C++ pass a function in a class by reference - c++

Question

I'm trying to pass a reference to a function in a class but am having trouble figuring out how to do it. So say I have a class test defined as such
#include <iostream>
class test {
public:
test () {};
~test () {};
void setA (int);
int getA (void);
private:
int a;
};
void test::setA (int A) { a = A; }
int test::getA (void) { return a; }
using namespace std;
int main ()
{
test T;
T.setA(5);
cout << "a = " << T.getA() << endl;
return 0;
}
That works fine but if I want to pass the values by reference
#include <iostream>
class test {
public:
test () {};
~test () {};
void setA (int);
int & getA (void);
private:
int a;
};
void test::setA (int & A) { a = A; }
int & test::getA (void) { return a; }
using namespace std;
int main ()
{
test T;
T.setA(5);
cout << "a = " << T.getA() << endl;
return 0;
}
I cannot figure out how to configure setA to pass by reference.

Answer 1

There are two issues with the code. First, the definition of setA does not match the declaration. You must make the declaration take in a reference as a parameter.
Change this:
void setA (int);
To this:
void setA (int&);
The second issue is that you are trying to pass an r-value (5) as a reference. You must pass in an l-value. You can do that by creating an int first and then passing that by reference:
int i = 5;
T.setA(i);
Full example:
#include <iostream>
class test {
public:
test () {};
~test () {};
void setA (int&);
int & getA (void);
private:
int a;
};
void test::setA (int & A) { a = A; }
int & test::getA (void) { return a; }
using namespace std;
int main ()
{
test T;
int i = 5;
T.setA(i);
cout << "a = " << T.getA() << endl;
return 0;
}
When you pass something by reference to a function in C++, the function does not keep the parameter in memory automatically. Thus, you have to declare it before so that it stays in memory throughout the entire function.
The 5 you tried to pass as a reference would go out of scope and get destroyed as soon as the function starts. The declared i variable is instead destroyed at the end of the main function.

Answer 2

The reason is because in order to pass by reference, you must have an lvalue, which is a fancy way of saying something that persists beyond a single use.
If you created an int variable, you would be able to pass it in by reference. In the code above, you attempted to pass in a raw integer value (5), which fails, since the compiler is expecting a reference to an int, not a raw integer value.
The following code would work:
int main ()
{
test T;
int myVariable = 4; // Need an actual variable to pass by reference.
T.setA(myVariable);
cout << "a = " << T.getA() << endl;
return 0;
}
However, if you want your function to take raw integer values like you showed in your second example, you must have a function definition like your first example, where all the function takes is an integer. Hope this helps!

Answer 3

Maybe you could try this:
#include <iostream>
class test {
public:
test() {};
~test() {};
void setA(int&&); // requires at least C++11
void setA(int&);
int & getA(void);
private:
int a;
};
void test::setA(int && A) { a = A; }
void test::setA(int&A) { a = A; }
int & test::getA(void) { return a; }
using namespace std;
int main()
{
test T;
int i = 5;
T.setA(i);
cout << "a = " << T.getA() << endl;
T.setA(8);
cout << "a = " << T.getA() << endl;
return 0;
}
In the example, int& passes a l-value while int&& passes a r-value as a reference.