Why is x == (x = y) not the same as (x = y) == x?











up vote
15
down vote

favorite
2












Consider the following example:



class Quirky {

    public static void main(String args) {

        int x = 1;

        int y = 3;



        System.out.println(x == (x = y)); // false

        x = 1; // reset

        System.out.println((x = y) == x); // true

     }

}


I'm not sure if there is an item in the Java Language Specification that dictates loading the previous value of a variable for comparison with the right side (x = y) which, by the order implied by brackets, should be calculated first.



Why does the first expression evaluate to false, but the second evaluate to true? I would have expected (x = y) to be evaluated first, and then it would compare x with itself (3) and return true.






java variable-assignment operator-precedence jls







share|improve this question




















  • 14




    The left hand side is always evaluated before the right hand side. The brackets don't make a difference to that.
    – Louis Wasserman
    10 hours ago












  • @LouisWasserman The term evaluate is, in my opinion, inapplicable here because x does not need to be evaluated, it just loaded from memory.
    – John McClane
    10 hours ago








  • 3




    Evaluating the expression x = y is certainly relevant, and causes the side effect that x is set to the value of y.
    – Louis Wasserman
    10 hours ago






  • 8




    Do yourself and your teammates a favor and don't mix state mutation into the same line as state examination. Doing so drastically reduces the readability of your code. (There are some cases where it's absolutely necessary because of atomicity requirements, but functions for those already exist and their purpose would be instantly recognized.)
    – jpmc26
    7 hours ago








  • 1




    The real question is why you want to write code like this.
    – Broman
    2 hours ago















up vote
15
down vote

favorite
2












Consider the following example:



class Quirky {

    public static void main(String args) {

        int x = 1;

        int y = 3;



        System.out.println(x == (x = y)); // false

        x = 1; // reset

        System.out.println((x = y) == x); // true

     }

}


I'm not sure if there is an item in the Java Language Specification that dictates loading the previous value of a variable for comparison with the right side (x = y) which, by the order implied by brackets, should be calculated first.



Why does the first expression evaluate to false, but the second evaluate to true? I would have expected (x = y) to be evaluated first, and then it would compare x with itself (3) and return true.






java variable-assignment operator-precedence jls







share|improve this question




















  • 14




    The left hand side is always evaluated before the right hand side. The brackets don't make a difference to that.
    – Louis Wasserman
    10 hours ago












  • @LouisWasserman The term evaluate is, in my opinion, inapplicable here because x does not need to be evaluated, it just loaded from memory.
    – John McClane
    10 hours ago








  • 3




    Evaluating the expression x = y is certainly relevant, and causes the side effect that x is set to the value of y.
    – Louis Wasserman
    10 hours ago






  • 8




    Do yourself and your teammates a favor and don't mix state mutation into the same line as state examination. Doing so drastically reduces the readability of your code. (There are some cases where it's absolutely necessary because of atomicity requirements, but functions for those already exist and their purpose would be instantly recognized.)
    – jpmc26
    7 hours ago








  • 1




    The real question is why you want to write code like this.
    – Broman
    2 hours ago













up vote
15
down vote

favorite
2









up vote
15
down vote

favorite
2






2





Consider the following example:



class Quirky {

    public static void main(String args) {

        int x = 1;

        int y = 3;



        System.out.println(x == (x = y)); // false

        x = 1; // reset

        System.out.println((x = y) == x); // true

     }

}


I'm not sure if there is an item in the Java Language Specification that dictates loading the previous value of a variable for comparison with the right side (x = y) which, by the order implied by brackets, should be calculated first.



Why does the first expression evaluate to false, but the second evaluate to true? I would have expected (x = y) to be evaluated first, and then it would compare x with itself (3) and return true.






java variable-assignment operator-precedence jls







share|improve this question















Consider the following example:



class Quirky {

    public static void main(String args) {

        int x = 1;

        int y = 3;



        System.out.println(x == (x = y)); // false

        x = 1; // reset

        System.out.println((x = y) == x); // true

     }

}


I'm not sure if there is an item in the Java Language Specification that dictates loading the previous value of a variable for comparison with the right side (x = y) which, by the order implied by brackets, should be calculated first.



