int i = 0;
boolean t = true;
boolean f = false, b;
b = (t || ((i++) == 0));
b = (f || ((i+=2) > 0));
System.out.println(i);
After the above code is executed, the print result is 2, not 3, why?
I find "i" was 0 not 1 after "b = (t || ((i++) == 0))" executed by debuging.Well,I'm confused why "i++" not changes "i".
Well,I'm confused why "i++" not changes "i".
Because i++ doesn't execute in the code you've provided.
In an expression of the form a || b, first a is evaluated, and if it's true (which it is in this case), the expression b isn't evaluated. This is known as short-circuiting.
This is described in JLS section 15.24.
If you change the code to use the non-short-circuited | operator instead, like this:
b = (t | ((i++) == 0));
... then it will evaluate both operands regardless.