Question

I am using this question as a guide to trimming a string in C. It works properly on a string bounded exclusively by whitespaces (' '), but on special whitespaces ('\r', '\n', '\t', etc.), it fails. Here’s an example:

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

size_t trim(char *out, size_t len, const char *str)
{
  if(len == 0)
    return 0;

  const char *end;
  size_t out_size;

  // Trim leading space
  while(isspace(*str)) str++;

  if(*str == 0)  // All spaces?
  {
    *out = 0;
    return 1;
  }

  // Trim trailing space
  end = str + strlen(str) - 1;
  while(end > str && isspace(*end)) end--;
  end++;

  // Set output size to minimum of trimmed string length and buffer size minus 1
  out_size = (end - str) < len-1 ? (end - str) : len-1;

  // Copy trimmed string and add null terminator
  memcpy(out, str, out_size);
  out[out_size] = 0;

  return out_size;
}

int main(){

    char *str = " \n\n  hello  \t    \r  ";
    char trimmed[strlen(str)];

    trim (trimmed, strlen(trimmed), str);
    printf("~%s~\n~%s~\n", str, trimmed);

    return 0;
}

produces the output:

~ 

  ~ello         
~~

can anyone fix the code to trim all whitespace characters properly?

Second Question: The first function in the referenced answer gives me a segfault. does anyone know why this is the case?

Answer 1

The way you called the function is incorrect.

char *str = " \n\n  hello  \t    \r  ";
char trimmed[strlen(str)+1]; // Note that you must +1 for the terminating \0.

// Use sizeof() instead of strlen() because trimmed is containing garbage.
// strlen() measures the length of the content while sizeof() measure the allocated size of the array.
trim (trimmed, sizeof(trimmed), str);