Square of a number being defined using #define

Question

I was just going through certain code which are frequently asked in interviews. I came up with certain questions, if anyone can help me regarding this?

I am totally

Answer 1

define is just a text macro

main()
{
      int i,j;
      i=4/ 4 * 4;  // 1 * 4
      j=64/4 * 4; // 16 * 4
      printf("\n %d",i);
      printf("\n %d",j);
      printf("\n %d",square(4));
      getch();
}

Answer 2

It's a macro! So it returns exactly what it substitutes.

i = 4/4*4;   Which is 4...
j = 64/4*4;   Which is 16...

Try this for your macro:

#define square(x) ((x)*(x))

Answer 3

Operator precedence is hurting you.

The macro is expanded by the pre-processor such that

  i=4/4*4;
  j=64/4*4;

which is equivalent to:

  i=(4/4)*4;
  j=(64/4)*4;

Answer 4

j = 4/square(4) == 4/4*4 == 1*4 == 4

Answer 5

square is under-parenthesized: it expands textually, so

#define square(x) x*x
   ...
i=4/square(4);

means

i=4/4*4;

which groups as (4/4) * 4. To fix, add parentheses:

#define square(x) ((x)*(x))

Still a very iffy #define as it evaluates x twice, so square(somefun()) calls the function twice and does not therefore necessarily compute a square but rather the product of the two successive calls, of course;-).

Answer 6

When you write i=4/square(4), the preprocessor expands that to i = 4 / 4 * 4.
Because C groups operations from left to right, the compiler interprets that as i = (4 / 4) * 4, which is equivalent to 1 * 4.

You need to add parentheses, like this:

#define square(x) ((x)*(x))

This way, i=4/square(4) turns into i = 4 / ((4) * (4)).
You need the additional parentheses around x in case you write square(1 + 1), which would otherwise turn into 1 + 1 * 1 + 1, which is evaluated as 1 + (1 * 1) + 1, or 3.