Macros vs. `const`

Definitions
Examples
Tricky macro question
1. Explanation
References

Definitions

Macros are used to nickname string expression to a reusable name.

No type-checking, expanded by substitution by the preprocessor

const is a type qualifier for variables

Prevents a variable from being modified after its first initialization
Often specified for arguments to enforce a “read-only” policy

Examples

#define SIZE1 10
const int SIZE2 = 10;

SIZE1 can easily be “a” without complaints (until you actually tried to use it as an integer later), since there is no type associated with macros.

// const to avoid unintentional changes to original
Obj *clone_obj(const Obj *original) {
    Obj *new = ...
    return new;
}

Tricky macro question

What is the value of r and b by the end of this program?

#define MAX(x, y) ((x) > (y) ? (x) : (y))

int main() {
    int a = 5, b = 8;
    int r = MAX(a++, b--);
    return 0;
}

Answer (click here)

r = 7, b = 6

Explanation

The preprocessor will expand r to:

r = (a++ > b-- ? a++ : b--);

Refresher on ternary operators: (condition) ? return if true : return if false

Since the increments/decrements are AFTER the variable name, this means “use the current value, and then increment/decrement”

Read more: K&R 2.8 Increment and Decrement Operators, page 46

Substituting the values of a and b (technically the right hand side too, but I will ignore those for now):

r = ((5)++ > (8)-- ? a++ : b--);

Since 5 is not > than 8, the return-if-false expression is returned.

Keep in mind that a needs to be incremented, and b needs to be decremented, so after the comparison, a = 6 and b = 7.

r = b--;

The post-decrement rule is applied again, so first r is set to the current value of b, which is 7, and then b is decremented to 6.

References

K&R, section 4.11.2: Macro Substitution (p.89 of the physical book)
K&R section A8.2 Type Specifiers (p. 211 of the physical book)

Macros vs. const

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