指针传染媒介对另一个传染媒介的元素的而不是克隆传染媒介

Question

我想有一个向量z,并能够访问它z[i]或*pz[i],其中*pz[i]指出z[i].所以有一组值,但有两种方法可以访问它.

这是我的代码:

std::vector<double> z;
std::vector<double*> pz;

for (int i = 0; i < 5; i++) {
  z.push_back(7+0.01*i);
  std::cout << i << "  z = " << z << std::endl;
  pz.push_back(&z.back());
  std::cout << i << "  pz = " << pz << std::endl;
  std::cout << i << "  *pz = " ;
  for (int j = 0; j < pz.size(); j++) {
    std::cout << "  " << *pz[j];
  }
  std::cout << std::endl;
}

z[1]=17.3;
std::cout << "z[1] = " << z[1] << std::endl;
std::cout << "*pz[1] = " << *pz[1] << std::endl;

*pz[2]=34.1;
std::cout << "z[2] = " << z[2] << std::endl;
std::cout << "*pz[2] = " << *pz[2] << std::endl;

输出:

0  z = vector(1) [ 7 ]
0  pz = vector(1) [ 0x1d00b80 ]
0  *pz =   7
1  z = vector(2) [ 7, 7.01 ]
1  pz = vector(2) [ 0x1d00b80, 0x1d00bc8 ]
1  *pz =   1.50254e-316  7.01
2  z = vector(3) [ 7, 7.01, 7.02 ]
2  pz = vector(3) [ 0x1d00b80, 0x1d00bc8, 0x1d00bf0 ]
2  *pz =   1.50254e-316  7.01  7.02
3  z = vector(4) [ 7, 7.01, 7.02, 7.03 ]
3  pz = vector(4) [ 0x1d00b80, 0x1d00bc8, 0x1d00bf0, 0x1d00bf8 ]
3  *pz =   1.50254e-316  7.01  7.02  7.03
4  z = vector(5) [ 7, 7.01, 7.02, 7.03, 7.04 ]
4  pz = vector(5) [ 0x1d00b80, 0x1d00bc8, 0x1d00bf0, 0x1d00bf8, 0x1d00c60 ]
4  *pz =   1.50254e-316  7.01  7.02  7.03  7.04
z[1] = 17.3
*pz[1] = 7.01
z[2] = 7.02
*pz[2] = 34.1

有3个问题:

1. *pz第二次推送后第一个元素未定义.
2. 当我给你z[1],*pz[1]是不是z[1]了.
3. 当我分配时*pz[2],它不再分配给它z[2]了.

(在我的实际程序中,我需要一个向量来访问偶数元素和一个向量来访问奇数元素,但我不想制作克隆并使用更多内存.)

Answer 1

两件事情:

1. 当您执行push_back()可能导致向量重新分配的操作或任何操作时,会使指向该向量元素的所有指针无效.
2. 在64位中,指针可能与a的大小相同,double因此您不会以这种方式保存任何内存.

如果你真的想要创建第二个向量,指向第一个向量的元素,你需要先完成构建第一个向量.否则,push_back()操作将使指针无效.