A conversion program is used to store data in a computer.

When I had nothing to do, I found a program design question to practice: Write an application and read an integer (that is, a binary integer) that only contains 0 and 1 ), then, the following decimal integer is printed. The modulo and division operators are required. So I wrote the following two methods:

Code:

1. Public static long readbinarynum ()
2. {
3. Repeated scan = new partition (system. In );
4. String binarynumstr = "";
5. Boolean available = false;
7. Do {
8. System. Out. Print ("enter a binary INTEGER :");
9. Binarynumstr = scan. nextline ();
10. Available = pattern. Matches ("[0-1] +", binarynumstr );
11. If (! Available)
12. {
13. System. Out. println ("non-binary data, this input is invalid! ");
14. Continue;
15. }
16. If (binarynumstr. Length ()> = 19)
17. {
18. System. Out. println ("the data is too big. This input is invalid! ");
19. Available = false;
20. Continue;
21. }
22. } While (! Available );
24. Return long. parselong (binarynumstr );
25. }
27. /**
28. * Convert the binary value to the corresponding decimal value.
29. * @ Param binarynum the binary value to be converted
30. * @ Return refers to the corresponding decimal value.
31. */
32. Public static long binarynum2decimalnum (long binarynum)
33. {
34. System. Out. println (binarynum );
35. Long decimalnum = 0;
36. Int position [] = new int [(INT) math. log10 (binarynum) + 1];
38. Position [0] = (INT) (binarynum/Math. Pow (10, position. Length-1 ));
39. For (INT I = 1; I <position. length; I ++)
40. {
41. Position [I] = (INT) (binarynum % (math. Pow (10, position. Length-I ))
42. /(Math. Pow (10, position. Length-I-1 )));
43. }
45. For (INT I = 0; I <position. length; I ++)
46. {
47. Decimalnum + = position [position. length-i-1] * Math. Pow (2, I );
48. System. Out. Print (position [I] + "");
49. }
50. System. Out. println ();
52. Return decimalnum;
53. }

During the test, I found that if the input data is 11111111111111111 (that is, 17 1), the program will encounter a problem during conversion, the result is 131072 (and the correct result should be 131071), so I tested it again. The final problem is found in the remainder operation:

Code:

1. Long n = 111111111111111l; // 17 1
2. System. Out. println (N % math. Pow (10, 1 ));
3. System. Out. println (N %10.0 );
4. System. Out. println (N % 10 );

The program runs and the output result is:

Code:

1. 2.0
2. 2.0
3. 1

When I see this, I think this should be related to the storage structure of the numeric value in the computer or the division operation mechanism, but I am not quite sure what is going on here! So I wrote this article to keep you updated. I hope you will discuss it and give me some advice!