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Now, according to this website, Unary post-increment has more precedence than Unary pre-increment. I presume it meant, that x++ is done before ++x.
However, when tried with the following expression, I've faced problems
int var1=10;
int b= --var1 + ++var1 + var1++;
Now, if var1++ executes first, then according to me,
b = --var1 + ++var1 + 10
Now, if I think it should do ++var1 before --var1, as it mentions that the Associativity is Right to Left. I suppose, it means do things on right, before the ones on the left
This, However, doesn't give the correct answer. The Correct Answer is 29, while I get 33. Why is it so ?
I presume it meant, that x++ is done before ++x.
No, it doesn't mean that. I've previously blogged about how I find viewing precedence in terms of execution order is problematic. It may well be valid in a computer-science-strict way, but it ends up confusing people.
I find it much easier to think about it as grouping. Likewise I think it's easiest to think of associativity as like precedence, but for operators of equal precedence. Evaluation order is always left-to-right.
So this:
int b = --var1 + ++var1 + var1++;
is grouped as if it were written:
int b = (--var1) + (++var1) + (var1++);
That's then equivalent to:
int b = ((--var1) + (++var1)) + (var1++);
(Due to binary + having left-to-right associativity.)
That's then effectively:
int tmp1 = --var1; // tmp1 = 9, var1 = 9
int tmp2 = ++var1; // tmp2 = 10, var1 = 10
int tmp3 = tmp1 + tmp2; // tmp3 = 19
int tmp4 = var1++; // tmp4 = 10, var1 = 11
int tmp5 = tmp3 + tmp4; // tmp5 = 29
int b = tmp5; // b = 29
... which confirms what you've observed.
Of course, you should always try to avoid having code as convoluted as this in the first place...
To demonstrate the evaluation order part, consider this expression:
int result = a() + b() * c();
Precedence means that's grouped as:
int result = a() + (b() * c());
That's equivalent to:
// Evaluate LHS of +, which is just a()
int tmp1 = a();
// Evaluate RHS of +, which is b() * c()
// First evaluate LHS of *, which is b()
int tmp2 = b();
// Next evaluate RHS of *, which is c()
int tmp3 = c();
// Now we can evaluate *:
int tmp4 = tmp2 * tmp3;
// Now we can evaluate +:
result = tmp1 + tmp4;
As we're executing methods, we can observe that execution order really easily:
public class Test {
public static void main(String[] args) throws Exception {
int result = a() + b() * c();
}
public static int a() {
System.out.println("a()");
return 3;
}
public static int b() {
System.out.println("b()");
return 4;
}
public static int c() {
System.out.println("c()");
return 5;
}
}
That prints:
a()
b()
c()
... which confirms the execution order shown in my expansion.
The sum --var1 + ++var1 + var1++ translates to:
9 + 10 + 10
since the various vars are changed as the sum is processed; obviously var1++ increments after the value is used, and ++var increments before it is used. The final value of var1 would be 11.
Please someone explain the ("x = " + x) part of the code.
public class While-With-Nested-If {
public static void main(String [] args) {
int x = 1;
while(x < 100) {
System.out.println("x = " + x);
if(x % 2 == 0) {
x++;
} else {
x *= 2;
}
}
}
}
In this case the operator is used to concatenate a string with the string representation of x.
It depends on the operand type.
For String type operands it creates a new String instance (String objects are immutable) and assigns to it the concatenation of two operands.
For numeric types it works as addition operator.
In this case, it concatenates a String with the string representation of x. For example: x = 42;
The "+" operator acts as syntactic sugar for the concatenation operator in regards to String operations.
It is used to concatenate the two strings in your case.
When you write
String a = b + c + d;
then it gets converted into:
String a = new StringBuilder(b).append(c).append(d).toString();
You may refer Oracle docs for more details
In this code block while loop will iterate till x is less than 100 in order to print all the values of x during "while" loop executes the System.out.println("x = " + x); is used.
Here java will send each value of x to output console by appending it to the "x = " string(text) so on each iteration of while loop you will get output on console like
x = 1
x = 2
x = 3
and so on...
x = 99
The + operator concatenates if the operands' type is string, and it performs summing if the operands are ints, floats or doubles.
Here is the out put:
X = 1, X = 2, X = 3, X = 6, X = 7, X = 14, X = 15, X = 30, X = 31, X = 62, X = 63...
