[英]Properly passing pointer from function C++
它在s_r->info
上崩潰,因為s_r
沒有指向obj2
。 函數search_recurs()
應該找到指針並返回結果。 它應該找到正確的指針,因為結果指向obj2
。
但是返回結果時會發生一些事情,因為s_r
(其中s_r=search_recurs()
返回結果)並不指向與結果相同的對象,應該從搜索返回。
輸出:
W Funkcji search:
Name: zmiana2d
Parameter_a : 5
Parameter_d : 6
adres rzutowanie: 00AFFC50
adres result: 00AFFC50
main:
adres obj2: 00AFFC50
adres s_r: 00AFFB04 //<======= Why is it not 00AFFC50 ?
碼:
#include "stdafx.h"
using namespace std;
class Node {
private:
public:
string name;
Node *parent;
vector <Node*> children;
Node() { name = "noname", parent = NULL; }
Node(string _name) { name = _name, parent = NULL; }
Node(Node *_parent) { parent = _parent; }
Node(string _name, Node *_parent) { name = _name, parent = _parent; }
Node(Node *_parent, vector <Node*> _children) { parent = _parent, children = _children; }
Node(string _name, Node *_parent, vector <Node*> _children) { name = _name, parent = _parent, children = _children; }
virtual ~Node() { cout << "Base Destructor called\n"; }
void add_parent(Node *wsk) {
parent = wsk;
}
void add_children(Node *child) {
children.push_back(child);
}
void info_node() {
cout << "Name: " << name << endl;
cout << "Address: " << this << endl;
cout << "Parent: " << parent << endl;
}
};
class A {
private:
string name;
int parameter_a;
protected:
A() { name = "untitled"; parameter_a = 1; }
A(string _name) { name = _name, parameter_a = 1; }
A(string _name, int _parameter_a) : name(_name), parameter_a(_parameter_a) {};
string info_name() {
return name;
}
int info_parameter_a()
{
return parameter_a;
}
public:
char info_type() {
return 'A';
}
friend class Leaf;
friend A* search_recurs_node(Node* root, string name);
virtual void info() = 0;
};
class Leaf : public Node {
private:
public:
vector<A*> objects;
Leaf() { name = "noname", parent = NULL; }
Leaf(string _name) { name = _name, parent = NULL; }
Leaf(Node *_parent) { parent = _parent; }
Leaf(string _name, Node *_parent) { name = _name, parent = _parent; }
Leaf(Node *_parent, vector <Node*> _children) { parent = _parent, children = _children; }
Leaf(string _name, Node *_parent, vector <Node*> _children) { name = _name, parent = _parent, children = _children; }
void add_objects_leaf(A* obj) {
objects.push_back(obj);
}
};
class X : public A, public Leaf {
private:
int parameter_x;
public:
X() : A("dziedziczone_w_X_z_A", 98), parameter_x(99) {};
X(string _name_x, int _parameter_a, int _parameter_x) : A(_name_x, _parameter_a), parameter_x(_parameter_x) {};
char info_type() {
return 'X';
}
void info() {
cout << "Name: " << A::info_name() << endl;
cout << "Parameter_a : " << A::info_parameter_a() << endl;
cout << "Parameter_d : " << parameter_x << endl;
}
};
A* search_recurs_node(Node* root, string name) {
A* result;
Leaf* rzutowanie;
if ((root->children.size()) > 0) {
for (int i = 0; i < (root->children.size()); ++i) {
search_recurs_node(root->children[i], name);
}
}
else if (rzutowanie = dynamic_cast<Leaf*>(root)) {
for (int i = 0; i < rzutowanie->objects.size();++i) {
if (rzutowanie->objects[i]->info_name() == name) {
cout << "W Funkcji search: " << endl;
rzutowanie->objects[i]->info();
cout << endl << "adres rzutowanie: " << rzutowanie->objects[i] << endl;
result = (rzutowanie->objects[i]);
cout << "adres result: " << result << endl;
cout << endl;
return result;
}
}
}
//return NULL;
};
int main()
{
//
Node A_node("node_A");
Leaf X_node("node_X", &A_node);
A_node.add_children(&X_node);
X obj1("name d1", 1, 2), obj2("zmiana2d", 5, 6);
X_node.add_objects_leaf(&obj1);
X_node.add_objects_leaf(&obj2);
A* s_r;
s_r = search_recurs_node(&A_node, "zmiana2d");
cout << "main: " << endl;
cout << "adres obj2: " << &obj2 << endl;
cout << "adres s_r: " << s_r << endl;
s_r->info();
cout << endl << "The cause of 90% of programming errors sits in front of the screen" << endl;
return 0;
}
初步評論
obj2
是X
類型,它繼承自A
和Leaf
。
您返回指向A
的指針。 但A
子對象嵌入X
與Leaf
; 這就是為什么你可以得到另一個地址。 您應該將其轉換為X*
以確保具有完全相同的地址, 無論您的類的布局如何 :
X* x_r = dynamic_cast<X*>(s_r); // you can, because there's a virtual fct
if (x_r)
cout << "adres x_r: "<<x_r<<endl;
else cout << "couldn't cast"<<endl;
順便說一下,想為任何具有虛函數的類添加虛擬析構函數。
問題的根本原因
您的遞歸搜索功能不起作用:您以遞歸方式調用它,但是您沒有對結果執行任何操作。 改變如下:
...
if ((root->children.size()) > 0) {
for (int i = 0; i < (root->children.size()); ++i) {
auto a= search_recurs_node(root->children[i], name); // <== keep response returned
if (a) // <== if value already found
return a; // <== return it
}
}
...
在函數結束時返回nullptr
是明智的(為什么要將它注釋掉?)
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