如何使用基于范围的 for 循环修改地图中的值?

我有一个基于范围的 for 循环来迭代 foobar 中的元素，如下所示:

#include 

此代码产生以下输出:

{1, 2} {2, 3} {3, 4}{1, 1} {2, 2} {3, 3}

第一行被修改并打印在 for 循环中，第二行应该打印相同的修改值.为什么输出不匹配?对 std::map 的更改是否仅在循环范围内有效?有没有办法不仅可以访问而且可以修改这些值?

可以在 cpp.sh 上找到 此代码的运行版本.

为了清楚起见，此处给出的示例经过修改以匹配接受的答案.

解决方案

你可以把 auto 变成 auto& 如果你想改变/修改容器，例如:

#include 

编译和输出

<前>{1, 2} {2, 3} {3, 4}

现场示例

I have a range based for loop to iterate over elements in foobar as follows:

#include <map>
#include <iostream>

int main()
{
  std::map<int, int> foobar({{1,1}, {2,2}, {3,3}});

  for(auto p : foobar) 
  {
    ++p.second;
    std::cout << "{" << p.first << ", " << p.second << "} ";
  }
  std::cout << std::endl;

  for(auto q : foobar) 
  {
    std::cout << "{" << q.first << ", " << q.second << "} ";
  } 
  std::cout << std::endl;
}

This code produces the following output:

{1, 2} {2, 3} {3, 4}
{1, 1} {2, 2} {3, 3}

The first line is modified and printed inside a for loop and the second line supposedly prints the same modified values. Why don't the outputs match? Are changes to std::map only effective in the scope of the loop? Is there a way I can not only access but modify these values?

A running version of this code can be found on cpp.sh.

EDIT: The example given here was modified to match the accepted answer for clarity.

解决方案

You can turn auto into auto& if you want to mutate/modify the container, for instance:

#include <map>
#include <iostream>

int main()
{
  std::map<int, int> foobar({{1,1}, {2,2}, {3,3}});
  for(auto& p : foobar) {
    ++p.second;
    std::cout << '{' << p.first << ", " << p.second << "} ";
  }
  std::cout << std::endl;
}

compiles ands outputs

{1, 2} {2, 3} {3, 4} 

live example