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this is undefined behaviour...

ready my comment (SHAN) on comment id:14259670

It's Simple,

Keep one thing in mind, calculation is done from right to left, you will get anwer for first question.

For second question, int division results to 1, incrementing that we will get 2.

Dont join that compant who ask these questions.. :)
Google Undefined Behaviour!!!

please someone explain

Undefined Behavior.....
It is explained very well here....

stackoverflow.com/questions/4176328/undefined-behavior-and-sequence-points
and
stackoverflow.com/questions/367633/what-are-all-the-common-undefined-behaviour-that-a-c-programmer-should-know-ab

These are the expression involved with sideeffect.
Preincrement is a unary operator and has highest precedence, post increment is done after assignment.So output is 2 in both cases.How ever o/p is compiler dependent.Modern gcc compiler will output 2 in both cases.

the first exp can be written as (x++)/(++x) //precedence of operator
now from right the value of x becomes 2 but the value of x on the left does not change as its a post fix so we have x/x resulting in 1 now the postfix takes effect so value becomes 2

the 2nd exp same thing follows (++x)/(x++)
from right being a post fix value remains one, but in the left value chages to 2
so x/x results in 2/2 i.e 1 and with post fix x again becomes 2

Undefined Behaviour - Explained very well in "C FAQ's" by Steve Summit.

Please refer this question (question No. 3.1)

" Q: Why doesn't this code:

a[i] = i++;

work?

Ans - The subexpression i++ causes a side effect--it modifies i's value--which leads to undefined behavior since i is also referenced elsewhere in the same expression. There is no way of knowing whether the reference will happen before or after the side effect--in fact, neither obvious interpretation might hold "

Acc to sequence point rule in 2 cosecutive sequence points if the value of a variable is modified more than once then it is voilation in c. The value to above ques wil b compiler dependent. ;,&&?: these r the eg of sequence points.

for x=x++/++x;
its evaluated in this order:
1.++x makes x =2.
2.now since x++ is post increment, the exp becomes x/x++ ie 1/2 = 0(since x is int).
3.now assignment operator(=) has low precedence than post increment(++), x=1 is incremented to 2 and then its assigned to x.


for x=++x/x++;
its evaluated in this order:
1.++x increments x =2
2.now exp becomes x = 2/(1)++ which is evaluated to 2/1 = 2
3.x is still 1 and its incremented to 2(due to post increment) and assigned to x making it 2.

its very simple ........in evaluation it will take first only expression then increment operator means
for both the expressions x=x/x now it will put the value of x =2 because of pre increment operator ..... so we will get the result 1 and then again incremented by 1 because of post increment......

here first evalution continue from right,so the ++x becomes 2 and now x=2 so
x=x/2,2/2,1 now the value of x is 1 but remember postincrement is waiting for increase the value(x++) so the value of x is become 2(ans).
here compiler tereatesd all x same it can not diffrenciate them....

In my environment x is 1. @ Ajeet, ++x has the highest priority, so in the second one x will be 2 first and then x = 2/2 which 1.