问题
I\'m doing some research about Java and find this very confusing:
for (int i = 0; i < 10; i = i++) {
System.err.print(\"hoo... \");
}
This is never ending loop!
Anybody has good explanation why such thing happens?
回答1:
for (int i = 0; i < 10; i = i++) {
The above loop is essentially the same as: -
for (int i = 0; i < 10; i = i) {
the 3rd part of your for
statement - i = i++
, is evaluated as: -
int oldValue = i;
i = i + 1;
i = oldValue; // 3rd Step
You need to remove the assignment from there, to make it work: -
for (int i = 0; i < 10; i++) {
(On OP request from Comments)
Behaviour of x = 1; x = x++ + x++;
: -
As far as your issue as specified in the comment is concerned, the result of the following expression: -
x = 1;
x = x++ + x++;
is obtained as follows: -
Let's mark different parts of the second statement: -
x = x++ + x++;
R A B
Now, first the RHS part (A + B)
will be evaluated, and then the final result will be assignmed to x
. So, let's move ahead.
First A
is evaluated: -
old1 = x; // `old1 becomes 1`
x = x + 1; // Increment `x`. `x becomes 2`
//x = old1; // This will not be done. As the value has not been assigned back yet.
Now, since the assignment of A
to R
is not done here, the 3rd step is not performed.
Now, move to B
evaluation: -
old2 = x; // old2 becomes 2. (Since `x` is 2, from the evaluation of `A`)
x = x + 1; // increment `x`. `x becomes 3`.
// x = old2; // This will again not be done here.
Now, to get the value of x++ + x++
, we need to do the last assignment that we left in the evaluation of A
and B
, because now is the value being assigned in x
. For that, we need to replace: -
A --> old1
B --> old2 // The last assignment of both the evaluation. (A and B)
/** See Break up `x = old1;` towards the end, to understand how it's equivalent to `A = old1; in case of `x = x++`, considering `x++ <==> A` in this case. **/
So, x = x++ + x++
, becomes: -
x = old1 + old2;
= 1 + 2;
= 3; // Hence the answer
Break up of 3rd part of x = x++
, to see how it works in x = x++ + x++
case: -
Wonder why the replacement is done as A --> old1
and not x --> old1
, as in case of x = x++
.
Take a deep look at x = x++
part, specially the last assignment: -
x = oldValue;
if you consider x++
to be A
here, then the above assignment can be broken into these steps: -
A = oldValue;
x = A;
Now, for the current problem, it is same as: -
A = old1;
B = old2;
x = A + B;
I hope that makes it clear.
回答2:
You're using post-increment: i = i++;
, it means something like this:
temp = i;
i = i + 1;
i = temp;
because 15.14.2 Postfix Increment Operator ++:
The value of the postfix increment expression is the value of the variable before the new value is stored.
That is why you have the old value.
For-loop done right:
for (int i = 0; i < 10; i++) {
System.err.print("hoo... ");
}
回答3:
because of i=i++
for (int i = 0; i < 10; i++) {
System.err.print("hoo... ");
}
回答4:
i++ will report the value of i, and THEN increment. This also means that you don't need to set i equal to i++, just change to
for (int i = 0; i < 10; i++) {
回答5:
The problem is in the statement i=i++
, this statement does three operations in sequence to complete. these are two operations in i++
and one assignment( =
) operation. These are;
i++
does two operation before any operation
Returns
i
Increments
i
finally assignment operation
- Assigns returned value
So i
is incremented before assignment making it to have old value.
Lets see i=i++
in the first loop( where i = 0
) with the three operations
- returns
0
, it will not be assigned toi
until the next operation is done - increments
i
(i= 0 + 1 = 1
) - assigns returned value
0
toi
, (i = 0
), finallyi
having its old value.
In i++
first i
is returned, but immediately before any operation is done it is incremented(i
becomes i+1
), However we are yet left with assignment operation to complete so that the returned value will be assigned to i
making it to have old value, thereby entering infinite loop.
so replace i=i++
with i++
or i=i+1
来源:https://stackoverflow.com/questions/14571327/post-increment-operator-not-incrementing-in-for-loop