How to return an array in method decleration using C++? - c++

Question

I am trying to write C++ code suitable for object oriented programming.
I have two classes, namely, Student and Course. In the Student class, I have quiz_scores which is a 1-D array of 4 integers. I need both set and get methods, both are used in natural common way.
In the following, I implement setQuizScores method:
void Student :: setQuizScores(int* quizscores){
for(int i = 0; i<4; i++){
quiz_scores[i] = quizscores[i];
}
Where quizscores are my private members.
Now, next thing is that I want to return this quiz_scores array in the getQuizScores for each students of Student class.
However, the problem is that C++ does not allow us to return arrays directly. Instead, I want the structure of my code as following:
int Student :: getQuizScores(){
Do something;
return the elements of quiz_scores;
}
How can I do that efficiently?
I prefer not to use the Standard Template Library (STL), so I need to create my own arrays and access them according to the explanation above.

Answer 1

There are a few ways how you could return an array:
Pass in an array to copy to
void Student::getQuizScores(int* out) {
for(int i = 0; i < 4; i++)
out[i] = quiz_scores[i];
}
Student student;
int scores[4];
student.getQuizScores(scores);
// use scores[0], etc...
return a struct containing the array
struct Scores {
int values[4];
};
Scores Student::getQuizScores() {
Scores s;
for(int i = 0; i < 4; i++)
s.values[i] = quiz_scores[i];
return s;
}
Student student;
Scores s = student.getQuizScores();
// use s.values[0], etc...
return a reference to the quiz_scores array inside the class
using Scores = int[4];
Scores const& Student::getQuizScores() const {
return quiz_scores;
}
Student student;
Scores const& scores = student.getQuizScores();
// use scores[0], etc...

Answer 2

Just as setQuizScores() is able to take a pointer to an array, so too can getQuizScores() return a pointer to the quiz_scores member array, eg:
const int* Student::getQuizScores() const {
// do something...
return quiz_scores;
}
The caller can then access the array elements as needed, eg:
Student s;
...
const int *scores = s.getQuizScores();
for(int i = 0; i < 4; ++i){
cout << scores[i] << ' ';
}
Alternatively, since the array is fixed size, you can return a reference to the array instead, eg:
typedef int scoresArr[4];
scoresArr quiz_scores;
...
const scoresArr& Student::getQuizScores() const {
// do something...
return quiz_scores;
}
Student s;
...
const scoresArr &scores = s.getQuizScores();
for(int i = 0; i < 4; ++i){
cout << scores[i] << ' ';
}

Answer 3

You can return a pointer to the quiz_scores array through getQuizScores method as shown below:
Version 1: Using trailing return type
auto getQuizScores() -> int(*)[4]
{
//Do something;
return &quiz_scores;//NOTE THE & INFRONT OF quiz_scores
}
Now you can use this returned pointer to initialize other arrays. One possible example would be:
#include <iostream>
struct Student
{
int quiz_scores[4]= {1,2,3,4};
//getQuizScores returns a pointer to an array of size 4 with element of type int
auto getQuizScores() -> int(*)[4]
{
//Do something;
return &quiz_scores;//NOTE THE & INFRONT OF quiz_scores
}
void setQuizScores(int* quizscores)
{
for(int i = 0; i<4; i++)
{
quiz_scores[i] = quizscores[i];
}
}
};
int main()
{
Student s;
int arr[4];
for(int i = 0; i< 4; ++i)
{
arr[i] = (*s.getQuizScores())[i];
std::cout<<arr[i]<<std::endl;
}
return 0;
}
Version 2: Without using trailing return type
int (*getQuizScores())[4]
{
//Do something;
return &quiz_scores;//NOTE THE & INFRONT OF quiz_scores
}
Version 2 is the same as version 1 except that this time the getQuizScores method does not uses trialing return type.
There are other possibilities also like returning a reference to the quiz_scores array.