Adding int to short

 

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  • Why don't Java's +=, -=, *=, /= compound assignment operators require casting? 11 answers 
    • int i = 123456;
short x = 12;
x += i;

    is actually 

    int i = 123456;
short x = 12;
x = (short)(x + i);

    Whereas x = x + i is simply x = x + i . It does not automatically cast as a short and hence causes the error ( x + i is of type int ).

    A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T)((E1) op (E2)) , where T is the type of E1 , except that E1 is evaluated only once.

    - JLS §15.26.2

    Numbers are treated as int unless you specifically cast them otherwise. So in the second statement when you use a literal number instead of a variable, it doesn't automatically cast it to the appropriate type. 

    x = x + (short)1;

    ...should work.

    The + operator of integral types (int, short, char and byte) always returns an int as result.

    You can see that with this code: 

    //char x = 0;
//short x = 0;
//byte x = 0;
int x = 0;
x = x + x;

    It won't compile unless x is an int .

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