i have problem to step by step java expression calculation
System.out.println(++x + x++ * y-- - --y);
I know this precedence:
1. postfix unary
2. prefix unary
3. multiplicative
4. additive
but when i calculate with this precedence the result is below:
// 12 + 11 * 19 - 18
can some one help me
You can understand it from the example given below:
public class Main {
public static void main(String[] args) {
int x = 5, y = 10;
System.out.println(++x + x++ * y-- - --y);// 6 + 6 * 10 - 8 = 58
}
}
Steps in this calculation:
++x = 6
6 + x++ * y-- = 6 + 6 * 10 = 6 + 60 = 66 (after this y will become 9 because of y-- and x will become 7 because of x++ but this increased value of x has never been used in subsequent calculation)
66 - 8 = 58 (before y gets subtracted from 66, it will become 8 because of --y)
Postfix unary is applied after the variable evaluation, opposite to prefix, which is applied before evaluation, your expression can be rewrited:
int x_prefix = x + 1; // ++x
int y_prefix = y - 1; // --y
System.out.println(x_prefix + x * y - y_prefix);
int x = x + 1; // x++
int y = y - 1; // y--
You write operators precedence, it's right but every operator has own behavior, in case of postfix increment, of course has to be evaluate before others, but its behavior is return the current variable and after increment its value.
NOTE: I just rewrited your expression as is, if you use in the same expression the variable postfix incremented, the next access see the variable incremented:
int x = 1;
System.out.println(x++ + x++); // 1 + 2
System.out.println(x) // 3
For completeness:
int x = 1;
System.out.println(++x + ++x); // 2 + 3
System.out.println(x) // 3
Related
Operator precedence of ++ and -- is higher than + and has associativity right to left. So I think output of x should be (++i+4 then 5+4) 9 and output of y should be (i+++5 then 5+5) 10. For both it print 11 as output.
class First {
public static void main(String[] args) {
int i = 5;
int x = ++i + --i;
int y = i++ + i--;
System.out.println("x="+x);
System.out.println("y="+y);
}
}
Let's translate this:
int i = 5;
int x = ++i + --i;
int y = i++ + i--;
System.out.println("x=" + x);
System.out.println("y=" + y);
Into (step by step):
int i = 5;
int x = ++i + --i;
// ^
// (i = i + 1) + --i
// (i = 5 + 1) + --i
// (i = 6) + --i
// ^
// 6 + (i = i - 1)
// 6 + (i = 6 - 1)
// 6 + (i = 5)
// 6 + 5
// x = 11
int y = i++ + i--;
// ^
// (i = 5) + i--
// 5 + i--
// i = i + 1 // post evaluation, after i was evaluated to 5, now i increments its value and it is 6
// 5 + (i = 6)
// 5 + 6
// y = 11
// i = i - 1 // post evaluation, after i was evaluated to 6, now i decrements its value and it is 5
System.out.println("x=" + x); // x=11
System.out.println("y=" + y); // y=11
Section 15.7 of the JLS is clear on this:
The Java programming language guarantees that the operands of operators appear to be evaluated in a specific evaluation order, namely, from left to right.
It is recommended that code not rely crucially on this specification. Code is usually clearer when each expression contains at most one side effect, as its outermost operation, and when code does not depend on exactly which exception arises as a consequence of the left-to-right evaluation of expressions.
Whenever assignment operators and other binary operators are mixed in with increment or decrement operators, evaluating the expression is not intuitive, and one should never write such confusing code.
Even if the increment and decrement unary operators are right-to-left associative, the values of an expression are evaluated left-to-right.
int x = ++i + --i;
Here, ++i increments i to 6 and returns 6, then --i decrements i to 5 and returns 5, and the sum of 6 and 5 is 11.
int y = i++ + i--;
Here, i++ returns 5 and increments i to 6, then i-- returns 6 and decrements i to 5, and the sum of 5 and 6 is 11.
With these expressions, the right-to-left associativity does not figure into the order of operations, because associativity only controls which operations are executed first among operators of the same precedence. You have a + operator of lower precedence between them, so the associativity doesn't matter here.
Associativity dictates how operators with the same precedence are parsed.
For example, the Unary Minus (- ...) and Unary Bitwise NOT (~ ...) operators have the same precedence and have right-to-left associativity. Therefore:
int i = ~-1; // == 0
int j = -~1; // == 2
This question already has an answer here:
How are java increment statements evaluated in complex expressions
(1 answer)
Closed 8 years ago.
I know there are several question about the x++ operation, I know the difference between ++x and x++. But now I have to solve this:
int x = 5;
x += x++ * x++ * x++;
Well I know that this shouldn't be too difficult, but still, I need an explanation how this calculatino is done, step by step, I don't get it by myself..
Your code is equivalent to:
int x = 5;
int originalX = x;
int a = x++;
int b = x++;
int c = x++;
x = originalX + a * b * c;
System.out.println("x = " + x); //215
x += x++ * x++ * x++;
can be written as:
x = x+ x++ * x++ * x++;
how will it be evaluated?
x= 5+(5 * 6 * 7) because you are using postfix. So, the incremented value of x will be visible from the second time it is used.
So, final output = 5+ (5*6*7) == 215
x++ would mean read the value and use it in the place which is referenced and then increment it.
So in your question:-
int x = 5;
x = 5 + 5 * 6 * 7
x += x++ * x++ * x++;
x = 215
int x = 5;
x += x++ * x++ * x++;
First, set some brackets to better see the calculation sequence:
x += ((x++ * x++) * x++);
Then, replace the first occurance of x with it's value, calculate, and continiue replacing with updated values:
5 += ((x++ * x++) * x++);
5 += ((5 * x++) * x++);
5 += ((5 * 6) * x++);
5 += ((5 * 6) * 7);
5 += 210;
and now, it's plain math...
Result should be: 215
And my compile gives me: 215
So I think my explanation is correct. But i'm not 100% sure...
I don`t understand how Java is progressing this arithmetic expression
int x = 1;
int y = 1;
x += y += x += y;
System.out.println("x=" + x + " y=" + y);
With Java I get x = 4 and y = 3. But in C, Perl, Php I get x=5 and y = 3
On the paper I also get x = 5 and y = 3
This is the nasty part, obviously:
x += y += x += y;
This is executed as:
int originalX = x; // Used later
x = x + y; // Right-most x += y
y = y + x; // Result of "x += y" is the value stored in x
x = originalX + y; // Result of "y += x" is the value stored in y
So:
x y
(Start) 1 1
x = x + y 2 1
y = y + x 2 3
x = originalX + y 4 3
The important part is the use of originalX here. The compound assignment is treated as:
x = x + y
and the first operand of + is evaluated before the second operand... which is why it takes the original value of x, not the "latest" one.
From JLS section 15.16.2:
If the left-hand operand expression is not an array access expression, then:
First, the left-hand operand is evaluated to produce a variable. If this evaluation completes abruptly, then the assignment expression completes abruptly for the same reason; the right-hand operand is not evaluated and no assignment occurs.
Otherwise, the value of the left-hand operand is saved and then the right-hand operand is evaluated. If this evaluation completes abruptly, then the assignment expression completes abruptly for the same reason and no assignment occurs.
When in doubt, consult the language specification - and never assume that just because two languages behave differently, one of them is "wrong". So long as the actual behaviour matches the specified behaviour for each language, all is well - but it does mean you need to understand the behaviour of each language you work with, of course.
That said, you should clearly avoid horrible code like this in the first place.
it starts from right to left
1. x += y gives x = 2, y = 1
2. y += ( x +=y ) gives x = 2, y = 3
3. x += ( y += x += y ) gives x = 4, y = 3
so, it's all consistent
I have two similar questions about operator precedences in Java.
First one:
int X = 10;
System.out.println(X++ * ++X * X++); //it prints 1440
According to Oracle tutorial:
postfix (expr++, expr--) operators have higher precedence than prefix (++expr, --expr)
So, I suppose that evaluation order:
1) first postfix operator: X++
1.a) X++ "replaced" by 10
1.b) X incremented by one: 10+1=11
At this step it should look like: System.out.println(10 * ++X * X++), X = 11;
2) second POSTfix operator: X++
2.a) X++ "replaced" by 11
2.b) X incremented by one: 11+1=12
At this step it should look like: System.out.println(10 * ++X * 11), X = 12;
3) prefix operator: ++X
3.a) X incremented by one: 12+1=13
3.b) ++X "replaced" by 13
At this step it should look like: System.out.println(10 * 13 * 11), X = 13;
4) evaluating 10*13 = 130, 130*11 = 1430.
But Java seems to ignore PRE/POST ordering and puts them on one level. So the real order:
X++ -> ++X -> X++
what causes the answer to be (10 * 12 * 12) = 1440.
Second one:
Example from this question:
int a=1, b=2;
a = b + a++;
Part of accepted answer:
"By the time of assignment, ++ has already incremented the value of a to 2 (because of precedence), so = overwrites that incremented value."
OK, let's look step-by-step:
1) replacing "b" with 2
2) replacing "a++" with 1
3) incrementing "a" by 1 -> at this point a==2
4) evaluating 2+1 = 3
5) overwriting incremented value of "a" with 3
Seems everything is fine.
But let's make a little change in that code (replace "=" with "+=")
a += b + a++;
steps 1-4 should be same as above.
so, after step 4 we have something like that:
a += 3;
where a==2
And then I think: OK, a = 2+3, so a should be 5. BUT the answer is only 4
I'm really confused. I already spent couple of hours but still can't understand where I am wrong.
P.S. I know, that I shouldn't use this "style" in real applications. I just want to understand what is wrong in my thoughts.
The confusion stems from the fact that the operands are evaluated from left to right. This is done first, before any attention is paid to operator precedence/order of operations.
This behavior is specified in JLS 15.7.2. Evaluate Operands before Operation
So X++ * ++X * X++ is first evaluated as 10 * 12 * 12 which yields, as you saw, 1440.
To convince yourself of this, consider the following:
X = 10; System.out.println(X++ * ++X);
X = 10; System.out.println(++X * X++);
If X++ were done first, then ++X second, then multiplication, both should print the same number.
But they do not:
X = 10; System.out.println(X++ * ++X); // 120
X = 10; System.out.println(++X * X++); // 121
So how does this make sense? Well if we realize that operands are evaluated from left to right, then it makes perfect sense.
X = 10; System.out.println(X++ * ++X); // 120 (10 * 12)
X = 10; System.out.println(++X * X++); // 121 (11 * 11)
The first line looks like
X++ * ++X
10 (X=11) * (X=12) 12
10 * 12 = 120
and the second
++X * X++
(X=11) 11 * 11 (X=12)
11 * 11 = 121
So why are prefix and postfix increment/decrement operators in the table?
It is true that increment and decrement must be performed before multiplication. But what that is saying is that:
Y = A * B++
// Should be interpreted as
Y = A * (B++)
// and not
Y = (A * B)++
Just as
Y = A + B * C
// Should be interpreted as
Y = A + (B * C)
// and not
Y = (A + B) * C
It remains that the order of the evaluation of the operands occurs left-to-right.
If you're still not conviced:
Consider the following program:
class Test
{
public static int a(){ System.out.println("a"); return 2; }
public static int b(){ System.out.println("b"); return 3; }
public static int c(){ System.out.println("c"); return 4; }
public static void main(String[] args)
{
System.out.println(a() + b() * c());
// Lets make it even more explicit
System.out.println(a() + (b() * c()));
}
}
If the arguments were evaluated at the time they were needed, either b or c would come first, the other next, and lastly a. However, the program outputs:
a
b
c
14
a
b
c
14
Because, regardless of the order that they're needed and used in the equation, they're still evaluated left to right.
Helpful reading:
What are the rules for evaluation order in Java?
a += a++ * a++ * a++ in Java. How does it get evaluated?
Appendix A: Operator Precedence in Java
In short,
Precedence is like preparing the expression to be calculated by putting parentheses. Evaluation comes next from left to right considering each pair of parentheses as a separate operation.
For example,if i=2 then i+i++ becomes i+(i++) after precedence and evaluates to 2+2 = 4.
However, i+++i becomes (i++)+i and evaluates to 2+3 = 5.
Same to i+(i=5) which evaluates to 2+5 = 7.
In fact the postfix operators do have higher precedence than prefix operators. For example, i+++++i becomes ((i++)++)+i after precedence which gives a compile error (the second postfix operator needs a variable to operate on but a value is found instead!). If both postfix and prefix operators had had equal precedence then the expression would have become (i++)+(++i) and evaluates to 2+4 = 6.
If you need more explanation you can compile and run the following code and inspect the examples printed at the output.
