replace a character element of an array for another character entered by user input

Question

I have an array full of * characters. I need to replace any character from the array for another character that the user enters. Is it possible? I'll appreciate any suggestions. Thanks

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

    strreplace(char[],char,char);
    int main()
    {
        char s[17]="* * * *\n* * * * ";
        char chr,repl_chr;
        printf("%s", s);
        printf("\nEnter character to be replaced: ");
        chr=getchar();
        fflush(stdin);
        printf("\nEnter replacement character: ");
        repl_chr=getchar();
        printf("\nModified string after replacement is: \n");
        strreplace(s,chr,repl_chr);
        getch();
        return 0;
   }



   strreplace(char s[], char chr, char repl_chr)
   {
      int i=0;
      while(s[i]!=' ')
      {
           if(s[i]==chr)
           {
               s[i]=repl_chr;
           }
           i++;
      }
      puts(s);
      return 0;
  }

To understand you right: you want to have **all** occurrences of the first entered character replaced by the second entered character? So for example if the user enters `*` to be replaced by `-` then the program should output `"- - - -\n- - - - "`, yes? — Jireugi, Oct 08 '16 at 15:43
But there´s the catch: since the char array only contains `*` available to get replaced (`strreplace` ignores spaces obviously), any other character that the user enters (e. g. `a`, `b`, `c`, `-`, ...) cannot get found in that array, and therefore nothing can get replaced. — Jireugi, Oct 08 '16 at 15:52
Not really my display should look like two rows and four columns of those - symbols. The use can replace any character in the array. Let's say the user wants to replace the third - with a B then it would look like this - - B - \n - - - -. He has to enter which element he wants to replace in the array for another character that he desires. — FruitPunchSamurai, Oct 08 '16 at 15:53
That´s a critical information of course: so the user does not have to specify what character to replace, but what the coordinate of that character in the source string is. — Jireugi, Oct 08 '16 at 15:57
Yes! is it possible to do this way? or is there an alternative way to do it? — FruitPunchSamurai, Oct 08 '16 at 16:07
It is possible. But normally there are also plenty of alternatives. — Jireugi, Oct 08 '16 at 16:17
That is, `fflush(stdin)` might work on some systems, but is strictly undefined in the standard and so is not portable. You can read about it [here](http://stackoverflow.com/questions/2979209/using-fflushstdin). — ad absurdum, Oct 08 '16 at 17:32

score 0 · Accepted Answer · answered Oct 08 '16 at 16:12

Change your strreplace so that the parameter chr gets converted into an index.

int strreplace(char s[], char chr, char repl_chr)
{
    int i = 0;

    // Convert the input char into an index.
    // Note that 1 is subtracted to make the index zero-based.
    // Remember that the user entered a one-based index.
    int index = chr - '0' - 1;
    if ((index < 0) || (index >= 8))
    {
        // Out of bounds.
        return -1;
    }
    while (s[i] != '\0')
    {
        if ((i/2) == index)
        {
            s[i] = repl_chr;
        }

        // Note: since every second char is to get ignored,
        // the increment must actually be 2 here.
        i += 2;
    }
    puts(s);
    return 0;
}

Entering `3` as first character and `B` as second character yields the output `* * B *\n* * * * `. — Jireugi, Oct 08 '16 at 16:28

score 0 · Answer 2 · answered Oct 08 '16 at 16:48

Try changing while(s[i]!=' ') in your loop to
while (s[i] != '\0') to check if you've reached the end of your string

int strreplace(char s[], char chr, char repl_chr);

 int main(void)
 {
     char s[17] = "* * * *\n* * * * ";
     char chr, repl_chr;

     printf("%s", s);

     printf("\nEnter character to be replaced: ");
     scanf("%c", &chr);

     printf("\nEnter replacement character: ");
     scanf(" %c", &repl_chr);
     printf("\nModified string after replacement is: \n");

     strreplace(s, chr, repl_chr);

     return 0;
 }

 int strreplace(char s[], char chr, char repl_chr)
 {
     int i = 0;

     while (s[i] != '\0') {
         if (s[i] == chr)
             s[i] = repl_chr;
         i++;
     }

     puts(s);
     return 0;
 }

Output:

* * * *
* * * * 
Enter character to be replaced: *

Enter replacement character: B

Modified string after replacement is: 
B B B B
B B B B

replace a character element of an array for another character entered by user input

2 Answers2