Using Macros in a program that calculates the midrange of three values - c

Question

I am trying to create a program that uses macros for determining the mid-range for three values. The midrange is defined as:
midrange(a, b, c) = (min(a, b, c) + max(a, b, c)) / 2
For example:
midrange(3, 10, 1) = (min(3, 10, 1) + max(3, 10, 1)) / 2
= (1 + 10) / 2
= 11 / 2
= 5.5
I am still new to programming, and I am not sure if the syntax I am using for macro definitions is correct. My first question is, can I define a macro in the main function? My second question, should I use curly braces, normal parenthesis, or nothing at all for the body of the macro, that is, the replacement list?
This is what my program looks like for calculating the midrange of three integer values:
#include <stdio.h>
#define MIN(A, B) {(A) < (B) ? (A) : (B);}
#define MAX(A, B) {(A) > (B) ? (A) : (B);}
int main(){
//scans the three values
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
int min = MIN(MIN(a, b), c);
int max = MAX(MAX(a, b), c);
//defines new macro for calculating the midrange
//is this allowed/correct formatting and syntax?
#define MIDRANGE((min + max) / 2);
/* initializes variable mrange to equal to the value returned
function-like macro, MIDRANGE */
double mrange = MIDRANGE((min + max) / 2);
printf("The mid-range is: %lf", mrange);
return 0;
}
The reason why I declared the variable mrange as a double is because I want the value of the MIDRANGE macro to be precise, meaning that it will keep the decimal places after the division by 2 in the equation.
Any help is appreciated:)

Answer 1

You can declare a macro anywhere in the code, but unless you have a very specific reason to restrict its usage to a function, it typically appears in the beginning of the .c file or in a .h file.
Macros work by substitution. So if you have something similar to the following:
#define MIN(A, B) {(A) < (B) ? (A) : (B);}
int main(void)
{
int min = MIN(MIN(a, b), c);
}
It will be preprocessed into:
int min = {({(a) < (b) ? (a) : (b);}) < (c) ? ({(a) < (b) ? (a) : (b);}) : (c);};
You can see this by yourself, if you are using GCC, with the command gcc -E source.c. This is obviously not what you wanted, and it will give you a compilation error.
We never use ; at the end of macros, neither put them in blocks. To avoid unexpected operator association with complicated expressions, we guard macro parameters and the whole macro within parentheses.
Your MIN macro should look somewhat like this:
#define MIN(A, B) ((A) < (B) ? (A) : (B))
Finally, there are two things going on with your MIDRANGE macro. First, it should be taking arguments, like you did with MIN. Ideally, it should take three arguments; after all, the whole reason to make it a macro is so that the macro user won't have to calculate everything. Second, if you want the result to be a floating-point number, you should divide by 2.0, not 2.
Here's my first suggestion:
#define MIDRANGE(A, B, C) ((MIN(MIN(A, B), C) + MAX(MAX(A, B), C)) / 2.0)
However, why is this a macro? Make it a function. It's more readable, and easier to mantain. So here's my second suggestion:
double midrange(int a, int b, int c)
{
int min = MIN(MIN(a, b), c);
int max = MAX(MAX(a, b), c);
return (min + max) / 2.0;
}

Answer 2

regarding:
#define MIN(A, B) {(A) < (B) ? (A) : (B);}
#define MAX(A, B) {(A) > (B) ? (A) : (B);}
Lets remember that a macro is nothing more than a direct text replacement.
So looking at:
int min = MIN(MIN(a, b), c);
results in:
int min = { {(a) < (b) ? (a) : (b);} < (c);};
as is obvious from the above,
min is outside the braces '{' and '}' so will be difficult to assign to
the resulting statement contains two spurious ';' characters, which is what the compiler is complaining about
Suggest:
#define MIN( A, B ) ((A) < (B)) ? (A) : (B)
Similar considerations exist for the MAX macro