Why does the first expression evaluate to false, but the second evaluate to true? I would have expected (x = y) to be evaluated first, and then it would compare x with itself (3) and return true.






java variable-assignment operator-precedence jls




java variable-assignment operator-precedence jls






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 2 hours ago









Boann

36.6k1287121




36.6k1287121










asked 10 hours ago









John McClane

452114




452114








  • 14




    The left hand side is always evaluated before the right hand side. The brackets don't make a difference to that.
    – Louis Wasserman
    10 hours ago












  • @LouisWasserman The term evaluate is, in my opinion, inapplicable here because x does not need to be evaluated, it just loaded from memory.
    – John McClane
    10 hours ago








  • 3




    Evaluating the expression x = y is certainly relevant, and causes the side effect that x is set to the value of y.
    – Louis Wasserman
    10 hours ago






  • 8




    Do yourself and your teammates a favor and don't mix state mutation into the same line as state examination. Doing so drastically reduces the readability of your code. (There are some cases where it's absolutely necessary because of atomicity requirements, but functions for those already exist and their purpose would be instantly recognized.)
    – jpmc26
    7 hours ago








  • 1




    The real question is why you want to write code like this.
    – Broman
    2 hours ago














  • 14




    The left hand side is always evaluated before the right hand side. The brackets don't make a difference to that.
    – Louis Wasserman
    10 hours ago












  • @LouisWasserman The term evaluate is, in my opinion, inapplicable here because x does not need to be evaluated, it just loaded from memory.
    – John McClane
    10 hours ago








  • 3




    Evaluating the expression x = y is certainly relevant, and causes the side effect that x is set to the value of y.
    – Louis Wasserman
    10 hours ago






  • 8




    Do yourself and your teammates a favor and don't mix state mutation into the same line as state examination. Doing so drastically reduces the readability of your code. (There are some cases where it's absolutely necessary because of atomicity requirements, but functions for those already exist and their purpose would be instantly recognized.)
    – jpmc26
    7 hours ago








  • 1




    The real question is why you want to write code like this.
    – Broman
    2 hours ago








14




14




The left hand side is always evaluated before the right hand side. The brackets don't make a difference to that.
– Louis Wasserman
10 hours ago






The left hand side is always evaluated before the right hand side. The brackets don't make a difference to that.
– Louis Wasserman
10 hours ago














@LouisWasserman The term evaluate is, in my opinion, inapplicable here because x does not need to be evaluated, it just loaded from memory.
– John McClane
10 hours ago






@LouisWasserman The term evaluate is, in my opinion, inapplicable here because x does not need to be evaluated, it just loaded from memory.
– John McClane
10 hours ago






3




3




Evaluating the expression x = y is certainly relevant, and causes the side effect that x is set to the value of y.
– Louis Wasserman
10 hours ago




Evaluating the expression x = y is certainly relevant, and causes the side effect that x is set to the value of y.
– Louis Wasserman
10 hours ago




8




8




Do yourself and your teammates a favor and don't mix state mutation into the same line as state examination. Doing so drastically reduces the readability of your code. (There are some cases where it's absolutely necessary because of atomicity requirements, but functions for those already exist and their purpose would be instantly recognized.)
– jpmc26
7 hours ago






Do yourself and your teammates a favor and don't mix state mutation into the same line as state examination. Doing so drastically reduces the readability of your code. (There are some cases where it's absolutely necessary because of atomicity requirements, but functions for those already exist and their purpose would be instantly recognized.)
– jpmc26
7 hours ago






1




1




The real question is why you want to write code like this.
– Broman
2 hours ago




The real question is why you want to write code like this.
– Broman
2 hours ago












3 Answers
3






active

oldest

votes

















up vote
26
down vote













== is a binary equality operator.




The left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated.



Java 11 Specification > Evaluation Order > Evaluate Left-Hand Operand First







share|improve this answer



















  • 1




    @JohnMcClane "by the order implied by brackets" - the parentheses don't set any order, they form an operand for ==. Without the parentheses, neither of these expressions would be syntactically correct.
    – Andrew Tobilko
    8 hours ago




















up vote
15
down vote













As LouisWasserman sad, the expression is evaluated left to right. And java doesn't care what "evaluate" actually does, it only cares about generating a (non volatile, final) value to work with.



//the example values

x = 1;

y = 2;


So in quickReplaceAndCompare the following is done:



x == (x = y)

1 == (x = y)

1 == (x = 2) //assign 2 to x, returns 2

1 == 2

false


and in badReplaceAndCompare():



(x = y) == x

(x = 2) == x //assign 2 to x, returns 2

2 == x

2 == 2

true


note that badReplaceAndCompare() will always return true, because you are effectively comparing the assignment of a value to the variable it is assigned to, and a = b and b will, evaluated in that order, always be the same by definition.






share|improve this answer




























    up vote
    0
    down vote













    In the first test you're checking does 1 == 3.



    In the second test your checking does 3 == 3.



    (x = y) assigns the value and that value is tested. In the former example x = 1 first then x is assigned 3. Does 1 == 3?



    In the latter, x is assigned 3, and obviously it's still 3. Does 3 == 3?






