C++ 没有对象就不能调用成员函数
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cannot call member function without object
提问by trikker
This program has the user input name/agepairs and then outputs them, using a class.
Here is the code.
该程序具有用户输入name/age配对，然后使用类输出它们。这是代码。
#include "std_lib_facilities.h"
class Name_pairs
{
public:
bool test();
void read_names();
void read_ages();
void print();
private:
vector<string>names;
vector<double>ages;
string name;
double age;
};
void Name_pairs::read_names()
{
cout << "Enter name: ";
cin >> name;
names.push_back(name);
cout << endl;
}
void Name_pairs::read_ages()
{
cout << "Enter corresponding age: ";
cin >> age;
ages.push_back(age);
cout << endl;
}
void Name_pairs::print()
{
for(int i = 0; i < names.size() && i < ages.size(); ++i)
cout << names[i] << " , " << ages[i] << endl;
}
bool Name_pairs::test()
{
int i = 0;
if(ages[i] == 0 || names[i] == "0") return false;
else{
++i;
return true;}
}
int main()
{
cout << "Enter names and ages. Use 0 to cancel.\n";
while(Name_pairs::test())
{
Name_pairs::read_names();
Name_pairs::read_ages();
}
Name_pairs::print();
keep_window_open();
}
However, in int main()when I'm trying to call the functions I get "cannot call 'whatever name is' function without object."I'm guessing this is because it's looking for something like variable.testor variable.read_names. How should I go about fixing this?
但是，int main()当我尝试调用我得到的函数时，我"cannot call 'whatever name is' function without object."猜这是因为它正在寻找类似variable.testor 的东西variable.read_names。我应该如何解决这个问题？
回答by sth
You need to instantiate an object in order to call its member functions. The member functions need an object to operate on; they can't just be used on their own. The main()function could, for example, look like this:
您需要实例化一个对象才能调用其成员函数。成员函数需要一个对象来操作；它们不能单独使用。main()例如，该函数可能如下所示：
int main()
{
Name_pairs np;
cout << "Enter names and ages. Use 0 to cancel.\n";
while(np.test())
{
np.read_names();
np.read_ages();
}
np.print();
keep_window_open();
}
回答by Rob K
If you want to call them like that, you should declare them static.
如果你想这样调用它们，你应该声明它们是静态的。
回答by dimba
You are right - you declared a new use defined type (Name_pairs) and you need variable of that type to use it.
您是对的 - 您声明了一个新的使用定义类型（Name_pairs），并且您需要该类型的变量才能使用它。
The code should go like this:
代码应该是这样的：
Name_pairs np;
np.read_names()
回答by Ketan Vishwakarma
just add statickeyword at the starting of the function return type.. and then you can access the member function of the class without object:) for ex:
只需在函数返回类型的开头添加static关键字..然后您就可以访问没有对象的类的成员函数:)例如：
static void Name_pairs::read_names()
{
cout << "Enter name: ";
cin >> name;
names.push_back(name);
cout << endl;
}
