Ok, here is my problem. When I initialize x with 5, the while loop doesn't terminate but initializing it with an even number such as 6 terminates the loop:
int x = 6;
while(x != 0){
x = x + 2;
}
Surprisingly for me, this terminates even though I thought it would not. However,the following similar loop, with odd value for x loops infinitely:
int x = 5
while(x != 0){
x += 2;
}
It seems this happens only with the while loop in Java as I didn't get similar results with the for loop. Please help explain this to me as I'm currently confused. Thanks
Integer overflows at 2147483647 and if you add 1, it will goes to -2147483648 . So when you start from even number you will get back to zero again. But when you start from odd number you will never get back to zero.
Not going to much into depth here, but its because of the overflow. An integer is 32 bit, so its a finite amount of numbers. Since its signed it means that the numbers range from −2.147.483.648 to 2.147.483.647. So in this example: 2.147.483.647 + 1 = -2.147.483.648 is an overflow.
This means if u have x+1 in your while loop will always terminate. Because once the 0 is reached, the while loop condition is broken. If you add 2 it seems to work because the number x reaches 0 at some point.
Hope it helps mate.
You can see the same effect on a regular 12 hour wall clock. Pick any hour. Until you reach 12, add two hours.
You'll notice that if you start on an even hour, you'll always stop. For example, 6:
6 + 2 = 8
8 + 2 = 10
10 + 2 = 12 // stop
However, if you start on an odd hour, you'll never stop. For example:
5 + 2 = 7
7 + 2 = 9
9 + 2 = 11
11 + 2 = 1
1 + 2 = 3
3 + 2 = 5
5 + 2 = 7
...
A Java int works the same way, but instead of 12 hours, it has 2^32 = 4294967296 values. Instead of wrapping from 12 to 1, it wraps from 2147483647 to -2147483648.
Note that for loops do behave the same way: for(int x=6; x!=0; x+=2) {} does indeed stop while x=5 does not.
I tried googling this but Google doesn't handle "--n" well. I saw this in my professor's code:
f[--n];
f[n++];
where f is an array of double values.
My guess is that it returns the value of f[n] before reducing (or adding) to n.
f[--n]; means :
n = n -1;
f[n];
f[n++]; means :
f[n];
n = n + 1;
It's actually a type of operator called a pre-decrement, and it's part of a family of 4 operators (see table of java operators)
For an integer-type variable called n:
post-increment n++ is the equivalent of n = n + 1, the 'post' part means that if you see it in a line of code (ex. foo(n++);) then the line of code will be called Before n is incremented.
pre-increment ++n is also the same as n = n + 1 but it occurs Before the line of code it belongs in has been run.
post-decrement n-- is the equivalent of n = n - 1 and occurs After the current line of code has been run
pre-decrement --n is the equivalent of n = n - 1 and occurs Before the current line of code has been run
Example of post vs pre decrement:
int n = 5;
System.out.println(n--); //This prints 5
System.out.println(n); //This prints 4
System.out.println(--n); //This prints 3
System.out.println(n); //this prints 3
you can look it up under predecrement (--n) or postincrement (n++).
It works like this:
f[--n]: first n is reduced by 1 then the value of f[n] (this is already the reduced n) is returned
f[n++]: first the value of f[n] is returned then n is increased by 1
Example:
f {1,3,5}
n=1;
f[--n] returns 1
f[n++] returns 3
The code
f[--n];
f[n++];
Is the same as
n--;
f[n];
f[n];
n++;
I don't understand how this works. I know basic factoral recursion, but this is of a mixed type. Could someone explain step by step on what's going on with the output based on this exact code snippet? Even just a few of the first values (commented at the end of the code)?
Thanks :)
public class recursion1st
{
public static String recFun(int x)
{
if (x <= 0) return "/";
return recFun(x-3) + x + recFun(x-2) + x;
}
public static void main(String[] args)
{
System.out.println(recFun(8));
}
}
//Produces '/2/25/3/1/1358/3/1/136/1/14/2/2468 '(?)
3| function taking an int argument
5| this is your base case, return "/" if...
6| this is your recursion, call recFun(x-3) + x /*this is your current value */ recFun(x-2) + x. basically this is going to go until you're x reaches 0 or less. Note the right root recursive call will go through more recursions before it hits the base case because it is passing x - 2 everytime instead of x-3. Also realize this statement calls recursively twice.
Make a chart on a big piece of paper and map out passing 8 in:
8
root
return recFun(5) + 8 + recFun(6) + 8
1st branch
return recFun(2) + 5 + recFun(3) + 8
2nd branch
return recFun(3) + 6 + recFun(4) + 6
.......................
