I'm trying to get deeper with post and pre incrementors but am a bit stuck with the following expression :
public static void main(String[] args) {
int i = 0;
i = i+=(++i + (i+=2 + --i) - ++i);
// i = 0 + (++i + (i+=2 + --i) - ++i);
// i = 0 + (1 + (3 + 2) - 1);
// i = 0 + (6 - 1);
System.out.println(i); // Prints 0 instead of 5
}
I know I'm missing the logic somewhere but where?
What I've tried :
- Going from left to right (though I know it is not recommended)
- Going from the insidest bracket and starting from there.
Thanks for the help
PS : The comments are the details of my calculus
EDIT 1
I tried to change de hard coded value from the expression from 2 to something else and the result always gives 0
Look at this example :
int i = 0;
i = i+=(++i + (i+=32500 + --i) - ++i);
System.out.println(i); // Prints 0
This expression should logically be nowhere near 0 but somehow it does print it.
The same happens when I use a negative :
int i = 0;
i = i+=(++i + (i+=(-32650) + --i) - ++i);
System.out.println(i); // Prints 0
EDIT 2
Now, I changed the value of i to begin with :
int i = 1;
i = i+=(++i + (i+=2 + --i) - ++i);
System.out.println(i); // Prints 2
i = 2;
i = i+=(++i + (i+=10000 + --i) - ++i);
System.out.println(i); // Prints 4
i = 3;
i = i+=(++i + (i+=(-32650) + --i) - ++i);
System.out.println(i); // Prints 6
It gives the double of i each time, whatever the hard coded value is.
Quoting Java Language Specification, 15.7 Evaluation Order:
The Java programming language guarantees that the operands of operators appear to be evaluated in a specific evaluation order, namely, from left to right.
The left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated.
If the operator is a compound-assignment operator (§15.26.2), then evaluation of the left-hand operand includes both remembering the variable that the left-hand operand denotes and fetching and saving that variable's value for use in the implied binary operation.
So, essentially, i += ++i will remember the old value of i on the left side, before evaluating the right side.
Remember, evaluation order of operands and precedence of operators are two different things.
Showing evaluation order, step by step, with saved value in {braces}:
int i = 0;
i = i += (++i + (i += 2 + --i) - ++i); // i = 0
i{0} = i += (++i + (i += 2 + --i) - ++i); // i = 0
i{0} = i{0} += (++i + (i += 2 + --i) - ++i); // i = 0
i{0} = i{0} += (1 + (i += 2 + --i) - ++i); // i = 1
i{0} = i{0} += (1 + (i{1} += 2 + --i) - ++i); // i = 1
i{0} = i{0} += (1 + (i{1} += 2 + 0 ) - ++i); // i = 0
i{0} = i{0} += (1 + (i{1} += 2 ) - ++i); // i = 0
i{0} = i{0} += (1 + 3 - ++i); // i = 3
i{0} = i{0} += (4 - ++i); // i = 3
i{0} = i{0} += (4 - 4 ); // i = 4
i{0} = i{0} += 0 ; // i = 4
i{0} = 0 ; // i = 0
0 ; // i = 0
Followup to edits to question
If we name the initial value I and the constant N:
int i = I;
i = i += (++i + (i += N + --i) - ++i);
Then we can see that the values are:
i{I} = i{I} += ((I+1) + (i{I+1} += N + I) - ((I+1+N+I)+1));
i{I} = i{I} += (I + 1 + (I + 1 + N + I) - (I + 1 + N + I + 1));
i{I} = i{I} += (I + 1 + I + 1 + N + I - I - 1 - N - I - 1);
i{I} = i{I} += I;
i{I} = I + I;
i = 2 * I;
This is the logic taking into account your first edit (with an unknown X):
public static void main(String[] args) {
int i = 0;
i = i+=(++i + (i+=X + --i) - ++i);
// i = 0 += (++i + ((i += (X + --i)) - ++i));
// i = 0 += (1 + ((i += (X + --i)) - ++i)); // i = 1
// i = 0 += (1 + ((1 += (X + --i)) - ++i)); // i = 1 and i will then take the result of 1 += (X + --i)
// i = 0 += (1 + ((1 += (X + 0)) - ++i)); // i = 0 and i will then take the result of 1 += (X + 0)
// i = 0 += (1 + (X + 1 - ++i)); // i = X + 1
// i = 0 += (1 + (X + 1 - X - 2)); // i = X + 2
// i = 0 += (0); // i = X + 2
// i = 0;
System.out.println(i); // Prints 0
}
Tricks here:
+=is an assignement operator so it is right-associative: in the snippets, I added parenthesis to express this more clearly- The result of the assignment expression is the value of the variable after the assignment has occurred
- The postfix increment operator
++and postfix decrement operator--add or subtract 1 from the value and the result is stored back into the variable. - The
+additive operator first computes the left-hand operand then the right-hand operand.
