I need to print element of a struct that is itself in a map in c++

user_4571350 0
Viewed 73

I have in c++ an include like that:

struct my_struct
{
  time_t time;
  double a, b, c, d;
}
typedef std::map<std::string, std::vector<my_struct> Data;

In my code (for debugging issue) I want to print some of the values in Data for specific key. I don't remember the syntax and keep having errors.

Here the kind of syntax I tried without success:

for (const auto& [key, value] : inputData) 
{
     if (key=="mytest")
     {
        std::cout << '[' << key << "] = " << value.a << endl;
    }
}

I also tried:

  for(const auto& elem : inputData)
  {
      if (elem.first=="mytest")
      {
              cout<<elem.second.a>>endl;
      }
}

Thanks for your help

c++
2 Answers

Look at the type of the element of the map:

typedef std::map<std::string, std::vector<my_struct> Data;
                              ^^^^^^^^^^^^^^^^^^^^^^

The map contains vectors.

for (const auto& [key, value] : inputData)

Here, value is a value of an element in the map. It is a vector.

value.a

Vector doesn't have a member named a. You should be getting error messages that explain this.

There are many ways of accessing elements within vectors. Here is an example:

std::cout << value.at(0).a;
user_20793030

Look at the following line properly:

typedef std::map<std::string, std::vector<my_struct> Data;

As you can see the second element of the std::map is a list of my_struct. Now here:

for (const auto& [key, value] : inputData)
{
    if (key == "mytest")
    {
        std::cout << '[' << key << "] = " << value.a << std::endl;
    }
}

..value.a makes no sense because std::vector<my_struct>::a is not a thing.

So replace value.a with:

value[0].a; // Replace 0 with the index of element you want to access

Or print every element in value:

for (const auto& [key, value] : inputData)
{
    if (key == "mytest")
    {
        std::cout << '[' << key << "] = ";
        for (auto& i : value)
        {
            std::cout << i.a << " : ";
        }
        std::cout << std::endl;
    }
}

You can use any one of the 2 options as per your choice.

user_180812440
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