output of this program is

Question

int main(){
        int n=1,sum =0;
        while(n<=10){
                        sum += n++*n++;
               }

        printf("Sum = %d\n",sum);

       return 0

}

ans is 165 . i got 190.please explain??

Answer 1

sum += n++*n++;

Here, the variable 'n' is modified more than once without an intermediate sequence point. So, the output comes under "undefined behaviour" part and whatever code the compiler generate, we cannot say it is wrong. i.e., even if the program outputs "0" compiler is correct and programmer is wrong. 

Before writing C code one must know the rules of C. But people even write C book without knowing the rules of C :)

Answer 2

sum=0 n=1

sum=1 n=3

sum=10 n=5

sum=35 n=7

sum=84 n=9

sum=165 n=11

Answer 3

Now whenever you see a post increment Mark that place ,evaluate the expression and then come back and increment it.

so here. initially n=1,sum=0

while(n<=10)..this is satisfied so it enters the loop

sum=sum+n++*n++

here is the sequence -> sum=0+1*1=>sum=1,n=3

sum=1+3*3=>sum=10,n=5

sum=10+5*5=>sum=35,n=7

sum=35+7*7=>sum=84,n=9

sum=84+9*9=>sum=165,n=11->now at this point the it will exit the loop

so we have sum=165

Answer 4

sum=sum+((n++)*(n++))

++  or postfix is left to right ascociative. So it goes like this

 sum=0  n=1

sum=0+(1*2) =2  n=3

sum=2+(3*4) =14 n=5

sum=14+(5*6) =44 n=7

sum=44+(7*8)=100  n=9

sum=100+(9*10)=190 n=11
 

Loop ends.

Comments

i am getting the same but ans is 165

---

https://ideone.com/O6ZthM

190 is the correct ans

---

165 is the correct ans