how can i convert my alphabetical index back to Ascii code in C?

Question

string plaintext = get_string("plaintext;");
  
  for (size_t k = 0; k<strlen(plaintext); k++)
  {
    int n = atoi(argv[1]);
    int m = plaintext[k] -'a' + 1;
    printf("%d", (m + n) % 26);
  }

in this piece of code i'm trying to cipher text that is inputted by the user. this: (m + n) % 26 ( m is i'th char of plaintext & n is key) is the formula i'm supposed to use.

i tried printing this:

printf("%d", (m + n) % 26);

as a char, but it won't print out. when i tried printing it out as an int as shown above, it outputted a number that is not in Ascii value (for ex.: 1294). i figured out that converting my index back to Ascii value would solve the problem, but i don't know how to do that.

please elaborate in your answer!

What is the purpose of this? `int m = plaintext[k] -'a' + 1;` — Fiddling Bits, Jun 26 '20 at 17:49
`(m + n) % 26` will result in a value from `0` to `25`. In ASCII, none (or few) of these are printable. — Fiddling Bits, Jun 26 '20 at 17:50
You can take this out of the loop as it only needs to be performed once. `int n = atoi(argv[1]);` — Fiddling Bits, Jun 26 '20 at 17:51
You probably want to add `n` to `plaintext[k]` and then subtract `26` until it's within the valid range. — Fiddling Bits, Jun 26 '20 at 17:52

score 2 · Accepted Answer · answered Jun 26 '20 at 17:56

2

If this is a generic ROTN type thing then converting back is easy:

printf("%c", (m + n) % 26 + 'a' - 1);

Where you just have to "undo" what you did earlier.

answered Jun 26 '20 at 17:56

tadman

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how can i convert my alphabetical index back to Ascii code in C?

1 Answers1