increment and decrement operation in c

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#include<stdio.h>
#include<math.h>
int main()
{
    int a=2,b;
    b=a++ + a-- + ++a + --a;
    printf("%d %d",a,b);
}

output is 2 9 why?????

Posted 11-Jan-14 18:06pm

Sunil Kumar Das

Updated 25-Jul-22 21:07pm

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4 solutions

Solution 15

In this question a value is 2.First evaluate a++ then the a value is 2,then evaluate a-- then a value is 3 because after evaluating a++ the value is 3.Then evaluate ++a then the value of a is 3.Then evaluate --a is 2.
2+3+3+2=10 is the answer.

Posted 2-Jul-22 3:46am

Member 15694542

Comments

Richard MacCutchan 2-Jul-22 11:18am

Not necessarily, you should read Solution 1.

Solution 16

a++ + a-- + ++a + --a
first we solve (a++ + a--) then b value will be 5 then ++a + --a solve this
++a=3 and --a=2 and sum all values of b will be 10 and a will be 2

a++ + a-- + ++a + --a
(2 + 3 + 3 + 2)=10

Posted 25-Jul-22 21:07pm

Member 15717694

Updated 25-Jul-22 23:10pm

v3

Comments

CHill60 26-Jul-22 3:29am

Not necessarily - read Solution 1

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Ron Beyer · Accepted Answer · 2014-01-11T18:11:00

Ahh, I expected this one from your last one...

The pre-increment operator and the + operators have equal precedence. That means which one is done first is completely compiler dependent. This is called undefined behavior and these types of statements should be avoided at all costs.

Why does it happen in your compiler? Its simply the precedence that the compiler decided to apply the + and ++/-- prefix operators in. I didn't work through your equation but it can either do the addition first, or the subtraction first.

There has recently been a lively discussion on this here[^], read the thread to see what is said, and ignore the guy who doesn't get it :)

nv3 · Accepted Answer · 2014-01-11T23:25:00

The first result (2) is pretty obvious, because a is incremented two times and decremented two times. Hence a has afterwards the same value as before.

The second result (9) is the one that is not so easy to see. The value of b depends on the time when the compiler chooses to do the post-increment and post-decrement operations. Your compiler has chosen to them after the entire expression has been completed. Hence b is computed as:

C++

b = 2 + 2 + 3 + 2;
    // afterwards the post-increment and post-decrement operators
    // are performed

Note that with pre-increment and pre-decrement operations the compiler has no other choice than doing them right away. Hence the last two values are 3 and 2.

As Ron has correctly pointed out in solution 1, the result of this expression is actually undefined. The compiler could have chosen to perform the post-increment and post-decrement operations at an earlier time and that could have resulted in:

C++

b = 2 + 3 + 3 + 2; // == 10

It certainly is no good idea to write statements like this. The case here is a little constructed just to demonstrate what can happen if you abuse the post-increment/decrement operators.