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近來的push_back功能的實現,我讀了SGI STL源和滿足休耕question.Why SGI STL 使用重載函數向量,而不是默認參數的push_back。這是關於push_back的SGI STL源碼。在SGI STL,載體不使用默認參數
void push_back(const _Tp& __x) {
if (_M_finish != _M_end_of_storage) {
construct(_M_finish, __x);
++_M_finish;
}
else
_M_insert_aux(end(), __x);
}
void push_back() {
if (_M_finish != _M_end_of_storage) {
construct(_M_finish);
++_M_finish;
}
else
_M_insert_aux(end());
}
但我不知道爲什麼不使用默認參數作爲休閒。
void push_back(const _Tp& __x = _Tp()) {
if (_M_finish != _M_end_of_storage) {
construct(_M_finish, __x);
++_M_finish;
}
else
_M_insert_aux(end(), __x);
}
例外,_M_insert_aux函數也具有重載函數,如下所示。 * __ position = _Tp()使用默認構造函數。我不知道爲什麼不使用默認參數。如果類型沒有默認構造函數,這將使它不可能使用push_back
與這些類型的
template <class _Tp, class _Alloc>
void
vector<_Tp, _Alloc>::_M_insert_aux(iterator __position)
{
if (_M_finish != _M_end_of_storage) {
construct(_M_finish, *(_M_finish - 1));
++_M_finish;
copy_backward(__position, _M_finish - 2, _M_finish - 1);
*__position = _Tp();
}
else {
const size_type __old_size = size();
const size_type __len = __old_size != 0 ? 2 * __old_size : 1;
iterator __new_start = _M_allocate(__len);
iterator __new_finish = __new_start;
__STL_TRY {
__new_finish = uninitialized_copy(_M_start, __position, __new_start);
construct(__new_finish);
++__new_finish;
__new_finish = uninitialized_copy(__position, _M_finish, __new_finish);
}
__STL_UNWIND((destroy(__new_start,__new_finish),
_M_deallocate(__new_start,__len)));
destroy(begin(), end());
_M_deallocate(_M_start, _M_end_of_storage - _M_start);
_M_start = __new_start;
_M_finish = __new_finish;
_M_end_of_storage = __new_start + __len;
}
}
非常感謝您的回答。但是我的問題表明我遇到了有關_M_insert_aux函數的另一個問題。 「* __ position = _Tp()」使用默認構造函數。 –
您正在查看'_M_insert_aux'的錯誤超載。 –