Explain the output for the below question

Question

int main()
{
int i=5;
int a= ++i + ++i + ++i;
printf("%d",a);
return 0;
}

O/P :22

Answer 1

The answer is compiler dependent. It could be 24 , 22 or anything else.

you are modifying i multiple times before a defined sequence point. So the behavior will be undefined.
https://en.m.wikipedia.org/wiki/Undefined_behavior
https://en.m.wikipedia.org/wiki/Sequence_point

Answer 2

Second ++ operators result will be overwritten to the first operand.. even if we have n number of operands, that effect will take place only between the first and second operand.
For example if we have
I=5
A=++I + ++I + ++I + ++I + ++I
i.e.. 6+7+8+9+10 and 6 will be replaced by 7.. so 7+7+8+9+10=41.
Will result in 41

Answer 3

according to the definition of the pre increment operator 🤔 taking two terms at a time

/*O/P :22 */
/* a = 6+7 //2 terms at a time
a = 14 (13+1)+ 8 (7+1)= 22

https://code.sololearn.com/cexnFWwRkI1h/?ref=app

Answer 4

I think that it will be 18 because if i = 5 and i is incresing ++I will be a = 6 + 6 + 6