Related Articles
Output of C Program | Set 28
• Difficulty Level : Easy
• Last Updated : 27 Dec, 2016

Predict the output of following C program.

Question 1

 `#include `` ` `int` `main()``{``    ``char` `a = 30;``    ``char` `b = 40;``    ``char` `c = 10;``    ``char` `d = (a * b) / c;``    ``printf` `(``"%d "``, d);`` ` `    ``return` `0;``}`

At first look, the expression (a*b)/c seems to cause arithmetic overflow because signed characters can have values only from -128 to 127 (in most of the C compilers), and the value of subexpression ‘(a*b)’ is 1200. For example, the following code snippet prints -80 on a 32 bit little endian machine.

```    char d = 1200;
printf ("%d ", d);```

Arithmetic overflow doesn’t happen in the original program and the output of the program is 120. In C, char and short are converted to int for arithmetic calculations. So in the expression ‘(a*b)/c’, a, b and c are promoted to int and no overflow happens.

Question 2

 `#include``int` `main()``{``    ``int` `a, b = 10;``    ``a = -b--;``    ``printf``(``"a = %d, b = %d"``, a, b);``    ``return` `0;``}`

Output:

```a = -10, b = 9
```

The statement ‘a = -b–;’ compiles fine. Unary minus and unary decrement have save precedence and right to left associativity. Therefore ‘-b–‘ is treated as -(b–) which is valid. So -10 will be assigned to ‘a’, and ‘b’ will become 9.
Try the following program as an exercise.

 `#include``int` `main()``{``    ``int` `a, b = 10;``    ``a = b---;``    ``printf``(``"a = %d, b = %d"``, a, b);``    ``return` `0;``}`

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up