kmiklas kmiklas - 4 years ago 109
C++ Question

Why can't "PURE" be defined as a const set to 0 to identify pure virtual classes?

Trying to define keyword

as a const set to
, to be used as an identifier for classes that are an abstract data type. Why won't this compile? Per Meyers in "Essential C++, topic 1," I would prefer to use
as opposed to #define, like so:

const int PURE = 0;
virtual void myFunction() = PURE;

Alas, this throws an error (same on Apple LLVM 7.0 and gcc):

Initializer on function does not look like pure-specifier

Example follows, with three techniques labeled A, B, and C;

1. const int PURE = 0 will not compile
2. #define PURE 0 compiles and runs fine
3. Simply setting function = 0 (Stroustrup) works fine.

It's now set up use the
solution, and so will not compile. Simply comment/uncomment lines 4, 5, 11, and 12 appropriately to examine the three different methods:

#include <iostream>
#include <string>

const int PURE = 0; // Case B
// #define PURE 0 // Case C

const float PI = 3.14159;

class Shape {
// virtual float area() = 0; // Case A: Compiles
virtual float area() = PURE; // Case B: This does not compile
// Case C: Compiles
class Circle: public Shape {
Circle(float radius):radius_(radius) {}
float area() { return PI * radius_ * radius_; }
float radius_;
class Rectangle : public Shape {
Rectangle(float base, float height):base_(base),height_(height) {}
float area() { return base_ * height_; }
float base_;
float height_;
int main(int argc, const char * argv[]) {
Circle c(3);
Rectangle r(3,5);

std::cout << "Circle Area: \t" << c.area() << std::endl;
std::cout << "Rectangle Area: \t" << r.area() << std::endl;

std::cout << std::endl;
return 0;

Answer Source

The language grammar says:

    = 0

That is, only the tokens = 0 are allowed. You cannot put an identifier there.

#define PURE 0 works fine because macro replacement happens before translation.

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