README.md

Assignment Zero: Base Conversion

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 <decimal> <base>
$ ./conv 11111 65
INVALID BASE
  1. Your program should be able to handle all bases from 2 to 64 and 256.
  2. 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.
  3. 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).
  4. Attempt to handle errors. Another problem could be an invalid base (see example output).
  5. Use a buffer of 32 characters. This will give you 31 locations for characters and the final location should be a null.
  6. 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:

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!