Have a look at this code
Integer x = 5;
Integer y = 2;
Integer xy = x+y;
System.out.println("xy = " + xy); // outputs: 7
System.out.println("xy2 = " + xy2); // outputs: 7
x++;
System.out.println("xy = " + xy); // outputs: 7
System.out.println("xy2 = " + xy2); // outputs: 7
How can I get the code to output 8 without using a method that calculates it for you?
An Integer in Java is immutable. You can not change its value. In addition, it's a special autoboxing type to provide an Object wrapper for the int primitive.
In your code, for example, x++ does not modify the Integer object x is referencing. It un-autoboxes it to a primitive int, post-increments it, re-autoboxes it returning a new Integer object and assigns that Integer to x.
Edit to add for completeness: Autoboxing is one of those special things in Java that can cause confusion. There's even more going on behind the scenes when talking about memory / objects. The Integer type also implements the flyweight pattern when autoboxing. Values from -128 to 127 are cached. You should always use the .equals() method when comparing Integer objects.
Integer x = 5;
Integer y = 5;
if (x == y) // == compares the *reference (pointer) value* not the contained int value
{
System.out.println("They point to the same object");
}
x = 500;
y = 500;
if (x != y)
{
System.out.println("They don't point to the same object");
if (x.equals(y)) // Compares the contained int value
{
System.out.println("But they have the same value!");
}
}
See: Why aren't Integers cached in Java? for more info (and of course the JLS)
You need to update xy after modifying x. So:
x++;
xy = x + y;
xy2 = x + y;
In Java you'll need to update a variable yourself if you want the value to change. It's not like other languages where you express a relationship between two variables and this relationship is maintained whenever a variable changes.
The expression:
xy = x + y
doesn't mean that now xy is depends on the values of x and y (if they changed, xy is changed too). You can see it as follows: the value of the expression x + y is inserted into xy.
Therefore, you must increase the value of x (by x++), before your set the value of xy.
I'm new to java, and I'm not really sure of the context for this question, but if all you want to do is output 8, you can just make it xy++ instead of x++.
Integer x = 5;
Integer y = 2;
Integer xy = x+y;
int xy2 = x+y; // just testing to see if it makes a difference
System.out.println("xy = " + xy); // outputs: 7
System.out.println("xy2 = " + xy2); // outputs: 7
**xy++;**
System.out.println("xy = " + xy); // **outputs: 8**
System.out.println("xy2 = " + xy2); // outputs: 7
public void increment(){
int zero = 0;
int oneA = zero++; // Compiles
int oneB = 0++; // Doesn't compile
int oneC = getInt()++; // Doesn't compile
}
private int getInt(){
return 0;
}
They are all int's, why won't B & C compile? Is it to do with the way ++ operator differs from = 0 + 1;?
Invalid argument to operation ++/--
i++ is an assignment to a variable i.
In your case, zero++ is an equivalent to zero = zero + 1. So 0++ would mean 0 = 0 + 1, which makes no sense, as well as getInt() = getInt() + 1.
More accurately :
int oneA = zero++;
means
int oneA = zero;
zero = zero + 1; // OK, oneA == 0, zero == 1
int oneB = 0++;
means
int oneB = 0;
0 = 0 + 1; // wrong, can't assign value to a value.
int oneC = getInt()++;
means
int oneC = getInt();
getInt() = getInt() + 1; // wrong, can't assign value to a method return value.
From a more general point of view, a variable is a L-value, meaning that it refers to a memory location, and can therefore be assigned. L in L-value stands for left side of the assignment operator (i.e. =), even if L-values can be found either on the left side or the right side of the assignment operator (x = y for instance).
The opposite is R-value (R stands for right side of the assignment operator). R-values can be used only on the right side of assignment statements, to assign something to a L-value. Typically, R-values are literals (numbers, characters strings...) and methods.
Because as stated in JLS:
The result of the postfix expression must be a variable of a type that
is convertible (§5.1.8) to a numeric type, or a compile-time error
occurs.
getInt() is not int
getInt() returns int
++ operator does two things increment + assignment
So for ++ operator to work you need a variable to store the result of increment operation which 0 and getInt() both are not.
The pre- and post- operators only operate on variables or lvalues as they are called. lvalue is short for left value, i.e. something that can stand to the left in an assignment.
In your example:
zero = 1; // OK
0 = 1; // Meaningless
getInt() = 1; // Also meaningless
//jk
Both B and C make the compiler say:
unexpected type, required: variable, found: value
So you can't increment a value, only a variable.
Why doesn't the post increment operator work on a method that returns an int?
