to resolve what’s on the right of the equals:
1: obtain the value of y first which is 1 before decrimenting (since this is a post-decriment),
2: then decriment y by –1 to 0,
3: again since you are post-decrimenting, obtain the value of y first, which is now 0,
4: decriment y to –1;

so, y = 1 (from step 1 above) + 0 (from step 3 above) = 1

notice the last decriment had no effect on the equation.

Matt

Tse Wu
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Joined: Jan 05, 2012
Posts: 10

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Thanks Matt.
Y ended up not being -1 due to the fact that the final value was assigned as the sum of the expression. Essentially y was -1 before being reassigned to 1.. Do I have the jist of it?

matt love
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Joined: Jan 25, 2010
Posts: 65

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Hi Tse.

I didn't quite follow your response.

If you are willing, let's take it step-by-step, with separate emails for each step, and see if there is a particular step that you're having difficulty with.

int y = 1;
y = y-- + y--;

we will resolve the expression left-to-right, so we will first address the first y--.

i believe y-- is called a post-decriment operator because we will evaluate y (or another phrase might be to obtain the value of y) first and then decriment y by 1.

--y would be called a pre-decriment operator because we would decriment y before we would evaluate and obtain the value of y.

so, back to our expression: we have evaluated and retained the value of y as 1.

we are still working with the leftmost y-- and we haven't decrimented it yet.

ok so far?

Matt

Tse Wu
Greenhorn

Joined: Jan 05, 2012
Posts: 10

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Hi Matt,
From my understanding.

Step1: Since the first y is post decrement, the first y value is 1.
Step2: After the resolution of the first y, y is decreased to 0.
Step3: The second y is also post decrement, so the second y value is assigned to 0.
Step4: after the resolution of the second y, y is decreased to -1.
Step5: so we ended up with 1 + 0
Step6: At this point y is -1, but since 1 + 0 is 1 the assignment of y changed from -1 to 1 at this point.

I think this is how it goes with the explanation you provided which made sense. Let me know if one of the steps was wrong. Thanks again.

matt love
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Joined: Jan 25, 2010
Posts: 65

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You have it correct to the end where you're a little off.

int y = 1;
y = y-- + y--;

Step1: Since the first y is post decrement, the first y value is 1.

yes, y is evaluated as having value 1.

Step2: After the resolution of the first y, y is decreased to 0.

Yes

Step3: The second y is also post decrement, so the second y value is assigned to 0.

correct

Step4: after the resolution of the second y, y is decreased to -1.

yep

Step5: so we ended up with 1 + 0

yep

Step6: At this point y is -1, but since 1 + 0 is 1 the assignment of y changed from -1 to 1 at this point.

I think you have the concept down Tse, although y did not change from -1 to 1 at the end. It’s more a matter of snapshots in time. As far as the expression is concerned, y resolves to 1 at the end because of the progression you described.

After the expression, y is in fact –1, because of that final post-decrementer. That second y-- evaluated to 0 so the expression could be resolved (it is a post-decrementer) and then that second y-- decremented y by 1.

int y = 1;
y = y-- + y--;
System.out.println(y); // y is –1 at this point after the above entire expression is executed, including that final post-decrement;

That final decrement, --, has the final effect on the value of y after the expression was completely and finally completed. Remember, the value of y in the second y—was evaluated before it was decremented.

I know it looks weird. Be sure and let me know if you’re still hung up on something and I’ll respond tomorrow.