C macros and use of arguments in parentheses

Question

Example

#define Echo(a)  a
#define Echo(a) (a)

I realize there probably isn’t a significant difference here, but why would you ever want to include the a within parenthesis inside the macro body? How does it alter it?

Answer 1

Suppose you have

#define mul(x, y)  x * y

What happens if I say:

mul(a + 5, 6); /* a + 5 * 6 */

Now if I slighlty change the macro:

#define mul(x, y)  ((x) * (y))
mul(a + 5, 6); /* ((a + 5) * (6)) */

Remember, the arguments aren't evaluated or anything, only textual substitution is performed.

EDIT

For an explanation about having the entire macro in parentheses, see the link posted by Nate C-K.

Answer 2

Just for the record, I landed from Here How to fix mathematical errors while using macros and I will try to expand this Answer here to fit the Other one.

You are asking about the difference about:

#define Echo( a )  a
#define Echo( a ) ( a )

which is fine as long as you do not understand the macro it self (I am not an expert too :) ).

First of all you already (probably) know that there is Operator Precedence, so there is a huge difference of this two programs:

1):

#include <stdio.h>
#define ADD( a , b ) a + b

int main( void )
{
    auto const int a = 5;
    auto const int b = 10;

    auto const int c = ADD (  2 + a ,  2 + b );
    printf( "%d", c );
    return 0;
}

Output:

19

and:

#include <stdio.h>
#define ADD( a , b ) ( a ) + ( b )

int main( void )
{
    auto const int a = 5;
    auto const int b = 10;

    auto const int c = ADD ( a , b );
    printf( "%d", c );
    return 0;
}

Output:

15

Now lets preplace + with *:

#define ADD( a, b ) a * b

The compiler treats a * b like for example a == 5 and b == 10 which does 5 * 10.

But, when you say: ADD ( 2 + a * 5 + b ) Like here:

#include <stdio.h>
#define ADD( a , b ) ( a ) * ( b )

int main( void )
{
    auto const int a = 5;
    auto const int b = 10;

    auto const int c = ADD ( 2 + a , 5 + b );
    printf( "%d", c );
    return 0;
}

You get 105, because the operator precedence is involved and treats

2 + b * 5 + a

as

( 2 + 5 ) * ( 5 + 10 )

which is

( 7 ) * ( 15 ) == 105

But when you do:

#include <stdio.h>
#define ADD( a, b ) a * b

int main( void )
{
    auto const int a = 5;
    auto const int b = 10;

    auto const int c = ADD ( 2 + a , 5 + b );
    printf( "%d", c );
    return 0;
}

you get 37 because of

 2 + 5 * 5 + 10

which means:

2 + ( 5 * 5 ) + 10

which means:

2 + 25 + 10

Short answer, there is a big difference between:

#define ADD( a , b ) a * b

and

#define ADD( a , b ) ( a ) * ( a )