Macro expansion in context of arithmetic expression?

Question

I saw this below code in an website. I could not able to understsnd how the result is coming as 11, instead of 25 or 13.

Why I am thinking 25 because SQ(5) 5*5

or 13 because

SQ(2) = 4;

SQ(3) = 9;

may be final result will be 13 (9 + 4) But surprised to see result as 11. How the result is coming as 11?

using namespace std;
#define SQ(a) (a*a)
int main()
{
    int ans = SQ(2 + 3);
    cout << ans << endl;
system("pause");
}

Answer 1

The preprocessor does a simple text substitution on the source code. It knows nothing about the underlying language or its rules.

In your example, SQ(2 + 3) expands to (2 + 3*2 + 3), which evaluates to 11.

A more robust way to define SQ is:

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

Now, SQ(2 + 3) would expand to ((2 + 3)*(2 + 3)), giving 25.

Even though this definition is an improvement, it is still not bullet-proof. If SQ() were applied to an expression with side effects, this could have undesired consequences. For example:

• If f() is a function that prints something to the console and returns an int, SQ(f()) would result in the output being printed twice.
• If i is an int variable, SQ(i++) results in undefined behaviour.

For further examples of difficulties with macros, see Macro Pitfalls.

For these reasons it is generally preferable to use functions rather than macros.