What is the output of the following C program?\n```c\n# include <stdio.h>\nvoid main (void)\n{ \n int shifty;\n shifty = 0570;\n shifty = shifty >> 4;\n shifty = shifty << 6;\n printf("The value of shifty is %o", shifty);\n}\n```
- A. The value of shifty is 15 00
- B. The value of shifty is 4300
- C. The value of shifty is 5700
- D. The value of shifty is 2700
Answer: D. The value of shifty is 2700
Explanation: 0570 is octal.\nOctal 570 = Binary `101 111 000` (9 bits). $000 000 001 011 110 000$ (16 bit).\n`>> 4`: `000 000 000 001 011 110`.\n`<< 6`: `000 010 111 100 000 000`.\nConvert back to octal: Split 3 bits: `010 111 100 000`. \n2 7 4 0? \nWait. $0570_8$. In binary: 101 111 000.\nRight shift 4: `000 010 111` ($27_8$).\nLeft shift 6: `010 111 000 000`. \nOctal: 010 -> 2, 111 -> 7, 000 -> 0, 000 -> 0. Result $2700_8$.