Macros in C / C++

I define a function as a macro:

#define DIVIDE(x, y) x / y

And I call it in the code:

z = DIVIDE(1 + 1, 1 + 1);

As a result, I get:

z = 3;

And it should be:

z = 1;

What is the reason?

 0
c++
Author: Nicolas Chabanovsky, 2010-12-10
Source

3 answers

The fact is that the macro definition

#define DIVIDE(x, y) x / y

When calling

z = DIVIDE(1 + 1, 1 + 1);

Expanded by the preprocessor into the expression

z = 1 + 1/1 + 1;

Which gives

z = 3;

The macro is expanded first and then calculated, not the other way around.

 5
Author: stanislav, 2010-12-12 10:37:20

The problem is that you are using macros incorrectly.

  1. Take the contents of the macro in parentheses.
  2. Take all arguments in parentheses.

Then the macros will be executed correctly:

#define DIVIDE(x, y) ((x) / (y))

Get the result:

z = DIVIDE(1 + 1, 1 + 1) = ((1 + 1) / (1 + 1)) = (2 / 2) = 1;
 4
Author: Алексей Сонькин, 2011-02-24 22:37:49

And in general-it is better not to use macros, but to define your own template function for division.

template <class Type>
Type my_divide(Type x, Type y)
{
  return x / y;
}

Almost certainly, this function will be built-in, so that the overhead of calling it will be minimal. And at the same time, there will be no disadvantages inherent in macros.

 0
Author: gecube, 2011-06-12 10:45:15