# Output of C programs | Set 55 (Shift Operators)

**Prerequisite:** Shift operators

**Q.1** What Is The Output Of this program?

`#include <stdio.h>` `int` `main()` `{` ` ` `unsigned ` `int` `i = 0x80;` ` ` `printf` `(` `"%d "` `, i << 1);` ` ` `return` `0;` `}` |

**Option**

a) 0

b) 256

c) 100

d) 80

ans :-b

Explanation :- We know that 0x is hexa-decimal representation of number so 80 converted in decimal is 128 binary(10000000), its left shift 1 so it is (100000000)equal to 256.

**Q.2** What Is The Output Of this program?

`#include <stdio.h>` `int` `main()` `{` ` ` `unsigned ` `int` `a = 0xffff;` ` ` `unsigned ` `int` `k = ~a;` ` ` `printf` `(` `"%d %d\n"` `, a, k);` ` ` `return` `0;` `}` |

**Option**

a) 65535 -65536

b) -65535 65535

c) 65535 65535

d) -65535 -65535

ans :-a

Explanation :- This code 0x is Hexadecimal representation of number so ffff hexadecimal is converted into Decimal 65535 and stored in a. The negation of a -65535 is stored in k. k is an unsigned integer but it has a negative value in it. When k is being printed the format specifier is %d which represents signed integer thus -65536 gets printed as value of k. If the format specifier for k had been %u the value that would have been printed is 4294901760.

**Q.3** What Is The Output Of this program?

`#include <stdio.h>` `int` `main()` `{` ` ` `unsigned ` `int` `a = -1;` ` ` `unsigned ` `int` `b = 4;` ` ` `printf` `(` `"%d\n"` `, a << b);` ` ` `return` `0;` `}` |

**Option**

a) -1

b) 16

c) 8

d) -16

ans :-d

Explanation :- Here a integer is simply left-shifted 4 bit and print so it is -16.

**Q.4** What Is The Output Of this program?

`#include <stdio.h>` `int` `main()` `{` ` ` `unsigned ` `int` `m = 32;` ` ` `printf` `(` `"%d %d"` `, m >> 1, ~m);` ` ` `return` `0;` `}` |

**Option**

a) 16 5

b) 64 -32

c) 16 33

d) 16 -33

ans :-d

Explanation :- This program 32 is(10000) and we know that negative number in binary system contain a one’s complement so here take a complement of a number and only print this number so the negation of 32 is -33.

**Q.5** What Is The Output Of this program?

`#include <stdio.h>` `int` `main()` `{` ` ` `unsigned ` `int` `a = -1;` ` ` `unsigned ` `int` `b = 15;` ` ` `printf` `(` `"%d, "` `, a << 1);` ` ` `printf` `(` `"%d\n"` `, b >> 1);` ` ` `return` `0;` `}` |

**Option**

a) -2, 7

b) 2, 8

c) -1, 15

d) -2, 15

ans :-a

Explanation :- Here -1 binary(0001) if left shift 1 so it is -2(0010) and 15 binary(1111) is right shift 1 so it is (0111)so value 7.

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