Strip whitespace from a string in-place?

Question

I saw this in a "list of interview questions". Got me wondering.

Not limited to whitespace necessarily, of course, easily generalized to "removing some specific character from a string, in-place".

My solution is:

void stripChar(char *str, char c = ' ') {
  int x = 0;
  for (int i=0;i<strlen(str);i++) {
    str[i-x]=str[i];
    if (str[i]==c) x++;
  }
  str[strlen(str)-x] = '\0';
}

I doubt there is one more efficient, but is there a solution that is more elegant?

edit: totally forgot I left strlen in there, it is most definitely not efficient

Answer 1

C doesn't have default arguments, and if you're programming in C++ you should use std::string and remove_if from <algorithm>.

You definitely can make this more efficient, by eliminating the calls to strlen, which are turning an O(N) algorithm into an O(N²) algorithm, and are totally unnecessary -- you're scanning the string anyway, so just look for the NUL yourself.

You can also make this more C-idiomatic, by using two pointers instead of array indexing. I'd do it like this:

void strip_char(char *str, char strip)
{
    char *p, *q;
    for (q = p = str; *p; p++)
        if (*p != strip)
            *q++ = *p;
    *q = '\0';
}

Answer 2

First of all, i<strlen(str) is always an inefficient idiom for looping over a string. The correct loop condition is simply str[i], i.e. loop until str[i] is the null terminator.

With that said, here's the simplest/most concise algorithm I know:

for (size_t i=0, j=0; s[j]=s[i]; j+=!isspace(s[i++]));

Note: My solution is for the question as written in the subject (whitespace) as opposed to the body (particular character). You can easily adapt it if needed.