Why sometime the compound assignment operators don't work the way as expected?

The answer is not always. Let's start with the following example:

public class Program {
  public static void main(String [] args){ 
    int x = 1;
    int y = 1; 
    System.out.println(x += 1);
    System.out.println(y = y + 1); 
  }
}

output

2
2

In this case, x+=a is the same as x=x+a.

Let's look at another example:

public class Program {
  public static void main(String [] args){ 
    short x = 1; 
    short y = 1; 
    System.out.println(x += 1);
    System.out.println(y = y + 1); // compile error
  }
}

You get a compile time error at y=y+1. Why x+=1 compiles but not y=y+1? "short y =1; y = y + 1;" doesn't compile is because of the so called Binary Numeric Promotion. The operator '+' does the binary numeric promotion to its operands, using Widening Conversion to convert operands as necessary. For y = y + 1, 1 is int, so y is converted to an int, the result of y + 1 is also of type int. But y is declared as of type short, you can't assign an int to a short without explicit casting. This causes compile time error.

"short x =1; x += 1;" compiles is because that the compound-assignment operator, '+=', converts the result to the type of the left-hand variable implicitly. That is:

short x = 1;
x += 1; 

is actually

short x = 1;
x = (short)(x + 1);

You need cautiously use the compound-assignment operator, since the compiler does the implicit conversion without warning. In a Narrowing conversions case, the conversion may lose information about the overall magnitude of a numeric value and may also lose precision. Do you know what the output is from the following code?

public class Program {
  public static void main(String [] args){ 
    byte x = 100; 
    System.out.println(x += 100);
  }
}

see Xyzws Java FAQ: How do compound assignment operator work for byte primitives?

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