Passing array between structures C++ - c++

Question

I am new to C++ and there is a basic problem I am dealing with. The code below gives complier error:
#include <iostream>
using namespace std;
struct contact_info {
long number;
string name;
};
contact_info take(){
contact_info takein[2];
for (int i=0; i<2; i++) {
cout<<"what is the name"<<"\n";
getline(cin,takein[i].name);
cout<<"what is the phone number"<<"\n";
cin>>takein[i].number;
};
return takein;
};
void give(contact_info takein){
cout<<"Name:"<<takein.name<<"\n"<<"Number:"<<takein.number;
};
int main(int argc, const char * argv[])
{
contact_info takein;
takein=take();
give(takein);
return 0;
}
The error comes from function "take" and is "No viable conversion from 'contact_info[2]' to 'contact_info'"
The code is supposed to take two contact informations in a loop and then prints them on the screen.
I think I need to use pointers for that to pass the "takein" from "take" function to "main" function. Can anyone says if I can fix the code using array and not pointer?

Answer 1

#include <iostream>
struct contact_info
{
long number;
string name;
};
void take(contact_info takein[2])
{
for (int i=0; i<2; i++)
{
cout<<"what is the name"<<"\n";
getline(cin,takein[i].name);
cout<<"what is the phone number"<<"\n";
cin>>takein[i].number;
};
};
void give(contact_info takein)
{
cout<<"Name:"<<takein.name<<"\n"<<"Number:"<<takein.number;
};
int main()
{
contact_info takein[2];
take(takein);
for(int i=0;i<2;++i)
give(takein[i]);
return 0;
}

Answer 2

return takein;
but
contact_info takein[2];
Compiler is upset because you're trying to return an array, use return takin[0]; or return takin[1]; to return a specific contact_info
Addendum
Don't be afraid to learn Python first, higher level languages are a lot more forgiving and python is by no means a toy language, Python and C++ are two things I use daily and I love them both. I tolerate Java, it can be good when you want static typing and something more forgiving than C++, or in my case Android :P
If you edit your question to describe what you want to do I am happy to provide some annotated code to demonstrate, leave a comment to this to get my attention.

Answer 3

contact_info[] take(){
...
}
if you are trying to return an array.
for returning an element, use
return takein[0];
or
return takein[1];

Answer 4

If your goal is "to take two contact informations in a loop and then prints them on the screen", you must pass an array to get information and pass an array to prints it:
contact_info * take(contact_info *takein){
for (int i=0; i<2; i++) {
cout<<"what is the name"<<"\n";
getline(cin, takein[i].name);
cout<<"what is the phone number"<<"\n";
cin>>takein[i].number;
};
return takein;
};
void give(contact_info * takein){
for (int i = 0; i < 2; i++)
cout<<"Name:"<<takein[i].name<<"\n"<<"Number:"<<takein[i].number;
};
int main(int argc, const char * argv[])
{
contact_info takein[2];
// pass takein as an array
take(takein);
give(takein);
return 0;
}

Answer 5

Secret's out! Arrays are pointers. Or at least, they implicitly convert to them. Even it you did have the right type, you'd be using undefined behavior, as the array is allocated on a place on the stack that no longer exists.
I'll assume you actually want to 'return two things', otherwise Alec's answer is perfectly valid.
Try a vector, or a 2 element struct. People often struggle mentally with the notion of creating a struct only to serve as a return value, but it's worth it from a maintainability standpoint. You get the power of naming the type and naming its contents, and that self-documents your intent.

Answer 6

In the main function takein is a variable of type contact_info
and here takein=take(); you are trying assign an array to a normal variable.