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Order of Evaluation for pre and post increment operators

 
Ranch Hand
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Whats the order of evaluation?


Output :
5 3

What is the right output :
1+2+2=5
(or)
1+1+3=5
 
Greenhorn
Posts: 19
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Initially value of i=0;

now , int x=++i
i=0 initial
i=1 final

int x= ++i + i++
i=0 initial i=1 (initial)
i=1 final i=2(final)


int x= ++i + i++ + ++i
i=0 initial i=1 (initial) i=3(initial)
i=1 final i=1 (final; but i=2 will go) i=3 (final;i 's value)

1+1+3=5 (x's value)


....hope that helps

 
lowercase baba
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Mohnish Khiani wrote:

What is the right output :
1+2+2=5
(or)
1+1+3=5


Wouldn't it have been easier to run the code yourself and find out?
 
Mohnish Khiani
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i have executed the code and know the output is 5,but i wanted to know how?

That means evaluation is from left to right no matter pre or post operators?
 
Sheriff
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Correct. That code is equivalent to this:
But this also shows how dangerous it can be to use pre and post increment operators inside other statements. It's very easy to loose track of what's going on.
 
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