How to remove all occurrences of a given character from string in C?

Question

I'm attempting to remove a character from a string in C. The problem I am having with my code is that it removes the first instance of the character from the string but also wipes everything after that character in the string too. For example, removing 'l' from 'hello' prints 'he' rather than 'heo'

int i; char str1[30] = "Hello", *ptr1, c = 'l'; ptr1 = str1; for (i=0; i<strlen(str1); i++) {     if (*ptr1 == c) *ptr1 = 0;     printf("%c\n", *ptr1);     ptr1++; } 

I need to use pointers for this and would like to keep it as simple as possible since I'm a beginner in C. Thanks

Answer 1

You can do it like this:

void remove_all_chars(char* str, char c) {     char *pr = str, *pw = str;     while (*pr) {         *pw = *pr++;         pw += (*pw != c);     }     *pw = '\0'; }  int main() {     char str[] = "llHello, world!ll";     remove_all_chars(str, 'l');     printf("'%s'\n", str);     return 0; } 

The idea is to keep a separate read and write pointers (pr for reading and pw for writing), always advance the reading pointer, and advance the writing pointer only when it's not pointing to a given character.

Answer 2

If you remove the characters in place you will have to shift the rest of the string one place to the left every time you remove a character, this is not very efficient. The best way is to have a second array that takes the filtered string. For example you can change your code like this.

int i; char str1[30] = "Hello", *ptr1, c = 'l'; char str2[30] = {0}, *ptr2; ptr1 = str1; ptr2 = str2; for (i=0; i<strlen(str1); i++) {     if (*ptr1 != c) *ptr2++=*ptr1;     ptr1++; } printf("%s\n", str2);