How to save unknown size structure (for later retrieval) - c++

Question

My plan is to store a couple dozen rows with 2 items per row and both items will have a different data type. Not sure if this is the right approach and have heard about using vectors but I can't find any samples that will take in 2 items with different types with many rows (an unknown amount of rows) similar to what I'm trying to do here. The following doesn't compile
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
struct movies_t {
string title;
int year;
} myRecNo[];
void printmovie (movies_t movie);
int main ()
{
string mystr;
for (int i=0; i < 2; i++)
{
switch (i)
{
case 1:
myRecNo[i].title = "2001 A Space Odyssey";
myRecNo[i].year = 1968;
cout << "Auto entry is:\n ";
printmovie (myRecNo[i]);
break;
case 2:
myRecNo[i].title = "2002 A Space Odyssey";
myRecNo[i].year = 1978;
cout << "Auto entry is:\n ";
printmovie (myRecNo[i]);
break;
}
}
return 0;
}
void printmovie (movies_t movie)
{
cout << movie.title;
cout << " (" << movie.year << ")\n";
}
This is the error I get:
Test1.obj||error LNK2019: unresolved external symbol "struct movies_t * myRecNo" (?myRecNo##3PAUmovies_t##A) referenced in function _main|

Answer 1

There are a couple of bad practices in your code, if you are just asking for a way to modify the program so that it will compile and work, see the following:
Declare struct and create struct variables in your main function.
struct movies_t
{
string title;
int year;
};
then, in your main function, movies_t myRecNo[2];
Arrays start at index 0, not 1. so your switch should be
switch (i)
{
case 0:
myRecNo[i].title = "2001 A Space Odyssey";
myRecNo[i].year = 1968;
cout << "Auto entry is:\n ";
printmovie(myRecNo[i]);
break;
case 1:
myRecNo[i].title = "2002 A Space Odyssey";
myRecNo[i].year = 1978;
cout << "Auto entry is:\n ";
printmovie(myRecNo[i]);
break;
}
// the rest of the code..
After you modify these, your code should work.
However, for a better data structure to save an array of paired values, you can use std::vector<std::pair<string, int>> myReg to save your data.
the following code should be much much better, remember to #include <vector>
#include <iostream>
#include <string>
#include <sstream>
#include <vector>
void printmovie(std::vector<std::pair<std::string, int>>);
int main()
{
std::vector<std::pair<std::string, int>> myReg;
myReg.push_back({ "2001 A Space Odyssey", 1968 });
myReg.push_back({ "2002 A Space Odyssey", 1978 }); // <- if your compiler is not using c++11 standard or above, please change this line to myReg.push_back(std::pair<std::string, int>("name of the movie", int)); to use to older version of Initializer
printmovie(myReg);
return 0;
}
void printmovie(std::vector<std::pair<std::string, int>> movie)
{
for (auto itor = movie.begin(); itor != movie.end(); ++itor)
{
//first is the first data in the pair, which is the title
//second is the second data in the pair, which is the year
std::cout << (*itor).first << " (" << (*itor).second << ")\n";
}
}

Answer 2

Thanks everyone & #Zhou.
Zhou's code above might work on a newer version of the compiler but I'm using Code::Blocks IDE with MS Visual C++ 2010 compiler.
Here is the vector method that worked:
#include <iostream>
#include <string>
#include <sstream>
#include <vector>
void printmovie(std::vector<std::pair<std::string, int>>);
int main()
{
std::vector<std::pair<std::string, int>> myReg;
myReg.push_back(std::pair<std::string, int>("title of the movie", 1968));
myReg.push_back(std::pair<std::string, int>("title of the movie2", 1978));
//myReg.push_back({ "2001 A Space Odyssey", 1968 });
//myReg.push_back({ "2002 A Space Odyssey", 1978 });
printmovie(myReg);
//or to print a single element (the 2nd row) thanks #zhou
std::cout << myReg[1].first << " " << myReg[1].second << std::endl;
return 0;
}
void printmovie(std::vector<std::pair<std::string, int>> movie)
{
for (auto itor = movie.begin(); itor != movie.end(); ++itor)
{
//first is the first data in the pair, which is the title
//second is the second data in the pair, which is the year
std::cout << (*itor).first << " (" << (*itor).second << ")\n";
}
}