我正在学习如何使用向量,并正在编写一个简单的程序,该程序会获取一些信息并将其放在向量上,然后对其进行迭代。 我的来源是

int main ()
{

    int answer= 1;
    int decide;
    int vectCount = 0;
    vector<animal> pet;

    while(answer > 0)
    {    

        pet.push_back(animal());
        cout << "enter the name of the pet" << endl;
        getline(cin,pet[vectCount].name);

            cout << "Please enter the age of the pet" << endl;
            cin >> pet[vectCount].age;

         cout << "enter the weight of the pet" << endl;
         cin >> pet[vectCount].weight;


        do
        {
        cout << "Please enter the size of the pet S/M/L" << endl;
        cin >> pet[vectCount].size;
        }while(pet[vectCount].size != 'L' 
        && pet[vectCount].size != 'M' 
        && pet[vectCount].size != 'S');

        answer = question(decide);

    }
    vector<animal>::iterator i;
    for(i = pet.begin(); i != pet.end(); ++i)
    {

        cout << "The name of the pet is " << *i->name << endl;
        cout << "The age of the pet is " << *i->age << endl;
        cout << "The weight if the pet is " << *i->weight << endl;
        cout << "The size of your pet is " << *i->size;
        if(*i->size == 'S')
        cout << "(-): meow" <<endl;
        if(*i->size == 'M')
        cout << "(---): woof" <<endl;
        if(*i->size == 'L')
        cout << "(------): moooo" <<endl;            
    }
    cout << "Exiting the program" << endl;


    cin.get();
    return 0;
}

我得到的错误是:

no match for 'operator*' in '*(&i)->__gnu_cxx::__normal_iterator<_Iterator, _Container>::operator-> [with _Iterator = animal*, _Container = std::vector<animal, std::allocator<animal> >]()->animal::name'

谁能帮助我找到问题的根源?

===============>>#1 票数:4

这个:

*i->size

应该:

i->size

->运算符(您的情况下等于(*i).size )将自动引用i

===============>>#2 票数:0

您收到该错误,因为编译器正在尝试执行以下操作:

*(i->name)

这正在尝试取消引用i->name ,并且由于i是指向具有name的对象的指针,因此它将失败。

Wheras您想要的是:

(*i).name

要么

i->name

在将名称从结构中取出之前,这将取消对i引用。

===============>>#3 票数:0

您应该使用以下任一方法:

 i -> size

要么

 (*i).size

但不是您惯用的方式。

===============>>#4 票数:0

尝试通过更改删除“ *”

 cout << "The name of the pet is " << *i->name << endl;

 cout << "The name of the pet is " <<  i->name << endl;

要么

 cout << "The name of the pet is " <<  (*i).name << endl;

  ask by user2284606 translate from so

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