public class TestPrecedence {
public static void main(String[] str) {
int i = 0;
System.out.println("\n");
i = 2; System.out.println("i = " + i + "\n");
i = 2; System.out.println("i++ = " + i++ + "\n");
i = 2; System.out.println("++i = " + ++i + "\n");
i = 2; System.out.println("i++i = (i++)i TestPrecedence.java:8: error: ')' expected\n"+
" i++i\n"+
" ^\n");
i = 2; System.out.println("i+-i = i+(-i) = " + (i+-i) + "\n");
i = 2; System.out.println("++i++ = ++(i++) TestPrecedence.java:12: error: unexpected type\n"+
" ++i++ \n"+
" ^\n"+
" required: variable\n"+
" found: value\n");
i = 2; System.out.println("i+++++i = ((i++)++)+i TestPrecedence.java:17: error: unexpected type\n"+
" i+++++i\n"+
" ^\n"+
" required: variable\n"+
" found: value\n");
i = 2; System.out.println("i++ + ++i = " + (i++ + ++i) + "\n");
i = 2; System.out.println("i+(i=3) = " + (i+(i=3)) + " evaluates left to right\n");
i = 2; System.out.println("i+i++ precedence yields i+(i++) evaluates to 2+2 = " + (i+i++) + "\n");
i = 2; System.out.println("i+++i precedence yields (i++)+i evaluates to 2+3 = " + (i+++i) + "\n");
System.out.println("\n");
}
}
The reason why its 1440 is because
x is set to 10 i.e 1st term is FIXED(overall equation 10 *)
x is incremented by 1,x =11 now
x is pre-incremented by 1 x=12 and second term FIXED now (overall equation 10 * 12 *)
now x is set to 12 and third term FIXED(overall equation 10 * 12 *12)
x is increment now but is in this case not used for evaluation,
in short a term is FIXED when variable occurs which in this case is X
2nd case:
I'm not sure but I guess can be broken as,
a=b+a
a++
which I think is what is happening.
For the second one ->
int a=1, b=2;
a += b + a++;
Compiler will convert it to
a = a + b + a++;
Then apply your logic and you will find the reason why a is 4.
First step
1) first postfix operator: X++
1.a) X++ "replaced" by 10
1.b) X incremented by one: 10+1=11
At this step it should look like: System.out.println(10 * ++X * X++), X = 11;
2) prefix operator: ++X
2.a) X "replaced" by 11
2.b) X incremented by one: 11+1=12
At this step it should look like: System.out.println(10 * 12 * X++), X = 12;
3) second POSTfix operator: X++
3.a) X "replaced" by 12
3.b) X incremented by one: 12+1=13
At this step it should look like: System.out.println(10 * 12 * 12), X = 13;
This prints 10 * 12 * 12 = 1440
This is the bytecode for the first step
1. bipush 10
2. istore_0
3. getstatic java/lang/System/out Ljava/io/PrintStream;
4. iload_0
5. iinc 0 1
6. iinc 0 1
7. iload_0
8. imul
9. iload_0
10. iinc 0 1
11. imul
12. invokevirtual java/io/PrintStream/println(I)V
13. return
This do the following:
1. Push 10 to the stack
2. Pop 10 from the stack to X variable
3. Push X variable value (10) to the stack
5. Increment local variable X (11)
6. Increment local variable X (12)
7. Push X variable value (12) to the stack
8. Multiply top and subtop (10*12) Now Top = 120
9. Push X variable value (12) to the stack
10. Increment local variable X (13)
11. Multiply top and subtop (120*12) Now Top = 1440
Notice than the last increment (10.) is done after the push of X to the stack
Sorry, I couldn't think of a better title.
Anyway, when I compile the following code:
class Example{
public static void main(String[] args) throws Exception {
int x = 3;
x -= 2 + x++ == 4 ? x++ : ++x;
System.out.println("x = " + x);
}
}
I get the answer -2.
Now I would really like to know how it comes to this answer, as I can't find it.
It would be amazing if I could get an answer to this, as this may be on a exam that I'll have somewhat soon.
Note: Of course I wouldn't ever code out something like this myself, this is just an exercise.
Thanks so much if you can help me out!
Edit: answer to fge's answer:
Okay, thanks! but now if I change it up a little and make it like so:
x = 2;
x += 3 + ++x == 6 ? x-- : x--;
System.out.println("x = " + x);
I would think that you would get the following:
++x == 6
is not true, as x is now 3. this means we get x--, making it now 2 again.
then we get 3 + 2 = 5, so the expression can be evaluated to:
x += 5
because x is now 2, we get 2 + 5 = 7.
When you create such code example
int x = 3;
System.out.println(2 + x++ == 4 ? x++ : ++x);
you will see that the output is 5. It is because it is equivalent to
(2 + x++) == 4 ? x++ : ++x
so first Java will evaluate (2+ x++). Since we have x++ value of x will be returned first creating (2+3) and then incremented, so x will be now 4 after it.
Since (2+3) == 4 is false because (5 == 4) we will execute this part ++x, first incrementing x to 5 and then returning it.
Now lets go back to your example
int x = 3;
x -= 2 + x++ == 4 ? x++ : ++x;
You may know that x -= y is in fact x = x - y so it can be written as
x = x - (2 + x++ == 4 ? x++ : ++x);
So, since Java evaluates from left to right it will be
x = 3 - (2 + x++ == 4 ? x++ : ++x);
We know now that (2 + x++ == 4 ? x++ : ++x) is in fact 5 so
x = 3 - 5; // == -2