reference to non-static member function must be called
时间: 2024-06-13 08:06:51 浏览: 16
这个错误通常出现在尝试调用非静态成员函数时,因为非静态成员函数需要通过对象来调用。如果你尝试直接调用非静态成员函数,就会出现“reference to non-static member function must be called”的错误。解决这个问题的方法是要么将该函数定义为静态成员函数,要么通过对象来调用该函数。
以下是一个例子,演示了如何通过对象来调用非静态成员函数:
```cpp
#include <iostream>
#include <vector>
#include <algorithm>
class MyClass {
public:
void printVector(std::vector<int>& vec) {
std::for_each(vec.begin(), vec.end(), printInt);
}
private:
static void printInt(int i) {
std::cout << i << " ";
}
};
int main() {
std::vector<int> vec = {1, 2, 3, 4, 5};
MyClass obj;
obj.printVector(vec); // 通过对象来调用非静态成员函数
return 0;
}
```
相关问题
Reference to non-static member function must be called
This error occurs when you try to call a non-static member function without an instance of the class. In other words, you are trying to call a function that belongs to an object of a class, but you didn't create an instance of that class.
To fix the error, you need to create an instance of the class and call the function on that instance. Alternatively, you can make the function static, which means it belongs to the class itself and not to a specific instance.
Here's an example of how to create an instance of a class and call a member function:
```
class MyClass {
public:
void myFunction() {
// do something
}
};
int main() {
MyClass obj;
obj.myFunction(); // call member function on instance
return 0;
}
```
In this example, we create an instance of the `MyClass` class and call the `myFunction()` member function on that instance.
sort error: reference to non-static member function must be called
回答: 这个错误是因为在sort函数的第三个参数中引用了非静态成员函数,而在类定义中若不将成员函数声明为静态成员函数,会默认给成员函数添加一个this指针。而sort函数的第三个参数不允许有指针参数,所以会导致报错。解决这个问题的方法是将成员函数声明为静态成员函数,这样就不会添加this指针了。或者可以考虑将比较函数cmp定义为类外的普通函数,而不是成员函数。这样就可以解决该错误。<span class="em">1</span><span class="em">2</span><span class="em">3</span>
#### 引用[.reference_title]
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