Method RemoveLeadingString

extern/re2/re2/parse.cc:798–869

Removes the first n leading runes from the beginning of re. Edits re in place.

// Removes the first n leading runes from the beginning of re.
// Edits re in place.
void Regexp::RemoveLeadingString(Regexp* re, int n) {
  // Chase down concats to find first string.
  // For regexps generated by parser, nested concats are
  // flattened except when doing so would overflow the 16-bit
  // limit on the size of a concatenation, so we should never
  // see more than two here.
  Regexp* stk[4];
  size_t d = 0;
  while (re->op() == kRegexpConcat) {
    if (d < arraysize(stk))
      stk[d++] = re;
    re = re->sub()[0];
  }

  // Remove leading string from re.
  if (re->op() == kRegexpLiteral) {
    re->rune_ = 0;
    re->op_ = kRegexpEmptyMatch;
  } else if (re->op() == kRegexpLiteralString) {
    if (n >= re->nrunes_) {
      delete[] re->runes_;
      re->runes_ = NULL;
      re->nrunes_ = 0;
      re->op_ = kRegexpEmptyMatch;
    } else if (n == re->nrunes_ - 1) {
      Rune rune = re->runes_[re->nrunes_ - 1];
      delete[] re->runes_;
      re->runes_ = NULL;
      re->nrunes_ = 0;
      re->rune_ = rune;
      re->op_ = kRegexpLiteral;
    } else {
      re->nrunes_ -= n;
      memmove(re->runes_, re->runes_ + n, re->nrunes_ * sizeof re->runes_[0]);
    }
  }

  // If re is now empty, concatenations might simplify too.
  while (d > 0) {
    re = stk[--d];
    Regexp** sub = re->sub();
    if (sub[0]->op() == kRegexpEmptyMatch) {
      sub[0]->Decref();
      sub[0] = NULL;
      // Delete first element of concat.
      switch (re->nsub()) {
        case 0:
        case 1:
          // Impossible.
          LOG(DFATAL) << "Concat of " << re->nsub();
          re->submany_ = NULL;
          re->op_ = kRegexpEmptyMatch;
          break;

        case 2: {
          // Replace re with sub[1].
          Regexp* old = sub[1];
          sub[1] = NULL;