考虑以下来自《C++程序设计语言》(第四版)Bjarne Stroustrup的std::vector::reserve()实现:
template<class T, class A>
void vector<T,A>::reserve(size_type newalloc)
{
if (newalloc<=capacity()) return;
vector_base<T,A> b {vb.alloc,newalloc}; // get new storage
// (see PS of question for details on vb data member)
T* src = elem; // ptr to the start of old storage
T* dest = b.elem; // ptr to the start of new storage
T* end = elem+size(); // past-the-end ptr to old storage
for (; src!=end; ++src, ++dest) {
new(static_cast<void*>(dest)) T{move(*src)}; // move construct
src–>~T(); // destroy
}
swap(vb,b); // install new base (see PS if needed)
} // implicitly release old space(when b goes out of scope)
注意,在循环中,对于向量中的每个元素,至少会调用一次构造函数和析构函数(如果元素的类具有基类或如果该类或其基类具有具有构造函数的数据成员,则可能触发更多此类调用)。 (在书中,“for-loop”实际上是一个单独的函数,但我在这里为了简单起见将其注入到reserve()中。)
现在考虑我的替代建议:
template<class T, class A>
void vector<T,A>::reserve(size_type newalloc)
{
if (newalloc<=capacity()) return;
vector_base<T,A> b {vb.alloc,newalloc}; // get new space
memcpy(b.elem, elem, sz); // copy raw memory
// (no calls to ctors or dtors)
swap(vb,b); // install new base
} // implicitly release old space(when b goes out of scope)
在我看来,最终结果似乎是相同的,减少了对构造函数/析构函数的调用。
是否存在这种情况,这种替代方案会失败?如果有,请问哪里出现了缺陷?
P.S. 我认为这并不太相关,但以下是 vector
和 vector_base
类的数据成员:
// used as a data member in std::vector
template<class T, class A = allocator<T> >
struct vector_base { // memory structure for vector
A alloc; // allocator
T* elem; // start of allocation
T* space; // end of element sequence, start of space allocated for possible expansion
T* last; // end of allocated space
vector_base(const A& a, typename A::size_type n)
: alloc{a}, elem{alloc.allocate(n)}, space{elem+n}, last{elem+n} { }
~vector_base() { alloc.deallocate(elem,last–elem); } // releases storage only, no calls
// to dtors: vector's responsibility
//...
};
// std::vector
template<class T, class A = allocator<T> >
class vector {
vector_base<T,A> vb; // the data is here
void destroy_elements();
public:
//...
};
this
来访问它。而且由于this
指针并不存在 - 它是在调用时为调用方法的任何对象生成的 - 所以它不可能出错。 - underscore_d