C ++通过引用传递类中的函数

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.