// Community Downvotes

Question

//Declined due to community downvotes. Please review and resubmit
What is the output of this code?
#define sqr(x) (x*x)
int main()
{
cout<<sqr(5+2);
}

Seriously ? Why this code has more downvotes when it doesn't have complicated maths.
If you know the concept of #define this question isn't a problem at all. But why down voting it :/

People will like cout<<1+2; question but not like this type of questions.Solo Learn should not decline quiz based on likes :/

P.s: My English isn't good though!

Answer 1

for starters, just checked and it doesn't work well

i expected the result 49 and got 17 (5*2+5+2 perhaps???) o_O

only when used with another set of brackets sqr((5+2)) did i get what i expected

Answer 2

@burey , In c++ , #define used to replace the parameter before the compiler executes.
So in the program , #define sqr(x) (x*x)
will replace the (x*x) in sqr(x) before compiler executes so sqr(5+2) = (5+2*5+2) = 17

Your answer 49 will come if the #define sqr(x) (x)*(x) is defined like this way.

I hope you understood. :D

Answer 3

understood now thanks

could be that many of those who downvoted expected the same answer as me and when didn't get it just treated the quiz as faulty

Answer 4

You can do this
#include <iostream>
using namespace std;
#define sqr(x,y) ((x*x)+(2*x*y)+(y*y))
int main()
{
cout<<sqr(5,2);
return 0;
}

Answer 5

I think I didn't understand the question