For your second edit (with an unknown I added):
public static void main(String[] args) {
int i = I;
i = i+=(++i + (i+=X + --i) - ++i);
// i = I += (++i + ((i += (X + --i)) - ++i));
// i = I += (I+1 + ((i += (X + --i)) - ++i)); // i = I+1
// i = I += (I+1 + ((I+1 += (X + --i)) - ++i)); // i = I+1 and i will then take the result of I+1 += (X + --i)
// i = I += (I+1 + ((I+1 += (X + I)) - ++i)); // i = I and i will then take the result of I+1 += (X + I)
// i = I += (I+1 + (X+2*I+1 - ++i)); // i = X + 2*I + 1
// i = I += (I+1 + (X+2*I+1 - X-2*I-2)); // i = X + 2*I + 2
// i = I += (I); // i = X + 2*I + 2
// i = 2 * I;
System.out.println(i); // Prints 2 * I
}
I suggest the following: format the code differently, so that there's only 1 statement per line, e.g.
@Test
public void test() {
int i = 0;
i =
i+=
(
++i
+
(
i+=
2
+
--i
)
-
++i
);
System.out.println(i); // Prints 0 instead of 5
}
Then run it under the debugger and press F5 ("Step into") always. This will help you to understand in which order items get evaluated:
int i=0;i=: ... (needs to wait for result of calculation A)i+=... (needs to wait B)++i: i=1i+=... (needs to wait C)2+--i: i=0- ...: i=3 (result for wait C)
-++i: i=4 and operand of - is also 4- ...: i=0 (result for wait B)
- ...: i=0 (result for wait A)
Line 10 will always make the result of line 3 0, so the initial value of i will never be changed by the whole operation.
Ok, let's break down every thing:
int i = 0; // i = 0, no big deal
Then starting into the most inner parenthesis:
(i+=2 + --i)
- it first decrement
iand use the result (-1) - then add 2 (
-1+2=1) - and add the result to i (which is 0 at the start of the operation) (
0+1=1=i)
At the end, the first decrement is ignored by the reassignation.
Next parenthesis:
i+= (++i + previous_result - ++i)
- it increase the
i( with++i) at two points - then the operation becomes
(i+1) + previous_result - (i+2)(notice how the increment are not simultaneous) that gives2 + 1 - 3 = 0. - the result of the operation is added to the initial
i(0)
Again the increment will get discard by the reassignation.
finally:
i = previous_result
Which gives 0 :)
I traced the value of i and here are the results:
i = i+=(++i + (i+=2 + --i) - ++i);
initialization: i = 0;
++i: i = 1;(1st one) and store this value
(i+=2 + --i): In it
--i: i = 0;(i was changed by the previous ++i)
i += 2 + 0: i = 2;(value of the inner (i+=2 + --i), store it)
++i: i = 3;
1 + 2 -3: i = 0;
i += 0: i = 0;
The value of the 2nd i from the left is not zero, it is the value of i after all the operations to the right are finished.
Because of the highest precedence (...) will be evaluated first then ++ & -- and then remaining operators. Your expression is like
i = i += ( (++i) + (i+=2 + (--i)) - (++i) );
//i = i = i + ( (++i) + (i+=2 + (--i)) - (++i) );
//i = i = 0 + ( (++i) + (i+=2 + (--i)) - (++i) );
//i = i = 0 + ( (1) + (i+=2 + (--i)) - (++i) );
//i = i = 0 + ( (1) + (i+=2 + (0)) - (++i) );
//i = i = 0 + ( (1) + (2 + (0)) - (++i) );
//i = i = 0 + ( (1) + (2) - (++i) );
//i = i = 0 + ( (1) + (2) - (3) );
//i = i = 0 + ( 3 - 3 );
//i = i = 0 + ( 0 );
//i = 0
Please fully parenthesize your expression. You will understand evaluation order of expression.
For your EDIT 1
i = i+=( (++i) + (i+=32500 + (--i) ) - (++i) );
// 0 + ( (++i) + (i+=32500 + (--i) ) - (++i) ); // i = 0
// 0 + ( (1) + (i+=32500 + (--i) ) - (++i) ); // i = 1
// 0 + ( (1) + (i+=32500 + (0) ) - (++i) ); // i = 0
// 0 + ( (1) + (32500 + (0) ) - (++i) ); // i = 32500
// 0 + ( (1) + (32500) - (++i) ); // i = 32500
// 0 + ( (1) + (32500) - (32501) ); // i = 32501
// 0 + ( 32501 - 32501 ); // i = 32501
// 0 // i = 0
Paranthesis takes the precedence. order would be from inner most to outermost
(i+=2 + --i) =>i=i+=2==(2) --2 =0
(++i - ++i) =>1-1=0
i=i+ =0+0 =0
Each expression evaluates to 0
来源：https://stackoverflow.com/questions/33120663/incrementor-logic