How defMethod is working here?

Question

#include<bits/stdc++.h>
#define defMethod(i)(i*i)
using namespace std;
int main()
{
int a,b=5;
a=defMethod(b+5);
cout<<a;
return 0;
}
Output: 35

Answer 1

#define replaces text when compiled. So,
defMethod(b + 5);
will be replaced with
(b + 5 * b + 5);
If you calculate (5 + 5 * 5 + 5), it will be 35. The expression inside the parentheses isn't evaluated.

Answer 2

Well, of course it is a sequence, where
an = 6n + 5
(or just a1 = 11; d = 6)
That's just common sense

Answer 3

Airree I have tried it by changing the parameter while calling with b=1,2,3,4,5 the output seems like a series.
Yeah you told is absolutely correct.
Thanks for helping.☺️☺️