在C++中,如何获取存储类实例字段值的地址?

a->i 。以下是工作代码:

#include <stdio.h>
class A {
    public:
        int i;
};

int main() {
    A* a = new A();
    int* i = &a->i;
    printf("a: %p\na->i: %p\n", a, i);
    return 0;
}

a.i a 其中一种类型。您需要取消引用指针才能访问成员。

$ ./a.out
a: 40748
a->i: 40748

A

class A {
    public:         // note colon after public
        int i;
};                  // note semicolon after bracket

A *a = new A();     // note *a not a*

cout << ios::hex << &a->i << endl;

&

&a.i

class A 
{
   public: 
       int i;
};

...

A *a = new A();
int *i = &a->i;

class A {
   public:
      int i;
};
int main() {
   A *a = new A();
   int *x = &a->i;
   return 0;
}

class A
{
public:
    int m_i;
};

int
main()
{
    A* a = new A();
    int* i_ptr = &(a->m_i);

    return 0;
}

A* a = ... A a* = ...