why macros return what we are not expected

Question

Possible Duplicate:
Problem with Macros

Hi all

I have defined this macro:

#define SQ(a) (a*a)

and called it in this way:

std::cout << SQ(2+3) << '\n';

the output was 11. Why ?

Thanks

Answer 1

Because the expansion of SQ in your example gives:

std::cout << (2+3*2+3) << '\n';

A better way to define the macro would be

#define SQ(a) ((a)*(a))

which solves the precedence issue in this case.

Better still would be to use a function which avoids any issues with the passed expression being evaluated more than once.

E.g. as a template:

template<class T>
T SQ(T a) { return a * a; }

Answer 2

Macros do only simple text manipulation, i.e. they are very stupid that way in that they don't see 'code', only 'text' which are then sent to to the C/C++ parser for validation.

SQ(2+3) becomes (2+3*2+3)

That's why you should use templates, which are a bit smarter, they would do what you expected: first calculate 2+3, then do 5*5.

Answer 3

Fix your macro to:

#define SQ(a) ((a)*(a))

Answer 4

Others already have the answer. The "fix" (if you must use macros at all) is to wrap all your params with parens, e.g.

#define SQ(a) ((a)*(a))

In most cases, you're better off using templates as this will still perform compile-time expansion for speed, but also provides language support for syntax and type checking.