How to return to the beginning of the line?

Question

C++ / C#

We need to encrypt argv[1] with argv[2]. There are 2 user-supplied arguments, both arguments are strings. Accordingly, the first argument receives a conditional text that will need to be encrypted, for example "ENCODE ME", the second argument is the very key with which the encoding takes place, for example "KEY". So, I would like "KEY" to be applied character by character and not stop working after a single execution. Due to lack of experience, it is not possible to do exactly this, tell me how you can implement it. I tried a lot of options, tried to make a loop so that after the word ends, return to the null character and run it again, but it didn’t work out, because I can’t think of anything to replace in C, indexOf in JS.

for (int i = 0; i < argv[1][i]; i++)
  {
    if (isalpha(argv[1][i]))
    {
      if (argv[1][i] >= 'A' && argv[1][i] <= 'Z')
      {
        int shift = argv[2][i] - 65;

        printf("%c + %c '%.3d' == ", argv[1][i], argv[2][i], shift);

        char rightNumber = argv[1][i] + shift;

        if (rightNumber > 'Z')
          rightNumber -= 26;

        printf("%c\n", rightNumber);
      }
    }
  }

Answer 1

Replace all argv[2][i] with argv[2][i % strlen(argv[2])] or create a separate variable under a new index.
It's standard practice to go around in circles - taking the remainder of the index divided by the range.
PS. The condition in the loop i < argv[1][i] is very dubious.

Answer 2

#include <string.h>
#include <stdio.h>
#include <ctype.h>

int main(int argc, char *argv[])
{
    int incode_len = strlen(argv[1]);
    int key_len = strlen(argv[2]);
    for (int i = 0, k = 0; i < incode_len; i++, k++)
    {
        if( k >= key_len) k = 0;
        if (isalpha(argv[1][i]))
        {
            if (argv[1][i] >= 'A' && argv[1][i] <= 'Z')
            {
                int shift = argv[2][k] - 65;
                printf("%c + %c '%.3d' == ", argv[1][i], argv[2][k], shift);
                char rightNumber = argv[1][i] + shift;
                if (rightNumber > 'Z') rightNumber -= 26;
                printf("%c\n", rightNumber);
            }
        }
    }
}

Answer 3

Cyclic String Code Example

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
    const char *s, *start, *end, *cur;
    int n, i;

    if (argc == 3) {
        s = argv[1];
        n = atoi(argv[2]);
    } else {
        s = "12345";
        n = 50;
    }

    start = end = s;
    if (*end)
        while (*(end + 1))
            end++;

    cur = start;
    for (i = 0; i < n; i++) {
        putchar(*cur);
        cur = (cur < end) ? cur + 1 : start;
    }
    putchar('\n');

    return 0;
}

Conclusion

[[email protected] c]$ .ansi t.c -o t
[[email protected] c]$ ./t
12345123451234512345123451234512345123451234512345
[[email protected] c]$ ./t abc 80
abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcab
[[email protected] c]$ ./t a 10
aaaaaaaaaa
[[email protected] c]$ ./t ab 10
ababababab
[[email protected] c]$ ./t abc 10
abcabcabca
[[email protected] c]$

Answer 4

#include <string.h>
#include <stdio.h>

int main(int argc, char *argv[])
{
  unsigned char *text, *key;
  unsigned char shift, rightNumber;
  for (text = argv[1], key = argv[2]; *text; text++, key++) {
    if (0 == *key)
      key = argv[2];
    if (*text >= 'A' && *text <= 'Z') {
      shift = *key - 'A';
    rightNumber = *text + shift;
    if (rightNumber > 'Z')
      rightNumber -= ('Z' - 'A');
    printf(""%c + %c '%.3d' == "%c\n", *text, *key, shift, rightNumber);
  }
}