Because it is a getter method, and it doesn't make sense to change a value via getter.
int z = x + y++;
is equivalent to:
int z = x + y;
y = y + 1;
so it is not valid to have something like:
int z = x + getY()++;
which is equivalent to:
int z = x + getY();
getY() = getY() + 1; // invalid!
0++
It is equivalent to 0 = 0 + 1; and certainly it is not possible.
i.e. it has to be l-value to assign to it.
getInt()++;
Similar reason here.
Because 0 is a rValue (i.e. You can use it only from right of the assignment operator) not a lValue.
++ operator increments the value and sets it to itself therefore 0++ will give You an error.
My answer its kind of "out of the box".
When I have doubt about an operator usage, I think "which its the overloaded function equivalent" of this operator ?
I, know, that Java operators doesn't have operator overloading, its just an alternative way to make a solution.
In this case:
...
x++;
...
should be read as:
...
int /* function */ postincrement (/* ref */ int avalue)
{
int Result = avalue;
// reference value,
avalue = avalue + 1;
return Result;
}
...
postincrement(/* ref */ x);
...
And:
...
++x;
...
...
int /* function */ preincrement (/* ref */ int avalue)
{
// reference value,
avalue = avalue + 1;
int Result = avalue;
return Result;
}
...
preincrement(/* ref */ x);
...
So, both, versions of "++", work as a function that receives a variable parameter by reference.
So, a literal value like "0++" or a function result like "getInt()++", are not a variable references.
Cheers.
postincrement and preincrement can apply only with the help of variable.So the first case compile.
Since function return is RHS expression and pre/post increment/decrement operations can be applied to LHS expressions only.
Is there a difference between ++x and x++ in java?
++x is called preincrement while x++ is called postincrement.
int x = 5, y = 5;
System.out.println(++x); // outputs 6
System.out.println(x); // outputs 6
System.out.println(y++); // outputs 5
System.out.println(y); // outputs 6
yes
++x increments the value of x and then returns x
x++ returns the value of x and then increments
example:
x=0;
a=++x;
b=x++;
after the code is run both a and b will be 1 but x will be 2.
These are known as postfix and prefix operators. Both will add 1 to the variable but there is a difference in the result of the statement.
int x = 0;
int y = 0;
y = ++x; // result: x=1, y=1
int x = 0;
int y = 0;
y = x++; // result: x=1, y=0
Yes,
int x=5;
System.out.println(++x);
will print 6 and
int x=5;
System.out.println(x++);
will print 5.
In Java there is a difference between x++ and ++x
++x is a prefix form:
It increments the variables expression then uses the new value in the expression.
For example if used in code:
int x = 3;
int y = ++x;
//Using ++x in the above is a two step operation.
//The first operation is to increment x, so x = 1 + 3 = 4
//The second operation is y = x so y = 4
System.out.println(y); //It will print out '4'
System.out.println(x); //It will print out '4'
x++ is a postfix form:
The variables value is first used in the expression and then it is incremented after the operation.
For example if used in code:
int x = 3;
int y = x++;
//Using x++ in the above is a two step operation.
//The first operation is y = x so y = 3
//The second operation is to increment x, so x = 1 + 3 = 4
System.out.println(y); //It will print out '3'
System.out.println(x); //It will print out '4'
Hope this is clear. Running and playing with the above code should help your understanding.
I landed here from one of its recent dup's, and though this question is more than answered, I couldn't help decompiling the code and adding "yet another answer" :-)
To be accurate (and probably, a bit pedantic),
int y = 2;
y = y++;
is compiled into:
int y = 2;
int tmp = y;
y = y+1;
y = tmp;
If you javac this Y.java class:
public class Y {
public static void main(String []args) {
int y = 2;
y = y++;
}
}
and javap -c Y, you get the following jvm code (I have allowed me to comment the main method with the help of the Java Virtual Machine Specification):
public class Y extends java.lang.Object{
public Y();
Code:
0: aload_0
1: invokespecial #1; //Method java/lang/Object."<init>":()V
4: return
public static void main(java.lang.String[]);
Code:
0: iconst_2 // Push int constant `2` onto the operand stack.
1: istore_1 // Pop the value on top of the operand stack (`2`) and set the
// value of the local variable at index `1` (`y`) to this value.
2: iload_1 // Push the value (`2`) of the local variable at index `1` (`y`)
// onto the operand stack
3: iinc 1, 1 // Sign-extend the constant value `1` to an int, and increment
// by this amount the local variable at index `1` (`y`)
6: istore_1 // Pop the value on top of the operand stack (`2`) and set the
// value of the local variable at index `1` (`y`) to this value.
