为什么sizeof运算符不显示大小为20？

Question

#include "iostream" 
    using namespace std; 
    class Algebra 
    { 
    private: 
     int a, b; 
     const int c; 
     const int d; 
     static int s; 


    public: 
     //default constructor 
     Algebra() : c(0), d(0) 
     { 
      s++; 
      a = b = 0; 
      cout << "Default Constructor" << endl; 
     } 
     //parameterized (overloaded) constructor 
     Algebra(int a, int b, int c1, int d1) : c(c1), d(d1) 
     { 
      s++; 
      setA(a); 
      setB(b); 
      cout << "Parameterized Constructor" << endl; 
     } 
     //copy (overloaded) constructor 
     Algebra(const Algebra &obj) : c(obj.c), d(obj.d) 
     { 
      s++; 
      this->a = obj.a; 
      this->b = obj.b; 
      cout << "Copy Constructor" << endl; 
     } 
     //Destructor 
     ~Algebra() 
     { 
      s--; 
      cout << "Destructor Called" << endl; 
     } 
     //Setter for static member s 
     static void setS(int s) 
     { 
      Algebra::s = s; 
     } 

     //Getter for static member s 
     static int getS() 
     { 
      return s; 
     } 

     //Getter for constant data member c 
     int getC() const 
     { 
      return this->c; 
     } 

     //Setter for data member a 
     void setA(int a) 
     { 
      if(a < 0) 
       this->a = 0; 
      else 
       this->a = a; 
     } 
     //Setter for data member b 
     void setB(int b) 
     { 
      if(b < 0) 
       this->b = 0; 
      else 
       this->b = b; 
     } 
    }; 
    int Algebra::s = 90; 
    int main() 
    { 
     Algebra obj1, obj2(1, 2, 3,4), obj3(obj1); 
     cout << "Size of object = " << sizeof(obj1) << endl; 
     return 0; 
    } 

为什么sizeof运算符显示大小为16，其中我已经声明了int类型的5个数据成员，它应该将所有5个数据成员相加，并将结果设为20.我还检查了sizeof运算符的静态int单独变量，工作正常。

Answer 1

静态成员不会影响类实例的大小，因为您只需要一个副本用于整个程序，而不是每个实例一个。因此，大小由4个整数组成，在您的平台上是16个字节。