In java if “char c = 'a' ” why does “c = c + 1” not compile?











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I tried to compile the following code:



public static void main(String args){

    for (char c = 'a'; c <='z'; c = c + 1) {

        System.out.println(c);

    }

}


When I try to compile, it throws:




Error:(5, 41) java: incompatible types: possible lossy conversion from
int to char




The thing is, it does work if I write c = (char)(c + 1), c += 1 or c++.



I checked and the compiler throws a similar error when I try char c = Character.MAX_VALUE + 1; but I see no way that the value of 'c' can pass 'char' type maximum in the original function.






java casting compiler-errors char primitive-types







share|improve this question




























    up vote
    2
    down vote

    favorite












    I tried to compile the following code:



    public static void main(String args){
    
    for (char c = 'a'; c <='z'; c = c + 1) {
    
        System.out.println(c);
    
    }
    
}


    When I try to compile, it throws:




    Error:(5, 41) java: incompatible types: possible lossy conversion from
    int to char




    The thing is, it does work if I write c = (char)(c + 1), c += 1 or c++.



    I checked and the compiler throws a similar error when I try char c = Character.MAX_VALUE + 1; but I see no way that the value of 'c' can pass 'char' type maximum in the original function.






    java casting compiler-errors char primitive-types







    share|improve this question


























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I tried to compile the following code:



      public static void main(String args){
      
    for (char c = 'a'; c <='z'; c = c + 1) {
      
        System.out.println(c);
      
    }
      
}


      When I try to compile, it throws:




      Error:(5, 41) java: incompatible types: possible lossy conversion from
      int to char




      The thing is, it does work if I write c = (char)(c + 1), c += 1 or c++.



      I checked and the compiler throws a similar error when I try char c = Character.MAX_VALUE + 1; but I see no way that the value of 'c' can pass 'char' type maximum in the original function.






      java casting compiler-errors char primitive-types







      share|improve this question















      I tried to compile the following code:



      public static void main(String args){
      
    for (char c = 'a'; c <='z'; c = c + 1) {
      
        System.out.println(c);
      
    }
      
}


      When I try to compile, it throws:




      Error:(5, 41) java: incompatible types: possible lossy conversion from
      int to char




      The thing is, it does work if I write c = (char)(c + 1), c += 1 or c++.



      I checked and the compiler throws a similar error when I try char c = Character.MAX_VALUE + 1; but I see no way that the value of 'c' can pass 'char' type maximum in the original function.






      java casting compiler-errors char primitive-types




      java casting compiler-errors char primitive-types






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 12 at 11:01









      Mickael

      2,90521528




      2,90521528










      asked Nov 12 at 10:54









      רעי וייס ליפשיץ

      165




      165
























          2 Answers
          2






          active

          oldest

          votes

















          up vote
          8
          down vote



          accepted










          c + 1 is an int, as the operands undergo binary numeric promotion:





          • c is a char


          • 1 is an int


          so c has to be widened to int to make it compatible for addition; and the result of the expression is of type int.



          As for the things that "work":





          • c = (char)(c + 1) is explicitly casting the expression to char, so its value is compatible with the variable's type;


          • c += 1 is equivalent to c = (char) ((c) + (1)), so it's basically the same as the previous one.


          • c++ is of type char, so no cast is required.






          share|improve this answer























          • Good answer, only suggestion would be to mention why you can set an int to a char, but not a char to an int (possible lossy conversion due to assigning a bigger data type to a smaller data type)
            – Eamon Scullion
            Nov 12 at 11:06










          • First of all, thank you for the good answer. Second, if I understand correctly, that mean java takes 'c + 1' and instead of casting '1' to a char it casts 'c' to an int in order to add them together. Is this because converting an int to char has a potential of losing data?
            – רעי וייס ליפשיץ
            Nov 12 at 11:23






          • 1




            @רעיוייסליפשיץ yes. Note also that even if you add two chars together, they are still widened to int first. (Same is true of byte and short).
            – Andy Turner
            Nov 12 at 12:13










          • @רעיוייסליפשיץ "Note also that even if you add two chars together" so c + (char) 1 wouldn't have worked.
            – Andy Turner
            Nov 12 at 13:02












          • Thanks, but that leaves me with a question, why does char c = 97 + 1 work? wouldn't java see 97 + 1 as an int?
            – רעי וייס ליפשיץ
            Nov 12 at 22:41


















          up vote
          0
          down vote













          First you are declaring c as char than you are using it as an int






          share|improve this answer





















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            8
            down vote



            accepted










            c + 1 is an int, as the operands undergo binary numeric promotion:





            • c is a char


            • 1 is an int


            so c has to be widened to int to make it compatible for addition; and the result of the expression is of type int.



