Question

我有以下问题。在下面的代码中，对象变量p的地址与其第一个成员'a'的地址相同。但是当我打印p和a的值时，它们都是。为什么同一地址位置有两个不同的值？

class sample {
public:
    int a;
    int b;
    int c;

    //sample();
    sample(int x,int y, int z){
        a=x;
        b=y;
        c=z;
    }    

};


int main() {
    //sample p;
    //sample sample(2,3,4);
    sample p = sample(2,3,4);
    // cout<<"The value that is stored in the z is "<< p <<endl;
    printf("The value of the object handle p is %d \n",p); 
    printf("The address of the object handle p is %d \n",&p);
    //p->a =2;
    //p->b =3; This could be accessesd in the later part of the code.
    //p->c =4;

    printf("The address of a is %d\n",&(p.a));
    printf("The value stored in the a is %d\n",p.a);
    printf("The value stored in the b is %d\n",p.b);
    printf("The value stored in the c is %d\n",p.c);
}

上述代码的输出是：

The value of the object handle p is 2358832
The address of the object handle p is 2358848
The address of a is 2358848
The value stored in the a is 2
The value stored in the b is 2358852
The value stored in the c is 2358856

--------------------------------
Process exited after 0.2105 seconds with return value 0
Press any key to continue . . .

Answer 1

printf使用可变参数。虽然，我看到gcc中的扩展来检查格式字符串的可变参数，实际上没有真正的编译时间检查格式字符串的参数的类型匹配。

因此，不应在C ++中使用printf。类型安全的替代方案是流。

流操作符（实际上这些是重载的移位操作符）可用于内置类型以及标准库的某些类（例如std::string）。

为了支持自行设计类的流输出，必须为它输出流输出操作符。 E.g：

#include <iostream>

class Sample {
  friend std::ostream& operator << (std::ostream&, const Sample&);
  private:
    int _a, _b, _c;
  public:
  Sample(int a, int b, int c): _a(a), _b(b), _c(c) { }

};

std::ostream& operator << (std::ostream &out, const Sample &s)
{
  return out << s._a << ", " << s._b << ", " << s._c;
}

int main(int, char**)
{
  Sample sample = Sample(123, 234, 345);
  std::cout << "sample: " << sample << std::endl;
  return 0;
}

在cygwin上使用gcc进行编译和测试：

$ g++ -std=c++11 -o test-cout test-cout.cc

$ ./test-cout
sample: 123, 234, 345

我确实要a, b, c private来证明如果设为friend，流输出操作员可以访问它们。

相同的地址位置如何给出两个不同的值？

1 个答案: