Question

HI ,

Can some one help me in understanding why the value of SQUARE(x) is 49 ?

I am using Visual C++ 6.0 .

#define SQUARE(X) X * X

int main(int argc, char* argv[])
{
    int y = 5;

    printf("%d\n",SQUARE(++y));
    return 0;
}

Answer 1

Neil Butterworth, Mark and Pavel are right.

SQUARE(++y) expands to ++y * ++y, which increments twice the value of y.

Another problem you could encounter: SQUARE(a + b) expands to a + b * a + b which is not (a+b)*(a+b) but a + (b * a) + b. You should take care of adding parentheses around elements when needed while defining macros: #define SQUARE(X) ((X) * (X)) is a bit less risky. (Ian Kemp wrote it first in his comment)

You could instead use an inline template function (no less efficient at runtime) like this one:

template <class T>
inline T square(T value)
{
    return value*value;
}

You can check it works:

int i = 2;
std::cout << square(++i) << " should be 9" << std::endl;
std::cout << square(++i) << " should be 16" << std::endl;

(no need to write

square<int>(++i)

because the int type is implicit for i)