            As for the things that "work":





            • c = (char)(c + 1) is explicitly casting the expression to char, so its value is compatible with the variable's type;


            • c += 1 is equivalent to c = (char) ((c) + (1)), so it's basically the same as the previous one.


            • c++ is of type char, so no cast is required.






            share|improve this answer























            • Good answer, only suggestion would be to mention why you can set an int to a char, but not a char to an int (possible lossy conversion due to assigning a bigger data type to a smaller data type)
              – Eamon Scullion
              Nov 12 at 11:06










            • First of all, thank you for the good answer. Second, if I understand correctly, that mean java takes 'c + 1' and instead of casting '1' to a char it casts 'c' to an int in order to add them together. Is this because converting an int to char has a potential of losing data?
              – רעי וייס ליפשיץ
              Nov 12 at 11:23






            • 1




              @רעיוייסליפשיץ yes. Note also that even if you add two chars together, they are still widened to int first. (Same is true of byte and short).
              – Andy Turner
              Nov 12 at 12:13










            • @רעיוייסליפשיץ "Note also that even if you add two chars together" so c + (char) 1 wouldn't have worked.
              – Andy Turner
              Nov 12 at 13:02












            • Thanks, but that leaves me with a question, why does char c = 97 + 1 work? wouldn't java see 97 + 1 as an int?
              – רעי וייס ליפשיץ
              Nov 12 at 22:41















            up vote
            8
            down vote



            accepted










            c + 1 is an int, as the operands undergo binary numeric promotion:





            • c is a char


            • 1 is an int


            so c has to be widened to int to make it compatible for addition; and the result of the expression is of type int.



            As for the things that "work":





            • c = (char)(c + 1) is explicitly casting the expression to char, so its value is compatible with the variable's type;


            • c += 1 is equivalent to c = (char) ((c) + (1)), so it's basically the same as the previous one.


            • c++ is of type char, so no cast is required.






            share|improve this answer























            • Good answer, only suggestion would be to mention why you can set an int to a char, but not a char to an int (possible lossy conversion due to assigning a bigger data type to a smaller data type)
              – Eamon Scullion
              Nov 12 at 11:06










            • First of all, thank you for the good answer. Second, if I understand correctly, that mean java takes 'c + 1' and instead of casting '1' to a char it casts 'c' to an int in order to add them together. Is this because converting an int to char has a potential of losing data?
              – רעי וייס ליפשיץ
              Nov 12 at 11:23






            • 1




              @רעיוייסליפשיץ yes. Note also that even if you add two chars together, they are still widened to int first. (Same is true of byte and short).
              – Andy Turner
              Nov 12 at 12:13










            • @רעיוייסליפשיץ "Note also that even if you add two chars together" so c + (char) 1 wouldn't have worked.
              – Andy Turner
              Nov 12 at 13:02












            • Thanks, but that leaves me with a question, why does char c = 97 + 1 work? wouldn't java see 97 + 1 as an int?
              – רעי וייס ליפשיץ
              Nov 12 at 22:41













            up vote
            8
            down vote



            accepted







            up vote
            8
            down vote



            accepted






            c + 1 is an int, as the operands undergo binary numeric promotion:





            • c is a char


            • 1 is an int


            so c has to be widened to int to make it compatible for addition; and the result of the expression is of type int.



            As for the things that "work":





            • c = (char)(c + 1) is explicitly casting the expression to char, so its value is compatible with the variable's type;


            • c += 1 is equivalent to c = (char) ((c) + (1)), so it's basically the same as the previous one.


            • c++ is of type char, so no cast is required.






            share|improve this answer














            c + 1 is an int, as the operands undergo binary numeric promotion:





            • c is a char


            • 1 is an int


            so c has to be widened to int to make it compatible for addition; and the result of the expression is of type int.



            As for the things that "work":





            • c = (char)(c + 1) is explicitly casting the expression to char, so its value is compatible with the variable's type;


            • c += 1 is equivalent to c = (char) ((c) + (1)), so it's basically the same as the previous one.


