nexevis. Now I need to figure out which one to accept...

      – Pshemo
      7 hours ago











    • Since Nexevis was first to show place in specification which forces primary expression (here foo) to be evaluated regardless of its result I decided to accept his answer. Hope you don't mind.

      – Pshemo
      5 hours ago












    • @Pshemo I am absolutely fine with your decision :)

      – Andrew Tobilko
      5 hours ago


















    1
















    There is value in keeping the rules as simple as possible, and “don’t use a variable that might not have been initialised” is as simple as it gets.



    More to the point, there is an established way of calling static methods - always use the class name, not a variable.



    System.out.println(Foo.x);


    The variable “foo” is unwanted overhead that should be removed, and the compiler errors and warnings could be seen as helping leading towards that.






    share|improve this answer





























      Your Answer






      StackExchange.ifUsing("editor", function ()
      StackExchange.using("externalEditor", function ()
      StackExchange.using("snippets", function ()
      StackExchange.snippets.init();
      );
      );
      , "code-snippets");

      StackExchange.ready(function()
      var channelOptions =
      tags: "".split(" "),
      id: "1"
      ;
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function()
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled)
      StackExchange.using("snippets", function()
      createEditor();
      );

      else
      createEditor();

      );

      function createEditor()
      StackExchange.prepareEditor(
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader:
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/4.0/"u003ecc by-sa 4.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      ,
      onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      );



      );














      draft saved

      draft discarded
















      StackExchange.ready(
      function ()
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f57964026%2fwhy-cant-we-use-uninitialized-local-variable-to-access-static-content-of-its-ty%23new-answer', 'question_page');

      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5
















      §15.11. Field Access Expressions:




      If the field is static:




      The Primary expression is evaluated, and the result is discarded. If evaluation of the Primary expression completes abruptly, the field access expression completes abruptly for the same reason.





      Where earlier it states that field access is identified by Primary.Identifier.



      This shows that even though it seems to not use the Primary, it is still evaluated and the result is then discarded which is why it will need to be initialized. This can make a difference when the evaluation halts the access as stated in the quote.






      share|improve this answer





























        5
















        §15.11. Field Access Expressions:




        If the field is static:




        The Primary expression is evaluated, and the result is discarded. If evaluation of the Primary expression completes abruptly, the field access expression completes abruptly for the same reason.





        Where earlier it states that field access is identified by Primary.Identifier.



        This shows that even though it seems to not use the Primary, it is still evaluated and the result is then discarded which is why it will need to be initialized. This can make a difference when the evaluation halts the access as stated in the quote.






        share|improve this answer



























          5














          5










          5









          §15.11. Field Access Expressions:




          If the field is static:




          The Primary expression is evaluated, and the result is discarded. If evaluation of the Primary expression completes abruptly, the field access expression completes abruptly for the same reason.





          Where earlier it states that field access is identified by Primary.Identifier.



          This shows that even though it seems to not use the Primary, it is still evaluated and the result is then discarded which is why it will need to be initialized. This can make a difference when the evaluation halts the access as stated in the quote.






          share|improve this answer













          §15.11. Field Access Expressions:




          If the field is static:




          The Primary expression is evaluated, and the result is discarded. If evaluation of the Primary expression completes abruptly, the field access expression completes abruptly for the same reason.





          Where earlier it states that field access is identified by Primary.Identifier.



          This shows that even though it seems to not use the Primary, it is still evaluated and the result is then discarded which is why it will need to be initialized. This can make a difference when the evaluation halts the access as stated in the quote.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 7 hours ago









          NexevisNexevis

          2,0832 gold badges5 silver badges13 bronze badges




          2,0832 gold badges5 silver badges13 bronze badges


























              4

















              Chapter 16. Definite Assignment



              Each local variable (§14.4) and every blank final field (§4.12.4, §8.3.1.2) must have a definitely assigned value when any access of its value occurs.




              It doesn't really matter what you try to access via a local variable. The rule is that it should be definitely assigned before that.



              To evaluate a field access expression foo.x, the primary part of it (foo) must be evaluated first. It means that access to foo will occur, which will result in a compile-time error.




