problemTo this day, I always thought: i + = j is equivalent to i = i + j; But suppose there is: int i = 5;long J = 8; i = i + j cannot compile, but i + = J can be compiled. This suggests that there is a difference between the two. Does this mean that i + = j, which is actually equivalent to i= (type of i) (i + j)?
Essence Answer:This question, in fact, has been answered in the official documentation.please look here. §15.26.2 Compound Assignment Operators. And then copy the official documentation. For compound assignment expressions, E1 op= E2 (such as i + = J;I-=J, etc.) are actually equivalent to E1 = (t) ((E1) op (E2)), where T is the type of the element E1. For example, the following code
Shortx= 3;x+= 4.6;
Equals
Shortx= 3;x= ( Short)(x+ 4.6);
StackOverflow Link Http://stackoverflow.com/questions/8710619/java-operator
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"StackOverflow Good question" Java + = operator Essence
