CODEWITHSUNDEEP
c++inlinec-preprocessorcallbynamecall-by-value
Tabrock
Tabrock

Reputation: 1169

Call by Name/Call by Value

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?

Upvotes: 2

Views: 297

Answers (1)

erol yeniaras
erol yeniaras

Reputation: 3795

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!

Upvotes: 6

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