Keep going it will become clear.
In the main function, it is calling recFun() to print the data. In that function, which is recursive, the first if condition is to break out of recursion, when that condition is met (when x is negative or equal to zero).
Else it will return a string, in turn again calling itself.
Here, in the first return, it calls recFun(5) and and recFun(6) and likewise, if you analyse step by step, the cycle continues till the break condition is met. Basically the return statement is concatenation of strings, and it separates the data by placing "/" on breaking out of recursion.
EDIT:
Value returned by return statement is
rec(5)+8+rec(6)+8......(1)
where
rec(5) returns rec(2)+5+rec(3)+5......(2)
and
rec(6) returns rec(3)+6+rec(4)+6.......(3)
where
rec(3) returns "/"+3+rec(1)+3.......(4)
and
rec(4) returns rec(1)+4+rec(2)+4.....(5)
and
rec(2) returns "/" +2 +"/"+2.....(6)
and
rec(1) returns "/"+1+"/"+1 .......(7)
Substitute (7) and (6) in (5)
Substitute (7) in (4)
Substitute (4) and (5) in (3)
Substitute (4) and (6) in (2)
and substitute (2) and (3) in (1)
You will need a sheet for this. Try and work it out using this. Hope it helps. :)
take a paper and write down step by step what the function does:
recFun(5)
|
recFun(2) + 5 + recFun(3)+5
| |
recFun(-1)+2+recFun(0)+2 recFun(0)+3+recFun(1)+3
| | | |
"/" "/" "/" recFun(-2)+1+recFun(-1)+1
| |
"/" "/"
1 call recFun with 5
2 5 leads to recFun with 2 & recFun with 3
3 2 leads to recFun with -1 & recFun with 0 , 3 leads to recFun with 0 & recFun with 1
4 -1 leads to "/", 0 leads to "/", 0 leads to "/", 1 leads to recFun with -2 & recFun(-1)
5 -2 leads to "/", -1 leads to "/"
Until every singe call is finished executing (until the x <= 0 and a "/" is returned) no call made before will return anything. The function starts returning Strings from the end of that chain of calls (in this pyramid: bottom to top and from right to left)
/ 1 / 1
/ 2 / 2 / 3 3
5 5
-> / 2 / 2 5 / 3 / 1 / 1 3 5
I know, recursion is confusing. The only thing that helps me is: first make every last call of the recursive function and only when the last call returns anything then roll back. Sorry if this isn't any more precise :)
I have a question,
In Java, does Math.min bind tighter than ++?
Let me illustrate with an example and maybe someone can explain to me why I get the results I get.
Here's a method I run:
private static void testIncrement() {
int x=10;
System.out.println(x++);
System.out.println(x);
x=10;
System.out.println("-----------");
System.out.println(++x);
System.out.println(x);
x=10;
System.out.println("-----------\n"+x); //10
x=Math.min(255, x++);
System.out.println(x); **//x=10 WHY NOT x=11?**
x=10;
System.out.println("-----------\n"+x);
x=Math.min(255, ++x);
System.out.println(x);
}
The results are:
10
11
-----------
11
11
-----------
10
10
-----------
10
11
On the line where I put //x=10 WHY NOT x=11?
I wonder why x is 10 and not 11. Maybe someone can explain this to me.
It looks as if Math.min create a copy of x (which is 10 at this time) which it uses to do Math.min. Then the original x is incremented from 10 to 11, but the copy which is still 10 comes out of Math.min and overwrites the incremented one.
Does this make sense?
Does anyone have an explanation for why x is 10 and not 11 in this case?
Thanks
PS - I totally understand How do the post increment (i++) and pre increment (++i) operators work in Java?
Let's deconstruct this line:
x = Math.min(255, x++);
The x++ means "remember the original value of x; then increment x; then the value of the expression is the original value". All of this happens before the assignment. So it's equivalent to:
int tmp = x; // x = 10, tmp = 10
x = x + 1; // x = 11, tmp = 10
x = Math.min(255, tmp); // x = 10
Hopefully that should make it clear. In particular, this has nothing to do with Math.min itself - it's just behaving as a normal method invocation. See section 15.14.2 of the JLS for more details.
I have the following code:
public class Tests {
public static void main(String[] args) throws Exception {
int x = 0;
while(x<3) {
x = x++;
System.out.println(x);
}
}
}
We know he should have writen just x++ or x=x+1, but on x = x++ it should first attribute x to itself, and later increment it. Why does x continue with 0 as value?
