Step by Step solution about How to resolve the algorithm Selectively replace multiple instances of a character within a string step by step in the C programming language

This C code performs a series of character replacements on a given string using dynamic memory allocation and string manipulation functions.

• It includes the necessary C standard libraries <stdio.h>, <stdlib.h>, and <string.h> for input/output, memory management, and string handling.

• The main function defines a constant string string with the value "abracadabra". This string is the source text for the character replacements.

• It allocates memory for a new string replaced using malloc and copies the content of string into it. The size of replaced is determined by the length of string. This creates a mutable copy of the original string, allowing for modifications.

• It declares three null-terminated replacement character arrays: aRep, bRep, and rRep. These arrays contain the replacement characters for 'a', 'b', and 'r', respectively.

• The code enters a loop that iterates through each character c in the replaced string.

• Inside the loop, it uses a switch statement to check the current character *c.

• If *c is 'a', it checks if there are any characters left in aRep. If so, it replaces *c with the next character from aRep and advances aRep to the next character.

• Similarly, it checks for 'b' and 'r' and replaces them with characters from bRep and rRep, if available.

• After processing all characters in the replaced string, it prints the modified string using printf.

• Finally, it frees the dynamically allocated memory for replaced using free to prevent memory leaks.

When you run this program, it will output the modified string with 'a' replaced by 'ABaCD', 'b' by 'E', and 'r' by 'rF'. In this case, the output would be "AbrEcadrFbrF".

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

int main(void) {
  const char string[] = "abracadabra";

  char *replaced = malloc(sizeof(string));
  strcpy(replaced, string);

  // Null terminated replacement character arrays
  const char *aRep = "ABaCD";
  const char *bRep = "E";
  const char *rRep = "rF";

  for (char *c = replaced; *c; ++c) {
    switch (*c) {
    case 'a':
      if (*aRep)
        *c = *aRep++;
      break;
    case 'b':
      if (*bRep)
        *c = *bRep++;
      break;
    case 'r':
      if (*rRep)
        *c = *rRep++;
      break;
    }
  }

  printf("%s\n", replaced);

  free(replaced);
  return 0;
}