C macros and use of arguments in parentheses - c

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 )