            • c++ is of type char, so no cast is required.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Nov 12 at 11:04

























            answered Nov 12 at 10:56









            Andy Turner

            80k879133




            80k879133












            • Good answer, only suggestion would be to mention why you can set an int to a char, but not a char to an int (possible lossy conversion due to assigning a bigger data type to a smaller data type)
              – Eamon Scullion
              Nov 12 at 11:06










            • First of all, thank you for the good answer. Second, if I understand correctly, that mean java takes 'c + 1' and instead of casting '1' to a char it casts 'c' to an int in order to add them together. Is this because converting an int to char has a potential of losing data?
              – רעי וייס ליפשיץ
              Nov 12 at 11:23






            • 1




              @רעיוייסליפשיץ yes. Note also that even if you add two chars together, they are still widened to int first. (Same is true of byte and short).
              – Andy Turner
              Nov 12 at 12:13










            • @רעיוייסליפשיץ "Note also that even if you add two chars together" so c + (char) 1 wouldn't have worked.
              – Andy Turner
              Nov 12 at 13:02












            • Thanks, but that leaves me with a question, why does char c = 97 + 1 work? wouldn't java see 97 + 1 as an int?
              – רעי וייס ליפשיץ
              Nov 12 at 22:41


















            • Good answer, only suggestion would be to mention why you can set an int to a char, but not a char to an int (possible lossy conversion due to assigning a bigger data type to a smaller data type)
              – Eamon Scullion
              Nov 12 at 11:06










            • First of all, thank you for the good answer. Second, if I understand correctly, that mean java takes 'c + 1' and instead of casting '1' to a char it casts 'c' to an int in order to add them together. Is this because converting an int to char has a potential of losing data?
              – רעי וייס ליפשיץ
              Nov 12 at 11:23






            • 1




              @רעיוייסליפשיץ yes. Note also that even if you add two chars together, they are still widened to int first. (Same is true of byte and short).
              – Andy Turner
              Nov 12 at 12:13










            • @רעיוייסליפשיץ "Note also that even if you add two chars together" so c + (char) 1 wouldn't have worked.
              – Andy Turner
              Nov 12 at 13:02












            • Thanks, but that leaves me with a question, why does char c = 97 + 1 work? wouldn't java see 97 + 1 as an int?
              – רעי וייס ליפשיץ
              Nov 12 at 22:41
















            Good answer, only suggestion would be to mention why you can set an int to a char, but not a char to an int (possible lossy conversion due to assigning a bigger data type to a smaller data type)
            – Eamon Scullion
            Nov 12 at 11:06




            Good answer, only suggestion would be to mention why you can set an int to a char, but not a char to an int (possible lossy conversion due to assigning a bigger data type to a smaller data type)
            – Eamon Scullion
            Nov 12 at 11:06












            First of all, thank you for the good answer. Second, if I understand correctly, that mean java takes 'c + 1' and instead of casting '1' to a char it casts 'c' to an int in order to add them together. Is this because converting an int to char has a potential of losing data?
            – רעי וייס ליפשיץ
            Nov 12 at 11:23




            First of all, thank you for the good answer. Second, if I understand correctly, that mean java takes 'c + 1' and instead of casting '1' to a char it casts 'c' to an int in order to add them together. Is this because converting an int to char has a potential of losing data?
            – רעי וייס ליפשיץ
            Nov 12 at 11:23




            1




            1




            @רעיוייסליפשיץ yes. Note also that even if you add two chars together, they are still widened to int first. (Same is true of byte and short).
            – Andy Turner
            Nov 12 at 12:13




            @רעיוייסליפשיץ yes. Note also that even if you add two chars together, they are still widened to int first. (Same is true of byte and short).
            – Andy Turner
            Nov 12 at 12:13












            @רעיוייסליפשיץ "Note also that even if you add two chars together" so c + (char) 1 wouldn't have worked.
            – Andy Turner
            Nov 12 at 13:02






            @רעיוייסליפשיץ "Note also that even if you add two chars together" so c + (char) 1 wouldn't have worked.
            – Andy Turner
            Nov 12 at 13:02














            Thanks, but that leaves me with a question, why does char c = 97 + 1 work? wouldn't java see 97 + 1 as an int?
            – רעי וייס ליפשיץ
            Nov 12 at 22:41




            Thanks, but that leaves me with a question, why does char c = 97 + 1 work? wouldn't java see 97 + 1 as an int?
            – רעי וייס ליפשיץ
            Nov 12 at 22:41












            up vote
            0
            down vote













            First you are declaring c as char than you are using it as an int






            share|improve this answer

























              up vote
              0
              down vote













              First you are declaring c as char than you are using it as an int






              share|improve this answer























                up vote
                0
                down vote










                up vote
                0
                down vote









                First you are declaring c as char than you are using it as an int






                share|improve this answer












                First you are declaring c as char than you are using it as an int







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 12 at 10:57









                Brian

                527




                527






























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