最近,我阅读了SGI STL源并遇到了一个问题。为什么sgi stl 在vector的push_back上使用重载函数而不是默认参数。这是关于push_back的SGI STL源。
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());
}
但我不知道为什么不将默认参数用作休闲。
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函数也有重载函数作为fallowed。 * __ position = _Tp()使用默认构造函数。我不知道为什么不使用默认参数。
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;
}
}
答案 0 :(得分:1)
如果类型没有默认构造函数,则带有默认参数的版本将无效,这将导致无法对此类型使用push_back
。模板将具有替换失败。使用两个函数可以将替换失败与其中一个函数隔离开来。