Description
You are to write a simple C program that will take a decimal number and a base as command line arguments and output the decimal number in that specified base. For example:
$ ./conv 13 2
1101
$ ./conv 236709 8
716245
$ ./conv 1352275 11
843A91
$ ./conv 38572856 64
2J9Cu
$ ./conv 3232236286 256
192.168.2.254
$ ./conv
Usage: conv
$ ./conv 11111 65
INVALID BASE
1. Your program should be able to handle all bases from 2 to 64 and 256.
1. For bases that are powers of 2 (2, 4, 8, 16, 32, 64 and 256) you should use a **mask and shift** algorithm. See below.
1. Please name your code file `conv.c`. This will be the only file you submit to me (via the git server, we will get you connected to this on Thursday, _so come prepared_).
1. Attempt to handle errors. Another problem could be an invalid base (see example output).
1. Use a buffer of **32 characters**. This will give you 31 locations for characters and the final location should be a null.
1. Remember: you have to put the digits into the buffer _backwards_ and when the algorithm is finished, you have to _printf from the correct location in that buffer_. See the example we did in class for how to accomplish that.
### Converting Integers to Characters
Think about how to use this character array:
“`c
char *ascii = “0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz+/”;
“`
You can declare this at the top of your file along with your buffer. How can you use this to accomplish a conversion to bases 11 to 64? Think about how to do this! It is just an array of characters …
### Masking & Shifting
Along with the algorithm we saw in class of modulo followed by division, you can accomplish the same result on bases that are powers of two _by masking off the number of bits you need and then shifting the bits down_. Remember how the `&` operator works? What if you are trying to convert the number 7 into base 4? How many bit patterns are there for base 4: `00`, `01`, `10` and `11` which is 0, 1, 2 and 3. Let’s see how masking works:
0111 = 7
& 0011 = 3
__________
11 = 3
So, you’ve “masked off” the first digit by using the `&` (bitwise and) operator instead of modulo. Now you can shift those bits to the right instead of using the division operator like so:
0111 >> 2 = 01
And continue with the algorithm:
01 = 1
& 11 = 3
________
01 = 1
So the number 7 in base 4 is: 13!
_You must use this algorithm for the specified bases!_
