Can any one tell me how it's giving answer " 83 " in linux platform..? I am not able to judge how post/pre increment works in multiplication scenario.

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C

#include < stdio.h >
void main()
{
int i=2;
i=i++ * ++i * i * i++;
printf("%d\n\n",i);
}

Posted 6-Feb-14 12:09pm

kumar_11

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bling · Accepted Answer · 2014-02-06T12:47:00

Solution 1

The trick here is the pre-increment happens before expression is evaluated and
post-increments happen *after* the expression is evaluated and assigned.

i=i++ * ++i * i * i++;

There is one pre-increment and two post-increments.

Above has four terms t1, t2, t3, and t4.

After applying the one pre-increment, i becomes 3.

t1, t2, t3, t4 = i so t1*t2*t3*t4 = 3^4 = 81.

Now the two post-increments are applied to i.
i plus 2 (the two post increments) becomes 83.

I do NOT recommend writing code this way.

Posted 6-Feb-14 12:47pm

bling

v2

Comments

[no name] 6-Feb-14 19:49pm

That's it.

kumar_11 7-Feb-14 3:39am

That's great.
What about this if I write like..... ++i * ++i * i++ * ++i
Can You tell me the answer.

[no name] 7-Feb-14 13:45pm

Assuming i starts at 2, there are three pre-increments so i+3=5.
5*5*5*5=625. Add the post-increment so i=626. At this point, you should be able to see the pattern.