假设我有一个构造成本昂贵的对象,并且需要reset()
函数将其重置为初始状态:
struct Example
{
// option 1: efficient, but need to duplicate initialization logic of reset()
Example(int param) : expensive_member_(param), param_(param)
{
}
/* option 2: less code, but initializes expensive_member_ twice
Example(int param) : param_(param)
{
reset();
}*/
void reset()
{
expensive_member_ = ClassWithExpensiveConstructor(param_);
}
ClassWithExpensiveConstructor expensive_member_;
int param_;
}
是否有更好的/ pattern方式可以有效地将对象重置为初始状态而不会复制初始化列表和reset
函数中的初始化逻辑?
编辑:如果没有达到我想要的通用方式,这也是这个问题的可接受结果!
答案 0 :(得分:1)
如果ClassWithExpensiveConstructor
是构造/重置费用昂贵的一个类,那么优化操作应该是它。
另一种选择是在const
成员中保留初始值的副本,假设复制构造/分配并不昂贵。这会占用更多内存,但如果碰巧大量调用Example::reset()
,可以提高性能。
struct Example
{
Example(int param)
: expensive_member_backup_(param)
, expensive_member_(expensive_mamber_backup)
, param_(param)
{
}
void reset()
{
expensive_member_ = expensive_member_backup_;
}
const ClassWithExpensiveConstructor expensive_member_backup_;
ClassWithExpensiveConstructor expensive_member_;
int param_;
}
答案 1 :(得分:1)
您可以将ExpensiveMember
作为指针,在这种情况下,您的选项2不会在ExpensiveMember
构造函数中调用Example
构造函数,除非您明确地调用它:
struct Example
{
Example(int param) : expensive_member_(), param_(param)
{
reset();
}
~Example() {
delete expensive_member_; // better use unique_ptr etc
}
// Also a copy constructor and assignment operator
// are needed to avoid memory problems such as double-delete.
// Also, better use unique_ptr etc.
// This answer does not use unique_ptr just to make the main idea more clear.
void reset()
{
delete expensive_member_; // better use unique_ptr etc
expensive_member_ = new ClassWithExpensiveConstructor(param_);
}
ClassWithExpensiveConstructor* expensive_member_; // better use unique_ptr etc
int param_;
}
答案 2 :(得分:1)
一个简单的解决方案是使用(智能或常规)指针,以便初始化成员(即指针)的成本变小,而实际对象仅在调用reset()
时初始化:
struct Example
{
Example(int param) : param_(param)
{
reset();
}
void reset()
{
p.reset(new ClassWithExpensiveConstructor(param_));
}
unique_ptr<ClassWithExpensiveConstructor> p;
int param_;
}