Post increment operator not incrementing in for loop

Question

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?

Answer 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 3^rd part of your for statement - i = i++, is evaluated as: -

int oldValue = i;  i = i + 1; i = oldValue;  // 3rd Step  

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You need to remove the assignment from there, to make it work: -

for (int i = 0; i < 10; i++) { 

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(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 

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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.