7: return
}
Thus, we finally have:
0,1: y=2
2: tmp=y
3: y=y+1
6: y=tmp
When considering what the computer actually does...
++x: load x from memory, increment, use, store back to memory.
x++: load x from memory, use, increment, store back to memory.
Consider:
a = 0
x = f(a++)
y = f(++a)
where function f(p) returns p + 1
x will be 1 (or 2)
y will be 2 (or 1)
And therein lies the problem. Did the author of the compiler pass the parameter after retrieval, after use, or after storage.
Generally, just use x = x + 1. It's way simpler.
Yes.
public class IncrementTest extends TestCase {
public void testPreIncrement() throws Exception {
int i = 0;
int j = i++;
assertEquals(0, j);
assertEquals(1, i);
}
public void testPostIncrement() throws Exception {
int i = 0;
int j = ++i;
assertEquals(1, j);
assertEquals(1, i);
}
}
Yes, using ++X, X+1 will be used in the expression. Using X++, X will be used in the expression and X will only be increased after the expression has been evaluated.
So if X = 9, using ++X, the value 10 will be used, else, the value 9.
If it's like many other languages you may want to have a simple try:
i = 0;
if (0 == i++) // if true, increment happened after equality check
if (2 == ++i) // if true, increment happened before equality check
If the above doesn't happen like that, they may be equivalent
Yes, the value returned is the value after and before the incrementation, respectively.
class Foo {
public static void main(String args[]) {
int x = 1;
int a = x++;
System.out.println("a is now " + a);
x = 1;
a = ++x;
System.out.println("a is now " + a);
}
}
$ java Foo
a is now 1
a is now 2
OK, I landed here because I recently came across the same issue when checking the classic stack implementation. Just a reminder that this is used in the array based implementation of Stack, which is a bit faster than the linked-list one.
Code below, check the push and pop func.
public class FixedCapacityStackOfStrings
{
private String[] s;
private int N=0;
public FixedCapacityStackOfStrings(int capacity)
{ s = new String[capacity];}
public boolean isEmpty()
{ return N == 0;}
public void push(String item)
{ s[N++] = item; }
public String pop()
{
String item = s[--N];
s[N] = null;
return item;
}
}
Yes, there is a difference, incase of x++(postincrement), value of x will be used in the expression and x will be incremented by 1 after the expression has been evaluated, on the other hand ++x(preincrement), x+1 will be used in the expression.
Take an example:
public static void main(String args[])
{
int i , j , k = 0;
j = k++; // Value of j is 0
i = ++j; // Value of i becomes 1
k = i++; // Value of k is 1
System.out.println(k);
}
The Question is already answered, but allow me to add from my side too.
First of all ++ means increment by one and -- means decrement by one.
Now x++ means Increment x after this line and ++x means Increment x before this line.
Check this Example
class Example {
public static void main (String args[]) {
int x=17,a,b;
a=x++;
b=++x;
System.out.println(“x=” + x +“a=” +a);
System.out.println(“x=” + x + “b=” +b);
a = x--;
b = --x;
System.out.println(“x=” + x + “a=” +a);
System.out.println(“x=” + x + “b=” +b);
}
}
It will give the following output:
x=19 a=17
x=19 b=19
x=18 a=19
x=17 b=17
public static void main(String[] args) {
int a = 1;
int b = a++; // this means b = whatever value a has but, I want to
increment a by 1
System.out.println("a is --> " + a); //2
System.out.println("b is --> " + b); //1
a = 1;
b = ++a; // this means b = a+1
System.out.println("now a is still --> " + a); //2
System.out.println("but b is --> " + b); //2
}
With i++, it's called postincrement, and the value is used in whatever context then incremented; ++i is preincrement increments the value first and then uses it in context.
If you're not using it in any context, it doesn't matter what you use, but postincrement is used by convention.
There is a huge difference.
As most of the answers have already pointed out the theory, I would like to point out an easy example:
int x = 1;
//would print 1 as first statement will x = x and then x will increase
int x = x++;
System.out.println(x);
Now let's see ++x:
int x = 1;
//would print 2 as first statement will increment x and then x will be stored
int x = ++x;
System.out.println(x);
Try to look at it this way:
from left to right do what you encounter first. If you see the x first, then that value is going to be used in evaluating the currently processing expression, if you see the increment (++) first, then add one to the current value of the variable and continue with the evaluation of the expression. Simple