String in C: replace characters with each other

Question

For example, I choose a string "sdoeoeeoasd". I should replace all the 'o' with 'e' and vice versa: "sdeoeooeasd". I intend to do this with the string lib.

I found a pretty similar question (Standard function to replace character or substring in a char array?), but I don't get how to make the code work according to my condition. It happens that the first occurence of the character is replaced, and then the only one character replaces:

char* replace_char(char* str, char find, char replace){
    char temp = find;
    char *current_pos = strchr(str,find); 
    char *current_posc2 = strchr(str,replace); 

    while (current_pos) {
        *current_pos = replace;
        current_pos = strchr(current_pos,find);
    }
        while (current_posc2) {
        *current_posc2 = find; 
        current_posc2 = strchr(current_posc2, replace);
    }
    return str;
}

With c1='e' and c2='o' I get:

I have a thought about adding the third temp variable, but my suggestions of its implementation were wrong and didn't work.

try one loop: `char *replace_char(char *str, char find, char replace) { char *str2 = str; while (*str2) { if (*str2 == find) *str2 = replace; else if (*str2 == replace) *str2 = find; str2++; } return str; }` — pmg, May 06 '20 at 09:38
Please, write your comment in the form of answer lower, in order to mark your code as the one that resolved my question. — Darya Balu, May 06 '20 at 12:14

score 3 · Answer 1 · answered May 06 '20 at 09:37

3

Another solution:

int i=0;
while(ch[i])
{
if(ch[i] == 'e')
ch[i]='o';
else if(ch[i] == 'o')
ch[i]='e';
i++;
}

answered May 06 '20 at 09:37

Bouraoui Al-Moez L.A

679
6
16

score 1 · Answer 2 · answered May 06 '20 at 09:42

Your code doesn't work because it replaces all of the first character with the second character, then replaces all of the second character with the first character. It completely undoes the work of the first step in the second step. It's much simpler to just iterate through the string one character at a time, like in Bouraoui Al-Moez L.A's code.

score 1 · Accepted Answer · answered May 06 '20 at 12:21

Do one loop:

char *replace_char(char *str, char find, char replace) {
    char *str2 = str; // save str for return value
    while (*str2) {
        if (*str2 == find) *str2 = replace;
        else if (*str2 == replace) *str2 = find;
        str2++;
    }
    return str;
}

score 0 · Answer 4 · answered May 10 '20 at 09:34

there is no need to use two loops

char *replace_char(char *str, char find, char replace) {
    char *ptr = str;
    while (*ptr) {
        if (*ptr == find) *ptr = replace;
        else if (*ptr == replace) *ptr = find;
        ptr++;
    }
    return str;
}

also you can use syntax array

int j=0;
while(str[j] !='\0') // will loop on your string till the end of it
{
if(str[j] == 'e')  // if char=e will be o
str[j]='o';
else if(str[j] == 'o')  // if char=o will be e
str[j]='e';
j++;
}
```````````````````````

String in C: replace characters with each other

4 Answers4