Why don t Java s - compound assignment operators require casting

Question

Until today  I thought that for example   i    j    Was just a shortcut for   i   i   j    But if we try this   int i   5  long j   8    Then i   i   j  will not compile but i    j  will compile fine   Does it mean that in fact i    j  is a shortcut for something like this i    type of i   i   j

User · Answer

Very good question  The Java Language specification confirms your suggestion      For example  the following code is correct   short x   3  x    4 6        and results in x having the value 7 because it is equivalent to   short x   3  x    short  x   4 6

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you need to cast from long to int explicitly in case of i   i   l  then it will compile and give correct output  like   i   i    int l    or  i    int   long i   l      this is what happens in case of      dont need  long  casting since upper casting is done implicitly    but in case of    it just works fine because the operator implicitly does the type casting from type of right variable to type of left variable so need not cast explicitly

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Yes   basically when we write  i    l     the compiler converts this to   i    int  i   l     I just checked the  class file code   Really a good thing to know

User · Answer

Java Language Specification defines E1 op  E2 to be equivalent to E1    T    E1  op  E2   where T is a type of E1 and E1 is evaluated once   That s a technical answer  but you may be wondering why that s a case  Well  let s consider the following program   public class PlusEquals       public static void main String   args            byte a   1          byte b   2          a   a   b          System out println a             What does this program print   Did you guess 3  Too bad  this program won t compile  Why  Well  it so happens that addition of bytes in Java is defined to return an int  This  I believe was because the Java Virtual Machine doesn t define byte operations to save on bytecodes  there is a limited number of those  after all   using integer operations instead is an implementation detail exposed in a language   But if a   a   b doesn t work  that would mean a    b would never work for bytes if it E1    E2 was defined to be E1   E1   E2  As the previous example shows  that would be indeed the case  As a hack to make    operator work for bytes and shorts  there is an implicit cast involved  It s not that great of a hack  but back during the Java 1 0 work  the focus was on getting the language released to begin with  Now  because of backwards compatibility  this hack introduced in Java 1 0 couldn t be removed

User · Answer

As always with these questions  the JLS holds the answer  In this case   15 26 2 Compound Assignment Operators  An extract      A compound assignment expression of the form E1 nbsp op  nbsp E2 is equivalent to E1 nbsp   nbsp  T   E1  nbsp op nbsp  E2    where T is the type of E1  except that E1 is evaluated only once    An example cited from   15 26 2           the following code is correct   short x   3  x    4 6        and results in x having the value 7 because it is equivalent to   short x   3  x    short  x   4 6      In other words  your assumption is correct

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A good example of this casting is using    or     byte b   10  b    5 7  System out println b      prints 57   or  byte b   100  b    2 5  System out println b      prints 40   or  char ch    0   ch    1 1  System out println ch      prints  4    or  char ch    A   ch    1 5  System out println ch      prints  a

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The main difference is that with a   a   b  there is no typecasting going on  and so the compiler gets angry at you for not typecasting  But with a    b  what it s really doing is typecasting b to a type compatible with a  So if you do   int a 5  long b 10  a  b  System out println a     What you re really doing is    int a 5  long b 10  a a  int b  System out println a

User · Answer

Subtle point here     There is an implicit typecast for i j when j is a double and i is an int  Java ALWAYS converts an integer into a double when there is an operation between them   To clarify i  j where i is an integer and j is a double can be described as  i    lt int gt   lt double gt i   j    See  this description of implicit casting  You might want to typecast j to  int  in this case for clarity

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Sometimes  such a question can be asked at an interview   For example  when you write   int a   2  long b   3  a   a   b    there is no automatic typecasting  In C   there will not be any error compiling the above code  but in Java you will get something like Incompatible type exception   So to avoid it  you must write your code like this   int a   2  long b   3  a    b    No compilation error or any exception due to the auto typecasting

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In Java type conversions are performed automatically when the type of the expression on the right hand side of an assignment operation can be safely promoted to the type of the variable on the left hand side of the assignment  Thus we can safely assign      byte -  short -  int -  long -  float -  double      The same will not work the other way round  For example we cannot automatically convert a long to an int because the first requires more storage than the second and consequently information may be lost  To force such a conversion we must carry out an explicit conversion  Type - Conversion

User · Answer

The problem here involves type casting   When you add int and long     The int object is casted to long  amp  both are added and you get long object  but long object cannot be implicitly casted to int  So  you have to do that explicitly    But    is coded in such a way that it does type casting  i  int  i m

[java] Why don't Java's +=, -=, *=, /= compound assignment operators require casting?

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