x += x--;
This is equivalent to:
x = x + x--;
Which is equivalent to:
int a1 = x; // a1 = 10, x = 10
int a2 = x--; // a2 = 10, x = 9
x = a1 + a2; // x = 20
So x is 20 afterwards - and that's guaranteed by the spec.
- 1) subexpressions are always evaluated left to right. Period. Evaluation of a subexpression may induce a side effect.
2) execution of operators is always done in the order indicated by parentheses, precedence and associativity. Execution of operators may induce a side effect.
The "x" to the left of the += is the leftmost subexpression, and therefore rule (1) applies. Its value is computed first -- 10.
The x-- to the right of the += is the next one in left-to-right order, so it is evaluated next. The value of x-- is 10, and the side effect is that x becomes 9. This is as it should be, because -- is of higher precedence than +=, so its side effect runs first.
Finally, the side effect of += runs last. The two operands were 10 and 10, so the result is to assign 20 to x.
I get questions about this all the time. Remember, the rules are very straightforward: subexpressions left-to-right, operators in precedence order, period.
- a difference between --x and x-- here.
If you had written x += --x;
this would be equivalent to x = x + --x;
then you would get 19.
This is because the value of x is decremented and the resulting value is used in the expression
(unlike x-- where the original value of x is used in the expression).
This expression x = x + --x + x will give 28 because the third timefourth time (see comments) x is evaluated it is 9.
http://stackoverflow.com/questions/2299437/int-x-10-x-x-in-net-why
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