--update
Here's the bytecode:
public class Tests extends java.lang.Object{
public Tests();
Code:
0: aload_0
1: invokespecial #1; //Method java/lang/Object."<init>":()V
4: return
public static void main(java.lang.String[]) throws java.lang.Exception;
Code:
0: iconst_0
1: istore_1
2: iload_1
3: iconst_3
4: if_icmpge 22
7: iload_1
8: iinc 1, 1
11: istore_1
12: getstatic #2; //Field java/lang/System.out:Ljava/io/PrintStream;
15: iload_1
16: invokevirtual #3; //Method java/io/PrintStream.println:(I)V
19: goto 2
22: return
}
I'll read about the instructions to try to understand...
Note: Originally I posted C# code in this answer for purposes of illustration, since C# allows you to pass int parameters by reference with the ref keyword. I've decided to update it with actual legal Java code using the first MutableInt class I found on Google to sort of approximate what ref does in C#. I can't really tell if that helps or hurts the answer. I will say that I personally haven't done all that much Java development; so for all I know there could be much more idiomatic ways to illustrate this point.
Perhaps if we write out a method to do the equivalent of what x++ does it will make this clearer.
public MutableInt postIncrement(MutableInt x) {
int valueBeforeIncrement = x.intValue();
x.add(1);
return new MutableInt(valueBeforeIncrement);
}
Right? Increment the value passed and return the original value: that's the definition of the postincrement operator.
Now, let's see how this behavior plays out in your example code:
MutableInt x = new MutableInt();
x = postIncrement(x);
postIncrement(x) does what? Increments x, yes. And then returns what x was before the increment. This return value then gets assigned to x.
So the order of values assigned to x is 0, then 1, then 0.
This might be clearer still if we re-write the above:
MutableInt x = new MutableInt(); // x is 0.
MutableInt temp = postIncrement(x); // Now x is 1, and temp is 0.
x = temp; // Now x is 0 again.
Your fixation on the fact that when you replace x on the left side of the above assignment with y, "you can see that it first increments x, and later attributes it to y" strikes me as confused. It is not x that is being assigned to y; it is the value formerly assigned to x. Really, injecting y makes things no different from the scenario above; we've simply got:
MutableInt x = new MutableInt(); // x is 0.
MutableInt y = new MutableInt(); // y is 0.
MutableInt temp = postIncrement(x); // Now x is 1, and temp is 0.
y = temp; // y is still 0.
So it's clear: x = x++ effectively does not change the value of x. It always causes x to have the values x0, then x0 + 1, and then x0 again.
Update: Incidentally, lest you doubt that x ever gets assigned to 1 "between" the increment operation and the assignment in the example above, I've thrown together a quick demo to illustrate that this intermediate value does indeed "exist," though it will never be "seen" on the executing thread.
The demo calls x = x++; in a loop while a separate thread continuously prints the value of x to the console.
public class Main {
public static volatile int x = 0;
public static void main(String[] args) {
LoopingThread t = new LoopingThread();
System.out.println("Starting background thread...");
t.start();
while (true) {
x = x++;
}
}
}
class LoopingThread extends Thread {
public #Override void run() {
while (true) {
System.out.println(Main.x);
}
}
}
Below is an excerpt of the above program's output. Notice the irregular occurrence of both 1s and 0s.
Starting background thread...
0
0
1
1
0
0
0
0
0
0
0
0
0
0
1
0
1
x = x++ works in the following way:
First it evaluates expression x++. Evaluation of this expression produces an expression value (which is the value of x before increment) and increments x.
Later it assigns the expression value to x, overwriting incremented value.
So, the sequence of events looks like follows (it's an actual decompiled bytecode, as produced by javap -c, with my comments):
8: iload_1 // Remember current value of x in the stack
9: iinc 1, 1 // Increment x (doesn't change the stack)
12: istore_1 // Write remebered value from the stack to x
For comparison, x = ++x:
8: iinc 1, 1 // Increment x
11: iload_1 // Push value of x onto stack
12: istore_1 // Pop value from the stack to x
This happens because the value of x doesn't get incremented at all.
x = x++;
is equivalent to
int temp = x;
x++;
x = temp;
Explanation:
Let's look at the byte code for this operation. Consider a sample class:
class test {
public static void main(String[] args) {
int i=0;
i=i++;
}
}
Now running the class disassembler on this we get:
$ javap -c test
Compiled from "test.java"
class test extends java.lang.Object{
test();
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_0
1: istore_1
2: iload_1
3: iinc 1, 1
6: istore_1
7: return
}
Now the Java VM is stack based which means for each operation, the data will be pushed onto the stack and from the stack, the data will pop out to perform the operation. There is also another data structure, typically an array to store the local variables. The local variables are given ids which are just the indexes to the array.
Let us look at the mnemonics in main() method:
iconst_0: The constant value 0
is pushed on to the stack.
istore_1: The top element of the
stack is popped out and stored in the
local variable with index 1 which
is x.
iload_1 : The value at the
location 1 that is the value of x
which is 0, is pushed into the stack.
iinc 1, 1 : The value at the
memory location 1 is incremented by 1. So x now becomes
1.
istore_1 : The value at the top of
the stack is stored to the memory location1. That is 0 is assigned
to x overwriting its incremented value.
Hence the value of x does not change resulting in the infinite loop.
Prefix notation will increment the variable BEFORE the expression is evaluated.
Postfix notation will increment AFTER the expression evaluation.
However "=" has a lower operator precedence than "++".
So x=x++; should evaluate as follows
x prepared for assignment (evaluated)
x incremented
Previous value of x assigned to x.
None of the answers where quite spot on, so here goes:
When you're writing int x = x++, you're not assigning x to be itself at the new value, you're assigning x to be the return value of the x++ expression. Which happens to be the original value of x, as hinted in Colin Cochrane's answer .
For fun, test the following code:
public class Autoincrement {
public static void main(String[] args) {
int x = 0;
System.out.println(x++);
System.out.println(x);
}
}
The result will be
0
1
The return value of the expression is the initial value of x, which is zero. But later on, when reading the value of x, we receive the updated value , that is one.
It has been already explained well by other. I just include the links to the relevant Java specification sections.
x = x++ is an expression. Java will follow the evaluation order.
It will first evaluate the expression x++, which will increment x and set result value to the previous value of x.
Then it will assign the expression result to the variable x. At the end, x is back at its previous value.
This statement:
x = x++;
evaluates like this:
Push x onto the stack;
Increment x;
Pop x from the stack.
So the value is unchanged. Compare that to:
x = ++x;
which evaluates as:
Increment x;
Push x onto the stack;
Pop x from the stack.
What you want is:
while (x < 3) {
x++;
System.out.println(x);
}
The answer is pretty straightforward. It has to do with the order things are evaluated. x++ returns the value x then increments x.
Consequently, the value of the expression x++ is 0. So you are assigning x=0 each time in the loop. Certainly x++ increments this value, but that happens before the assignment.
From http://download.oracle.com/javase/tutorial/java/nutsandbolts/op1.html
The increment/decrement operators can
be applied before (prefix) or after
(postfix) the operand. The code
result++; and ++result; will both end
in result being incremented by one.
The only difference is that the prefix
version (++result) evaluates to the
incremented value, whereas the
postfix version (result++) evaluates
to the original value. If you are
just performing a simple
increment/decrement, it doesn't really
matter which version you choose. But
if you use this operator in part of a
larger expression, the one that you
choose may make a significant
difference.
To illustrate, try the following:
int x = 0;
int y = 0;
y = x++;
System.out.println(x);
System.out.println(y);
Which will print 1 and 0.
You don't really need the machine code to understand what's happending.
According the definitions:
The assignment operator evaluates the right-hand side expression, and stores it in a temporary variable.
1.1. The current value of x is copied into this temporary variable
1.2. x is incremented now.
The temporary variable is then copied into the left-hand side of the expression, which is x by chance! So that's why the old value of x is again copied into itself.
It is pretty simple.
You're effectively getting the following behavior.
grab the value of x (which is 0) as "the result" of the right side
increment the value of x (so x is now 1)
assign the result of the right side (which was saved as 0) to x (x is now 0)
The idea being that the post-increment operator (x++) increments that variable in question AFTER returning its value for use in the equation it's used in.
Edit: Adding a slight bit because of the comment. Consider it like the following.
x = 1; // x == 1
x = x++ * 5;
// First, the right hand side of the equation is evaluated.
==> x = 1 * 5;
// x == 2 at this point, as it "gave" the equation its value of 1
// and then gets incremented by 1 to 2.
==> x = 5;
// And then that RightHandSide value is assigned to
// the LeftHandSide variable, leaving x with the value of 5.
This is because it never gets incremented in this case. x++ will use the value of it first before incrementing like on this case it will be like:
x = 0;
But if you do ++x; this will increase.