    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',
      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/3.0/"u003ecc by-sa 3.0 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%2f53749841%2fwhy-is-x-x-y-not-the-same-as-x-y-x%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








      up vote
      26
      down vote













      == is a binary equality operator.




      The left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated.



      Java 11 Specification > Evaluation Order > Evaluate Left-Hand Operand First







      share|improve this answer



















      • 1




        @JohnMcClane "by the order implied by brackets" - the parentheses don't set any order, they form an operand for ==. Without the parentheses, neither of these expressions would be syntactically correct.
        – Andrew Tobilko
        8 hours ago

















      up vote
      26
      down vote













      == is a binary equality operator.




      The left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated.



      Java 11 Specification > Evaluation Order > Evaluate Left-Hand Operand First







      share|improve this answer



















      • 1




        @JohnMcClane "by the order implied by brackets" - the parentheses don't set any order, they form an operand for ==. Without the parentheses, neither of these expressions would be syntactically correct.
        – Andrew Tobilko
        8 hours ago















      up vote
      26
      down vote










      up vote
      26
      down vote









      == is a binary equality operator.




      The left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated.



      Java 11 Specification > Evaluation Order > Evaluate Left-Hand Operand First







      share|improve this answer














      == is a binary equality operator.




      The left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated.



      Java 11 Specification > Evaluation Order > Evaluate Left-Hand Operand First








      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 9 hours ago

























      answered 9 hours ago









      Andrew Tobilko

      24.4k84182




      24.4k84182








      • 1




        @JohnMcClane "by the order implied by brackets" - the parentheses don't set any order, they form an operand for ==. Without the parentheses, neither of these expressions would be syntactically correct.
        – Andrew Tobilko
        8 hours ago
















      • 1




        @JohnMcClane "by the order implied by brackets" - the parentheses don't set any order, they form an operand for ==. Without the parentheses, neither of these expressions would be syntactically correct.
        – Andrew Tobilko
        8 hours ago










      1




      1




      @JohnMcClane "by the order implied by brackets" - the parentheses don't set any order, they form an operand for ==. Without the parentheses, neither of these expressions would be syntactically correct.
      – Andrew Tobilko
      8 hours ago






      @JohnMcClane "by the order implied by brackets" - the parentheses don't set any order, they form an operand for ==. Without the parentheses, neither of these expressions would be syntactically correct.
      – Andrew Tobilko
      8 hours ago














      up vote
      15
      down vote













      As LouisWasserman sad, the expression is evaluated left to right. And java doesn't care what "evaluate" actually does, it only cares about generating a (non volatile, final) value to work with.



      //the example values
      
x = 1;
      
y = 2;


      So in quickReplaceAndCompare the following is done:



      x == (x = y)
      
1 == (x = y)
      
1 == (x = 2) //assign 2 to x, returns 2
      
1 == 2
      
false


      and in badReplaceAndCompare():



      (x = y) == x
      
(x = 2) == x //assign 2 to x, returns 2
      
2 == x
      
2 == 2
      
true


      note that badReplaceAndCompare() will always return true, because you are effectively comparing the assignment of a value to the variable it is assigned to, and a = b and b will, evaluated in that order, always be the same by definition.






      share|improve this answer

























        up vote
        15
        down vote













        As LouisWasserman sad, the expression is evaluated left to right. And java doesn't care what "evaluate" actually does, it only cares about generating a (non volatile, final) value to work with.



        //the example values
        
x = 1;
        
y = 2;


        So in quickReplaceAndCompare the following is done:



        x == (x = y)
        
1 == (x = y)
        
1 == (x = 2) //assign 2 to x, returns 2
        
1 == 2
        
false


        and in badReplaceAndCompare():



        (x = y) == x
        
(x = 2) == x //assign 2 to x, returns 2
        
2 == x
        
2 == 2
        
true


        note that badReplaceAndCompare() will always return true, because you are effectively comparing the assignment of a value to the variable it is assigned to, and a = b and b will, evaluated in that order, always be the same by definition.






        share|improve this answer























          up vote
          15
          down vote










          up vote
          15
          down vote









          As LouisWasserman sad, the expression is evaluated left to right. And java doesn't care what "evaluate" actually does, it only cares about generating a (non volatile, final) value to work with.



          //the example values
          
x = 1;
          
y = 2;


          So in quickReplaceAndCompare the following is done:



          x == (x = y)
          
1 == (x = y)
          
1 == (x = 2) //assign 2 to x, returns 2
          
1 == 2
          
false


          and in badReplaceAndCompare():



          (x = y) == x
          
(x = 2) == x //assign 2 to x, returns 2
          
2 == x
          
2 == 2
          
true


          note that badReplaceAndCompare() will always return true, because you are effectively comparing the assignment of a value to the variable it is assigned to, and a = b and b will, evaluated in that order, always be the same by definition.






          share|improve this answer












          As LouisWasserman sad, the expression is evaluated left to right. And java doesn't care what "evaluate" actually does, it only cares about generating a (non volatile, final) value to work with.



