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# Output of C Program | Set 28

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.

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