c++ - 打印整数指针 vector 的值

Question

我创建了一个整数 vector ，如何在 vector 中取消引用以打印值

#include <iostream>
#include <vector>

int main ()
{
  std::vector<int*> myvector;
  for (int i=1; i<=5; i++){
    myvector.push_back(&i);
  }
  std::cout << "myvector contains:";
  for (std::vector<int*>::iterator it = myvector.begin() ; it !=myvector.end(); ++it){
    std::cout << ' ' << (**it);//?????????//how to print value of vector
  }
  std::cout << '\n';
  return 0;
}

Answer 1

myvector.push_back(&i); 并不像您想象的那样。变量 i 只存在于 for 循环的范围内，因此存储它的地址不会按预期工作，地址很可能是相同的。这是未定义的行为:

在我的平台上输出是:

 6 6 6 6 6 

代替

 1 2 3 4 5

并且您错过了 *(请参阅下面代码中的注释)。所以你可能想要这个:

#include <iostream>
#include <vector>

int main()
{
  std::vector<int*> myvector;
  for (int i = 1; i <= 5; i++) {

    myvector.push_back(new int(i));  // new int() allocated a new pointer to int
                                     // and puts i there
  }
  std::cout << "myvector contains:";

  for (std::vector<int*>::iterator it = myvector.begin(); it != myvector.end(); ++it) {
                   // ^<<<<<<<<<<<< you missed this *
    std::cout << ' ' << **it;
  }
  std::cout << '\n';

  // and now we need to delete the allocted pointers

  for (std::vector<int*>::iterator it = myvector.begin(); it != myvector.end(); ++it) {
    delete *it;
  }

  return 0;
}

但是拥有一个指向整数的指针 vector 是毫无意义的，您可能只需要一个 int vector 。那么你需要这个:

#include <iostream>
#include <vector>

int main()
{
  std::vector<int> myvector;
  for (int i = 1; i <= 5; i++) {

    myvector.push_back(i);
  }
  std::cout << "myvector contains:";

  for (std::vector<int>::iterator it = myvector.begin(); it != myvector.end(); ++it) {
    std::cout << ' ' << *it;
  }
  std::cout << '\n';

  return 0;
}

在 C++11 中，你可以这样写来遍历 vector :

for (auto & i: myvector) {
  std::cout << ' ' << i;
}

这样就简单多了。