#define a b+c confusion..I did not get the output that I expected

Question

   #include <iostream>
   using namespace std;
   # define p 5+3
   int main() {
    //code

    int i=p*p;
   cout<<i;

    return 0;
     }

expected output:64 actual output:23 howwW?? I dont understand

Welcome to SO! Please specify which output you are getting instead of the expected one. — nicolaus-hee, Sep 05 '15 at 16:42
Very good sample why you shouldn't use macros, especially when not using parentheses as necessary. — πάντα ῥεῖ, Sep 05 '15 at 16:45
[The need for parentheses in macros in C](http://stackoverflow.com/q/10820340/995714) — phuclv, Sep 05 '15 at 17:13

score 7 · Accepted Answer · answered Sep 05 '15 at 16:43

7

p will be replaced by 5+3. So the line int i=p*p; is int i=5+3*5+3;. * has an higher rank than +, so the result is 23.

You need to paranthese your define:

#define p (5+3)

answered Sep 05 '15 at 16:43

mch

score 4 · Answer 2 · answered Sep 05 '15 at 16:42

4

 int i=p*p;

p*p is replaced as 5+3*5+3 ,thus gives 23 (5+15+3=23).As you know * will be evaluated before + .

Define macro as follows -

#define p (5+3)

answered Sep 05 '15 at 16:42

ameyCU

2 Answers2