Xcode Question posted in
Whether you're new to Xcode or an experienced developer, our archive has everything you need to know about this integrated development environment (IDE). From basic functionalities to advanced features, our archive covers a wide range of Xcode-related questions and answers. Browse our archive now and find solutions to your Xcode questions, and take your app development skills to the next level

Xcode – How can I print a list of structs that contain two members?

I created a struct and put two member variables in it. Then initialized a list with the mentioned struct type.

struct job {
    string name;
    int time;
};

list<job> jobList;

string line;
ifstream file("input.txt");  // the direction of my input file.
if (file.is_open()) {
    while (getline(file, line)) {
        string col1;
        int col2;
        istringstream ss(line);
        ss >> col1 >> col2;
        job a;
        a.name = col1;
        a.time = col2;

        jobList.push_back(a);
    }
}

Now I want to know, how can I print my jobList? Whatever I do, does not work and I get errors.

I used the following function to print my list contents:

for (auto v : jobList)
    std::cout << v << "n";
}

The error I get is:

Invalid operands to binary expression (‘std::__1::ostream’ (aka ‘basic_ostream<char>’) and ‘job’)

I think that is because my struct has two variables (name and time) but I am not sure. The purpose of printing jobList is to make sure that I have created my list properly.

After printing, the output should be as follows:

MPL 3
JOB 1
CORE 50
DISK 10
CORE 150
DISK 0
CORE 100
DISK 0
Tags:

2

Answers


  1. 0

    You can print the individual struct members in your range-for loop:

    for (auto const& j : jobList)
    std::cout << j.name << ' ' << j.time << 'n';
    Login or Signup to reply.
  2. 0

    The problem with:

    for (auto v : jobList)
    std::cout << v << "n";
}

    is that v is a job structure, and you have no stream output operator overloaded for job.

    There are two possible solutions:

    1. Output the members one by one:

      for (auto const& v : jobList) {
    std::cout << v.name << " - " << v.time << 'n';
}

    2. Create an overloaded operator for outputting the structure:

      std::ostream& operator<<(std::ostream& os, job const& j)
{
    return << v.name << " - " << v.time;
}

...

for (auto const& v : jobList) {
    std::cout << v << 'n';
}
    Login or Signup to reply.
Please signup or login to give your own answer.

Back To Top
Search