              For every access of a local variable or blank final field x, x must be definitely assigned before the access, or a compile-time error occurs.







              share|improve this answer



























              • I think correct answer is combination of yours and Nexevis. Now I need to figure out which one to accept...

                – Pshemo
                7 hours ago











              • Since Nexevis was first to show place in specification which forces primary expression (here foo) to be evaluated regardless of its result I decided to accept his answer. Hope you don't mind.

                – Pshemo
                5 hours ago












              • @Pshemo I am absolutely fine with your decision :)

                – Andrew Tobilko
                5 hours ago















              4

















              Chapter 16. Definite Assignment



              Each local variable (§14.4) and every blank final field (§4.12.4, §8.3.1.2) must have a definitely assigned value when any access of its value occurs.




              It doesn't really matter what you try to access via a local variable. The rule is that it should be definitely assigned before that.



              To evaluate a field access expression foo.x, the primary part of it (foo) must be evaluated first. It means that access to foo will occur, which will result in a compile-time error.




              For every access of a local variable or blank final field x, x must be definitely assigned before the access, or a compile-time error occurs.







              share|improve this answer



























              • I think correct answer is combination of yours and Nexevis. Now I need to figure out which one to accept...

                – Pshemo
                7 hours ago











              • Since Nexevis was first to show place in specification which forces primary expression (here foo) to be evaluated regardless of its result I decided to accept his answer. Hope you don't mind.

                – Pshemo
                5 hours ago












              • @Pshemo I am absolutely fine with your decision :)

                – Andrew Tobilko
                5 hours ago













              4














              4










              4










              Chapter 16. Definite Assignment



              Each local variable (§14.4) and every blank final field (§4.12.4, §8.3.1.2) must have a definitely assigned value when any access of its value occurs.




              It doesn't really matter what you try to access via a local variable. The rule is that it should be definitely assigned before that.



              To evaluate a field access expression foo.x, the primary part of it (foo) must be evaluated first. It means that access to foo will occur, which will result in a compile-time error.




              For every access of a local variable or blank final field x, x must be definitely assigned before the access, or a compile-time error occurs.







              share|improve this answer
















              Chapter 16. Definite Assignment



              Each local variable (§14.4) and every blank final field (§4.12.4, §8.3.1.2) must have a definitely assigned value when any access of its value occurs.




              It doesn't really matter what you try to access via a local variable. The rule is that it should be definitely assigned before that.



              To evaluate a field access expression foo.x, the primary part of it (foo) must be evaluated first. It means that access to foo will occur, which will result in a compile-time error.




              For every access of a local variable or blank final field x, x must be definitely assigned before the access, or a compile-time error occurs.








              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited 7 hours ago

























              answered 7 hours ago









              Andrew TobilkoAndrew Tobilko

              35.1k10 gold badges54 silver badges105 bronze badges




              35.1k10 gold badges54 silver badges105 bronze badges















              • I think correct answer is combination of yours and Nexevis. Now I need to figure out which one to accept...

                – Pshemo
                7 hours ago











              • Since Nexevis was first to show place in specification which forces primary expression (here foo) to be evaluated regardless of its result I decided to accept his answer. Hope you don't mind.

                – Pshemo
                5 hours ago












              • @Pshemo I am absolutely fine with your decision :)

                – Andrew Tobilko
                5 hours ago

















              • I think correct answer is combination of yours and Nexevis. Now I need to figure out which one to accept...

                – Pshemo
                7 hours ago











              • Since Nexevis was first to show place in specification which forces primary expression (here foo) to be evaluated regardless of its result I decided to accept his answer. Hope you don't mind.

                – Pshemo
                5 hours ago












              • @Pshemo I am absolutely fine with your decision :)

                – Andrew Tobilko
                5 hours ago
















              I think correct answer is combination of yours and Nexevis. Now I need to figure out which one to accept...

              – Pshemo
              7 hours ago





              I think correct answer is combination of yours and Nexevis. Now I need to figure out which one to accept...

              – Pshemo
              7 hours ago













              Since Nexevis was first to show place in specification which forces primary expression (here foo) to be evaluated regardless of its result I decided to accept his answer. Hope you don't mind.

