[Solved] Macro vs function ???

Question

Pls explain the result. I am much new to macro's or overloading in C.
https://code.sololearn.com/cHMXTEK229jp/?ref=app

Answer 1

hepy(z) (7*z)
y=2;
hepy(y+1);
the value pass to the macro will replace the expression
z=y+1
7*y+1;
y=2
7*2+1 //multiplication precede addition
14+1=15

Answer 2

It's not operator overloading and there is no operator overloading in c. It's about macros.

Answer 3

This is not called operator overloading though.

Answer 4

Can we say that "hapy (z)" is a function, "z" is an argument, and "(7 * z)" is the return value?

Answer 5

✳AsterisK✳ thank you. But why 7*z will transform to 7*y+1 while z = (y+1)? The case is what you wrote but I have little idea about macros (that's why I hate them).

Edit: Understood. Macro passes expressions as is unlike function.
So if I add one more () arround y+1, 21 is the result. Anyway thanks.

Answer 6

To_be_unnamed(_?_^@%∞)
and yet the analogy with the function can be used. For example:
#include <stdio.h>

int macro(int z) {
return (z+1);
}

int main() {
int y=2;
printf("%d",macro(7*y));
return 0;
}

Answer 7

what is the difference?
https://code.sololearn.com/cbrFD2qqFM64/?ref=app
Macros can be called as small functions that are not as overhead to process. If we have to write a function (having a small definition) that needs to be called recursively (again and again), then we should prefer a macro over a function.

Answer 8

I seem to have found a solution, look at the code please
https://code.sololearn.com/cbrFD2qqFM64/?ref=app

Answer 9

Difference.

https://code.sololearn.com/c2oE94SXX1e9/?ref=app

Answer 10

Sonic yeah it sould be macro

Answer 11

Alex Gyperkate if hapy(z) is a function then the result would be 21 not 15

Answer 12

Alex Gyperkate that's a different question.

Answer 13

This is elementary, but it is possible to make an analogy

The #define directive is used to create object-like macros for constants based on values or expressions.
#define can also be used to create function-like macros with arguments that will be replaced by the preprocessor.

A function:
• is a block of code that performs a specific task
• is reusable
• makes a program easier to test
• can be modified without changing the calling program