I'm trying to understand this block of code here:
#include <iostream>
using namespace std;
#define mymult(a, b) a*b
inline int mymult1(int a, int b) {return a*b;}
int main() {
cout << "mymult(2+2, 3+3) = " << mymult(2+2, 3+3) << "\n";
cout << "mymult1(2+2, 3+3) = " << mymult1(2+2, 3+3) << "\n";
}
mymult = 11, and mymult1 = 24. I know that '#define's essentially work via call by name, rather than call by value. However, I'm having trouble understanding why the value it returns is 11... and not 24. What causes this?
Option 1: In the case of:
#define mymult(a, b) a*b
a
and b
are treated like place holder strings and when you call mymult
, the parameters a
and b
are just copied as they were written. In other words:
mymult(2+2, 3+3) = 2+2*3+3
where a = 2+2, b = 3+3
.
Therefore you may call mymult
as follows:
mymult( (2+2), (3+3) )
where a = (2+2), b = (3+3)
.
This will be interpreted as:
mymult( (2+2), (3+3) ) = (2+2)*(3+3)
and return the value of 24
as expected.
Option 2:
If we are allowed to modify the #define
statement then an alternative way of doing this is defining it with the parentheses as follows:
#define mymult(a, b) (a)*(b)
This will give the same expected result since a
and b
will be put directly into the parentheses as they are. In other words:
mymult(2+2, 3+3) = (2+2)*(3+3) = 24
where a = 2+2, b = 3+3.
Option 3: Stick with the inline function as defined in OP:
inline int mymult(int a, int b) {return a*b;}
Good luck!