C中的引用是一个convneint结构,允许我们简化以下C代码：
f(object *p){
//do something
}
int main(){
object* p = (object*) calloc(sizeof(object));
f(p);
}
至
f(object& o){
//do something
}
int main(){
object o = object();
f(o);
}
共享指针是C中另一种简化内存管理的便利.但是,我不知道如何将shared_ptr传递给f(object& o)这样的函数,它通过引用接受参数？
f(object& o){
//do something
}
int main(){
shared_ptr<object> p (new object());
f(*p);
}
解决方法
按值获取shared_ptr,引用计数将增加.当你输入它时,这会更容易：
typedef boost:shared_ptr<object> object_ptr;
void foo(object_ptr obj)
{
obj->/* stuff*/;
obj.reset(); //only resets this local copy,that means:
// reduce reference count (back to 1),and
// set obj to point at null.
}
int main(void)
{
object_ptr obj(new object());
foo(obj);
}
请记住,引用是别名.当您通过引用传递时,您没有传递指针,副本等…,您正在别名另一个对象. (实际上它们是作为指针实现的)：
typedef boost:shared_ptr<object> object_ptr;
void foo(object_ptr& obj)
{
obj.reset(); // the references was never increased,since a copy has not
// been made,this *is* obj in main. so the reference
// goes to 0,and obj is deleted
}
int main(void)
{
object_ptr obj(new object);
foo(obj); // after this,obj has been reset!
}
永远记住要正确,以防止错误：
typedef boost:shared_ptr<object> object_ptr;
void foo(const object_ptr& obj)
{
obj.reset(); // cannot do!
}
int main(void)
{
object_ptr obj(new object);
foo(obj);
}
我认为您应该尽可能将智能指针作为引用传递,以避免无关的增量和减量(以及副本和诸如此类的东西).
