CareerCup

why ans is 2 for

int main() {
int x = 1;
x=x++/++x;
printf("%d\n",x);
return 0;
}

well if we convert above statement into instruction set then it ll execute as follow:
eg: Add a,b,c (in instruction set)
i.e add value at address of b with value at address of c and store at address of a. Remember address plays imp role here.
Now
x=x++/++x
It will be converted as follows:
(x++): take value at address of x and then increment later. (x is still 1)
(++x): first increment value at address of x then take its value. (x becomes 2)
finally x++/++x i.e (2/2. This is because in second instruction value has become 2).
so it is 1. now final execution of 1st instruction (i.e increment value of x) is still left. so x becomes 1+1 i.e 2.

For simplicity Look this instruction set conversion:

Div &x,&(x++),&(++x)
++x will cause value to change 2.
so x becomes 2.
x++ will first use this 2 and then increment x by 1.
so 2/2=1 and increment it. =2
Hope tht helps :)

i think this is right ...

-1 for wrong answer. Its undefined behavior.

See here: stackoverflow.com/questions/4176328/undefined-behavior-and-sequence-points

if we declare x as a volatile int x. then it will give result as 3 why?

why ans is 3 for

int main() {
volatile int x = 1;
x=x++/++x;
printf("%d\n",x);
return 0;
}

It's undefined behavior. See the answer that mentions sequence points.

Thank you Deven Bhooshan for your kind explanation. But i couldn't follow you. Just go through the link given above.

Read about what undefined behavior is.

Have you heard of "undefined behavior'? Undefined behavior means the answer doesn't have to be anything.