              – Pshemo
              5 hours ago






              Since Nexevis was first to show place in specification which forces primary expression (here foo) to be evaluated regardless of its result I decided to accept his answer. Hope you don't mind.

              – Pshemo
              5 hours ago














              @Pshemo I am absolutely fine with your decision :)

              – Andrew Tobilko
              5 hours ago





              @Pshemo I am absolutely fine with your decision :)

              – Andrew Tobilko
              5 hours ago











              1
















              There is value in keeping the rules as simple as possible, and “don’t use a variable that might not have been initialised” is as simple as it gets.



              More to the point, there is an established way of calling static methods - always use the class name, not a variable.



              System.out.println(Foo.x);


              The variable “foo” is unwanted overhead that should be removed, and the compiler errors and warnings could be seen as helping leading towards that.






              share|improve this answer































                1
















                There is value in keeping the rules as simple as possible, and “don’t use a variable that might not have been initialised” is as simple as it gets.



                More to the point, there is an established way of calling static methods - always use the class name, not a variable.



                System.out.println(Foo.x);


                The variable “foo” is unwanted overhead that should be removed, and the compiler errors and warnings could be seen as helping leading towards that.






                share|improve this answer





























                  1














                  1










                  1









                  There is value in keeping the rules as simple as possible, and “don’t use a variable that might not have been initialised” is as simple as it gets.



                  More to the point, there is an established way of calling static methods - always use the class name, not a variable.



                  System.out.println(Foo.x);


                  The variable “foo” is unwanted overhead that should be removed, and the compiler errors and warnings could be seen as helping leading towards that.






                  share|improve this answer















                  There is value in keeping the rules as simple as possible, and “don’t use a variable that might not have been initialised” is as simple as it gets.



                  More to the point, there is an established way of calling static methods - always use the class name, not a variable.



                  System.out.println(Foo.x);


                  The variable “foo” is unwanted overhead that should be removed, and the compiler errors and warnings could be seen as helping leading towards that.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 7 hours ago









                  Pshemo

                  99.7k17 gold badges141 silver badges199 bronze badges




                  99.7k17 gold badges141 silver badges199 bronze badges










                  answered 7 hours ago









                  racramanracraman

                  2,7281 gold badge10 silver badges10 bronze badges




                  2,7281 gold badge10 silver badges10 bronze badges































                      draft saved

                      draft discarded















































                      Thanks for contributing an answer to Stack Overflow!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid


                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.

                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function ()
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f57964026%2fwhy-cant-we-use-uninitialized-local-variable-to-access-static-content-of-its-ty%23new-answer', 'question_page');

                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      Invision Community Contents History See also References External links Navigation menuProprietaryinvisioncommunity.comIPS Community ForumsIPS Community Forumsthis blog entry"License Changes, IP.Board 3.4, and the Future""Interview -- Matt Mecham of Ibforums""CEO Invision Power Board, Matt Mecham Is a Liar, Thief!"IPB License Explanation 1.3, 1.3.1, 2.0, and 2.1ArchivedSecurity Fixes, Updates And Enhancements For IPB 1.3.1Archived"New Demo Accounts - Invision Power Services"the original"New Default Skin"the original"Invision Power Board 3.0.0 and Applications Released"the original"Archived copy"the original"Perpetual licenses being done away with""Release Notes - Invision Power Services""Introducing: IPS Community Suite 4!"Invision Community Release Notes

                      Canceling a color specificationRandomly assigning color to Graphics3D objects?Default color for Filling in Mathematica 9Coloring specific elements of sets with a prime modified order in an array plotHow to pick a color differing significantly from the colors already in a given color list?Detection of the text colorColor numbers based on their valueCan color schemes for use with ColorData include opacity specification?My dynamic color schemes

                      Ласкавець круглолистий Зміст Опис | Поширення | Галерея | Примітки | Посилання | Навігаційне меню58171138361-22960890446Bupleurum rotundifoliumEuro+Med PlantbasePlants of the World Online — Kew ScienceGermplasm Resources Information Network (GRIN)Ласкавецькн. VI : Літери Ком — Левиправивши або дописавши її