Question

（对不起我的英语，我和乌克兰语）我正在做一个实现了＃＆＃34;数字堆栈＆＃34; （控制台应用程序）。我想要在删除stac之后添加将被清理的动态内存的析构函数。当我在没有析构函数的情况下执行stac时all right，如果我添加析构函数 - 我有error。应用程序结束时调用析构函数，但程序调用第一个函数时出错。没有析构函数我没有这个错误。这是我的源代码，析构函数被注释掉了。

#include<iostream>
#include<ctime>
using namespace std;

struct Oneof
{
int num;
Oneof* next;
};

class Stac
{
private:
Oneof * first;
public:
/*~Stac();*/
Stac(); 
Stac(Stac &n);  
void New(int n);                 //Adding new element
int Remove();                    //Reading last element and removing it
int SetLast();                   //Reading last element without removing        
void Read();                     //Reading all stac
friend bool Eq(Stac a, Stac b);  //Equive two another stacs 
};

Stac::Stac(){first=NULL;}

Stac::Stac(Stac &n){first=n.first;}

/*Stac::~Stac()
{
    while(first!=NULL)
    {
        Oneof *temp=first;
        first=first->next;
        delete temp;
    }   
}
*/

void Stac::New(int n)                   //Adding new element
{
Oneof* temp=new Oneof;
temp->num=n;
temp->next=first;
first=temp;
}

int  Stac::Remove()                     //Reading last element and removing it
{
int a=first->num;
Oneof *temp=first;
first=first->next;
delete temp;
return a;
}

int  Stac::SetLast()                    //Reading last element without removing
{
return first->num;
}

void Stac::Read()                       //Reading all stac
{
Oneof* temp=NULL;
Oneof* save=NULL;
save=first;
while(first!=NULL)
{
    temp=first;
    cout<<temp->num<<" ";
    first=temp->next;
}
first=save;
}

bool Eq(Stac a, Stac b)                 //Equive two another stacs
{
Oneof* tempa=a.first;
Oneof* tempb=b.first;
while(tempa!=NULL && tempb!=NULL)
{
    if(tempa->num==tempb->num)
    {
        tempa=tempa->next;
        tempb=tempb->next;
    }
    else return false;
}
if(tempa==NULL && tempb==NULL)return true;
else return false;
}

int main()
{
Stac a;
srand(time(0));
for(int i=0; i<10; i++)
    {
        a.New(rand()%100);
    }
Stac b(a);

cout<<"Chek equive...\n";
bool equ=Eq(a,b);
if(equ==0)cout<<"First!=Second\n";
else cout<<"First==Second\n";

cout<<"\nReading without removing first number of fisrt stac...\n";
int n=a.SetLast();
cout<<n<<endl;

cout<<"\nReading first Stac...\n";
b.Read();

cout<<"\n\nReading second Stac...\n";
a.Read();

cout<<"\n\nAdding new number and reading first Stac...\n";
b.New(rand());
b.Read();

cout<<"\n\nRemoving number and reading second Stac...\n";
int last=a.Remove();
cout<<last<<endl;
a.Read();

cout<<"\n\nChek equive...\n";
bool equ1=Eq(a,b);
if(equ1==0)cout<<"First!=Second\n\n";
else cout<<"First==Second\n\n";

system("pause");
return 0;

}

Answer 1

您的函数Eq按值<{em> 获取两个Stac个对象

bool Eq(Stac a, Stac b)

因此，a和b将是您输入的功能本地副本。因此，在函数结束后，它们将超出范围并且将调用它们的析构函数。要避免制作本地副本，请按const&
传递对象
bool Eq(Stac const& a, Stac const& b)

添加析构函数后，我在程序调用它之前有错误

1 个答案: