Inability to understand the output of a c program

Question

I have come across a c program in a interview question .The program is as follow:

#include<stdio.h>
#define x 9+2/4*3-2*4+(5-4)*3
void main()
{

   int i,y,p;
   y=6+3*3/5;
   printf("%d",x);
   i=x*x+y;
   printf("%d",i);

}

The output of this program is 30.But I can't understand how the program produce 30.Can anyone explain ?

Answer 1

#define, in your specific case, does simple text replacement.

First expression (for x) is

9+2/4*3-2*4+(5-4)*3
9+ 0 *3-2*4+  1  *3
9+  0  - 8 +    3
  9    - 8 +    3
     1     +    3
          4      

Second expression (for i) is

9+2/4*3-2*4+(5-4)*3*9+2/4*3-2*4+(5-4)*3+y // y is 7
9+2/4*3-2*4+(5-4)*3*9+2/4*3-2*4+(5-4)*3+7 // doing parenthesis
9+2/4*3-2*4+  1  *3*9+2/4*3-2*4+  1  *3+7 // doing multiplications and divisions
9+ 0 *3- 8 +    3  *9+ 0 *3- 8 +    3  +7 // doing multiplications and divisions again
9+   0 - 8 +    27   +  0  - 8 +    3  +7 // doing addition and subtraction left to right
   9   - 8 +    27   +  0  - 8 +    3  +7
       1   +    27   +  0  - 8 +    3  +7
            28       +  0  - 8 +    3  +7
                28         - 8 +    3  +7
                      20       +    3  +7
                               23      +7
                                   30

When writing #defines with operations always enclose the whole thing inside an extra set parenthesis:

#define x (9+2/4*3-2*4+(5-4)*3)

Answer 2

Integer division is part of the answer, but the trickier part is the #define. Many people assume that a #define is evaluated when it is defined, but it is really just a simple text substitution. So when you are given x*x, you get

9+2/4*3-2*4+(5-4)*3*9+2/4*3-2*4+(5-4)*3

Note that is the middle of this you have 3*9, which you would not get if you assumed x was a simple numeric value. Heres the whole process:

  x * x
= 9+2/4*3-2*4+(5-4)*3*9+2/4*3-2*4+(5-4)
= 9 +  0 - 8 +    27   +  0  - 8 +  3
=        1   +    27   -     5
=              23

Then, since y=7, 23 + 7 = 30

Voila