 Open in App
Not now

# C | Operators | Question 10

• Difficulty Level : Medium
• Last Updated : 10 Nov, 2020

What is the output of following program?

 `#include ` ` `  `int` `main() ` `{ ` `   ``int` `a = 1; ` `   ``int` `b = 1; ` `   ``int` `c = a || --b; ` `   ``int` `d = a-- && --b; ` `   ``printf``(``"a = %d, b = %d, c = %d, d = %d"``, a, b, c, d); ` `   ``return` `0; ` `} `

(A) a = 0, b = 1, c = 1, d = 0
(B) a = 0, b = 0, c = 1, d = 0

(C) a = 1, b = 1, c = 1, d = 1
(D) a = 0, b = 0, c = 0, d = 0

Explanation: Let us understand the execution line by line.
Initial values of a and b are 1.

```   // Since a is 1, the expression --b
// is not executed because
// of the short-circuit property
// of logical or operator
// So c becomes 1, a and b remain 1
int c = a || --b;

// The post decrement operator --
// returns the old value in current expression
// and then updates the value. So the
// value of expression a-- is 1.  Since the
// first operand of logical and is 1,
// shortcircuiting doesn't happen here.  So
// the expression --b is executed and --b
// returns 0 because it is pre-increment.
// The values of a and b become 0, and
// the value of d also becomes 0.
int d = a-- && --b;
```
My Personal Notes arrow_drop_up
Related Articles