          //the example values
          
x = 1;
          
y = 2;


          So in quickReplaceAndCompare the following is done:



          x == (x = y)
          
1 == (x = y)
          
1 == (x = 2) //assign 2 to x, returns 2
          
1 == 2
          
false


          and in badReplaceAndCompare():



          (x = y) == x
          
(x = 2) == x //assign 2 to x, returns 2
          
2 == x
          
2 == 2
          
true


          note that badReplaceAndCompare() will always return true, because you are effectively comparing the assignment of a value to the variable it is assigned to, and a = b and b will, evaluated in that order, always be the same by definition.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 9 hours ago









          Poohl

          33319




          33319






















              up vote
              0
              down vote













              In the first test you're checking does 1 == 3.



              In the second test your checking does 3 == 3.



              (x = y) assigns the value and that value is tested. In the former example x = 1 first then x is assigned 3. Does 1 == 3?



              In the latter, x is assigned 3, and obviously it's still 3. Does 3 == 3?






              share|improve this answer

























                up vote
                0
                down vote













                In the first test you're checking does 1 == 3.



                In the second test your checking does 3 == 3.



                (x = y) assigns the value and that value is tested. In the former example x = 1 first then x is assigned 3. Does 1 == 3?



                In the latter, x is assigned 3, and obviously it's still 3. Does 3 == 3?






                share|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  In the first test you're checking does 1 == 3.



                  In the second test your checking does 3 == 3.



                  (x = y) assigns the value and that value is tested. In the former example x = 1 first then x is assigned 3. Does 1 == 3?



                  In the latter, x is assigned 3, and obviously it's still 3. Does 3 == 3?






                  share|improve this answer












                  In the first test you're checking does 1 == 3.



                  In the second test your checking does 3 == 3.



                  (x = y) assigns the value and that value is tested. In the former example x = 1 first then x is assigned 3. Does 1 == 3?



                  In the latter, x is assigned 3, and obviously it's still 3. Does 3 == 3?







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 1 hour ago









                  Michael Puckett II

                  3,67541436




                  3,67541436






























                      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.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • 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%2f53749841%2fwhy-is-x-x-y-not-the-same-as-x-y-x%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

                      數位音樂下載

                      Image
                      數位音樂下載 為非以「實體」方式來販賣音樂的相關產品，而使用數位格式如mp3、AAC等文件格式來進行銷售販賣的一种音乐产业形式。著名的數位音樂下載服務商有iTunes Store、RecoChoku、Google Play音乐和微软Zune等。数字音乐下载在二十一世纪初一度风靡全球， [1] [2] 使得传统唱片产业走向衰落，但其自身也在2010年代流媒体音乐服务如Apple Music、Spotify等的挤压下迅速失去市场份额。 [3] [4] 数字音乐下载有便于携带、传播迅速、制作简单成本低等优点，与流媒体音乐相比也能更好保障音乐人权益； [5] 但它同时很可能带来侵犯版权和滥用的情形， [6] 也给音乐产业带来了诸多不利影响。数字音乐下载为音乐人和乐迷提供了更加便捷直接的交流渠道，也一定程度上推动了网络音乐人的出现和独立音乐的迅速发展。 [7] 目录 1 历史 1.1 技术积淀（1987-2000） 1.2 出现与兴起（2000-2004） 1.3 唱片市场的颠覆者（2004-2007） 1.4 巅峰和衰落（2005-2013） 1.5 摇摇欲坠的现状与不可预知的明天（2014至今） 2 主要文件格式 2.1 PCM编码格式 2.1.1 WAV 2.1.2 APE 2.1.3 FLAC 2.2 苹果推出的无损文件格式 2.2.1 AIFF 2.2.2 ALAC 2.3 有损压缩格式 2.3.1 MP3 2.3.2 AAC 3 主要市场 3.1 iTunes Store 3.2 Google Play音乐商店 3.3 亚马逊音乐商店 3.4 部分其他音乐商店 4 影响与评价 4.1 对音乐产业的影响 4.2 对科技行业的影响 4.3 正面评价 4.4 负面评价 4.4.1 侵权问题 4.4.2 经营模式不明晰 4.4.3 不适应移动时代 4.5 文化现象 5 延伸阅读 6 参考资料 历史 苹果iTunes+iPod的软硬件搭配让数字音...
                      Read more

                      When can things happen in Etherscan, such as the picture below?

                      Image
                      2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  ...
                      Read more

                      格利澤436b

                      Image
                      格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP...
                      Read more