Re: [Groff] Readable tbl-source

[Ralph Corderoy]

Hi Werner,

> Indeed, a very nice idea, but no longer necessary: In the CVS, I've
> just added a new global option `nospaces' to make tbl ignore leading
> and trailing spaces in data fields.

Sounds good.  I'm not up to speed with C++ but something looks a bit
odd.

         1  char *string::extract() const
         2  {
         3    char *p = ptr;
         4    int n = len;
         5    int nnuls = 0;
         6    int i;
         7    for (i = 0; i < n; i++)
         8      if (p[i] == '\0')
         9        nnuls++;
        10    char *q = new char[n + 1 - nnuls];
        11    char *r = q;
        12    for (i = 0; i < n; i++)
        13      if (p[i] != '\0')
        14        *r++ = p[i];
        15    q[n] = '\0';
        16    return q;
        17  }

Does `new char[n]' give memory pre-filled with '\0'?  I guess not
because #15 puts the terminating '\0' in place.  But if on entry ptr ==
"a\0c\0" and len == 3 then #7 loops through "a\0c" and by #10 nnuls ==
1.  This means `n + 1 - nnuls' == 3 which is correct to hold "ac\0".
But won't #15 be writing to q[3] which is past the end of q?  Shouldn't
it be `*r = '\0';'?

Assuming I'm right that then...

        19  void string::remove_spaces()
        20  {
        21    int l = len - 1;
        22    while (l >= 0 && ptr[l] == ' ')
        23      l--;
        24    char *p = ptr;
        25    if (l > 0)
        26      while (*p == ' ') {
        27        p++;
        28        l--;
        29      }
        30    if (len - 1 != l) {
        31      if (l >= 0) {
        32        len = l + 1;
        33        char *tmp = new char[len];
        34        memcpy(tmp, p, len);
        35        a_delete ptr;
        36        ptr = tmp;
        37      }
        38      else {
        39        len = 0;
        40        if (ptr) {
        41          a_delete ptr;
        42          ptr = 0;
        43        }
        44      }
        45    }
        46  }

given ptr == "abc \0" and len == 4 by #33 len == 3 and #34 will copy
"abc" without a '\0' character on the end.

The two routines don't seem to match in terms of what ptr holds, a \0
terminated string or not.

As I said, I could be way off beam on this.

Cheers,


Ralph.