Question

I have a text file which contains lines, such that every line contains no more than 80 chars. every line contains (one or more words) divided by commas or spaces. I need lexicography sort the words. I want to use scanf with the symbol “%s” of strings, but it considers only spaces. So I guess I can’t. any smarter way to deal with the parsing the words than get char by char?

Here’s a sketch of what I was planning to do:

   char**arr;
    arr=calloc(Size, sizeof(char)*80);
    int m=0;

    while (!feof(file)) {

        char c=fgetc(file);
        while (c!='/n') {
             j=0;
            char* current;
            current=calloc(1,sizeof(char)*80);
            while (c!=','&& c!=' ' && c!='/n') {
                current[j]=c;
                j++;
                c=fgetc(c);
            }
            current[j]='\0';
            arr[m]=current;

            free(current);

        }
    }

Answer 1

Assuming you have an entire line in a suitable variable (line, below), you could use something like:

const char *ptr = line;
char token[80];

while(*ptr != '\0' && sscanf(ptr, "%79[^, ]", token) == 1)
{
  ptr += strlen(token);
  while(*ptr == ',' || *ptr == ' ')
    ++ptr;
}

This uses the (rather under-utilized, in my opinion) character group format %[] to grab characters until a comma or space is found, then skip past the parsed token, and any separators that follow.