在C++中引发NullPointerException

是的，你可以，但这不是一个好主意（无论如何，你都不应该处理指针，在现代C++中，指针被保存在管理其寿命的对象中）。

您可以始终定义一个包含指针的类。然后，当您尝试使用operator->()时，如果持有的指针是nullptr，它就会抛出。

template<typename T>
class ThrowingUniquePtr
{
     T*   ptr;
     public:
        // STUFF to create and hold pointer.
        T* operator->()
        {
            if (ptr) {
                return ptr;
            }
            throw NullPointerException; // You have defined this somewhere else.
        }
};
class Run
{
    public:
        void run() {std::cout << "Runningn";}
};
int main()
{
    ThrowingUniquePtr<Run>    x(new Run);
    x->run();  // will call run.
    ThrowingUniquePtr<Run>    y(nullptr);
    y->run();  // will throw.
}

另一种异常处理方式：使用NULL指针调用函数

#include <iostream>
#include <typeinfo>
using namespace std;
char str_NullPointer[25] = "NULL Pointer exception";
char str_Arithmetic[25] = "Arithmetic exception";
class A
{
public:
   int i = 20; 
public:
    int function()
    {
        printf("Function startn");
         try
        {
            if (this)
            {
                printf("value of i = %d n", i);  
            }
            else
            {
                throw str_NullPointer;          /* Exception thrown in Case 2 */
            }
        }
        catch(const int s)
        {
            printf("%dn", s);
        }
        catch(const char* s)
        {
            printf("%s in %sn", s, typeid(this).name());  /* Exception and Class Name */
        }
        printf("Function endnnn");
    }
};
int main() {
    //code
printf("Case 1: With Pointern");
A *obj = new A();
obj->i = 20;
obj->function();
printf("Case 2: With NULL Pointern");
delete obj;
obj = NULL;
obj->function();
return 0;
}

输出：

Case 1: With Pointer
Function start
value of i = 20 
Function end

Case 2: With NULL Pointer
Function start
NULL Pointer exception in P4abcd
Function end