The value stays at 0 because the value of x++ is 0. In this case it doesn't matter if the value of x is increased or not, the assignment x=0 is executed. This will overwrite the temporary incremented value of x (which was 1 for a "very short time").
This works how you expect the other one to. It's the difference between prefix and postfix.
int x = 0;
while (x < 3) x = (++x);
Think of x++ as a function call that "returns" what X was before the increment (that's why it's called a post-increment).
So the operation order is:
1: cache the value of x before incrementing
2: increment x
3: return the cached value (x before it was incremented)
4: return value is assigned to x
When the ++ is on the rhs, the result is returned before the number is incremented.
Change to ++x and it would have been fine.
Java would have optimised this to perform a single operation (the assignment of x to x) rather than the increment.
Well as far as I can see, the error occurs, due to the assignment overriding the incremented value, with the value prior to incrementation, i.e. it undoes the increment.
Specifically, the "x++" expression, has the value of 'x' prior to increment as opposed to "++x" which has the value of 'x' after incrementation.
If you are interested in investigating the bytecode, we will take a look at the three lines in question:
7: iload_1
8: iinc 1, 1
11: istore_1
7: iload_1 # Will put the value of the 2nd local variable on the stack
8: iinc 1,1 # will increment the 2nd local variable with 1, note that it leaves the stack untouched!
9: istore_1 # Will pop the top of stack and save the value of this element to the 2nd local variable
(You can read the effects of each JVM instruction here)
This is why the above code will loop indefinitely, whereas the version with ++x will not.
The bytecode for ++x should look quite different, as far as I remember from the 1.3 Java compiler I wrote a little over a year ago, the bytecode should go something like this:
iinc 1,1
iload_1
istore_1
So just swapping the two first lines, changes the semantics so that the value left on the top of stack, after the increment (i.e. the 'value' of the expression) is the value after the increment.
x++
=: (x = x + 1) - 1
So:
x = x++;
=> x = ((x = x + 1) - 1)
=> x = ((x + 1) - 1)
=> x = x; // Doesn't modify x!
Whereas
++x
=: x = x + 1
So:
x = ++x;
=> x = (x = x + 1)
=> x = x + 1; // Increments x
Of course the end result is the same as just x++; or ++x; on a line by itself.
x = x++; (increment is overriden by = )
because of above statement x never reaches 3;
I wonder if there's anything in the Java spec that precisely defines the behavior of this. (The obviously implication of that statement being that I'm too lazy to check.)
Note from Tom's bytecode, the key lines are 7, 8 and 11. Line 7 loads x into the computation stack. Line 8 increments x. Line 11 stores the value from the stack back to x. In normal cases where you are not assigning values back to themselves, I don't think there would be any reason why you couldn't load, store, then increment. You would get the same result.
Like, suppose you had a more normal case where you wrote something like:
z=(x++)+(y++);
Whether it said (pseudocode to skip technicalities)
load x
increment x
add y
increment y
store x+y to z
or
load x
add y
store x+y to z
increment x
increment y
should be irrelevant. Either implementation should be valid, I would think.
I'd be extremely cautious about writing code that depends on this behavior. It looks very implementation-dependent, between-the-cracks-in-the-specs to me. The only time it would make a difference is if you did something crazy, like the example here, or if you had two threads running and were dependent on the order of evaluation within the expression.
I think because in Java ++ has a higher precedence than = (assignment)...Does it?
Look at http://www.cs.uwf.edu/~eelsheik/cop2253/resources/op_precedence.html...
The same way if you write x=x+1...+ has a higher precedence than = (assignment)
The x++ expression evaluates to x. The ++ part affect the value after the evaluation, not after the statement. so x = x++ is effectively translated into
int y = x; // evaluation
x = x + 1; // increment part
x = y; // assignment
Before incrementing the value by one, the value is assigned to the variable.
It's happening because it's post incremented. It means that the variable is incremented after the expression is evaluated.
int x = 9;
int y = x++;
x is now 10, but y is 9, the value of x before it was incremented.
See more in Definition of Post Increment.
Check the below code,
int x=0;
int temp=x++;
System.out.println("temp = "+temp);
x = temp;
System.out.println("x = "+x);
the output will be,
temp = 0
x = 0
post increment means increment the value and return the value before the increment. That is why the value temp is 0. So what if temp = i and this is in a loop (except for the first line of code). just like in the question !!!!
The increment operator is applied to the same variable as you are assigning to. That's asking for trouble. I am sure that you can see the value of your x variable while running this program.... that's should make it clear why the loop never ends.