Solution

The correct answer is 90 (Option 2)

Key Points

• The expression on the right side of c = ( ... ) uses the comma operator. It evaluates sub-expressions from left to right and returns the value of the last sub-expression.
• Assignments inside that comma expression change a and b step by step; the final value (of the last assignment) is stored in c.
• --(*i) is a pre-decrement on the value pointed by i. Since i = &c, it decreases c by 1 and then prints it.

Given code (formatted):

void main() {
    int *i, a = 12, b = 2, c;
    c = ( a = a + b,    /* step 1 */
          b = a / b,    /* step 2 */
          a = a * b,    /* step 3 */
          b = a - b );  /* step 4 (value of comma expr) */
    i = &c;
    printf("%d", --(*i));
}

Step-by-step evaluation of the comma expression:

1. a = a + b → a = 12 + 2 = 14
2. b = a / b → b = 14 / 2 = 7 (integer division)
3. a = a * b → a = 14 * 7 = 98
4. b = a - b → b = 98 - 7 = 91
   The comma expression’s value is the value of this last assignment, i.e., 91. Therefore, c = 91.

Printing: i = &c so *i refers to c. --(*i) pre-decrements c to 90 and then prints it.

Important Points

• Comma operator: Evaluates operands left→right; only the last value is returned, but all side effects occur.
• Pre vs Post decrement: --x decrements then yields the new value; x-- would print the old value and decrement later.

Why other options are wrong:

• 91 – would be printed if there were printf("%d", *i--) or no decrement; but here we pre-decrement to 90.
• 98 – comes from the intermediate value of a (step 3), not what we store in c.
• 92 – results from an incorrect arithmetic/ordering assumption; correct final c is 91, then decremented to 90.