Taking address of rvalue warning

I am trying to pass in the memory address of the mpuTest class property to the fsVar.initialize() method, but I get this error:

main.cpp:38:42: error: taking address of rvalue [-fpermissive]
   38 |     fsVar.initialize(&test3Var.getMpuTest());

I have done a little reading and have learned that an rvalue is something that may not have storage or memory associated with it, like a simple integer. I also know that an lvalue is something that is stored in memory with an address, like a variable.

But I have noticed with the below code that if I have a getter in a class for a property of that class, and I try and use that return value’s memory address with &, it will give me the above error. If you don’t understand this example please view the below code example.

#include <iostream>

using namespace std;

class mpu {
public:
    void printSomething() {
        cout << "Hi" << endl;
    }
};

class fs {
private:
    mpu* t1;
public:
    void initialize(mpu *mpuPtr) {
        this->t1 = mpuPtr;
    }
    
    void printSomething2() {
        this->t1->printSomething();
    }
};

class Test3 {
private:
    mpu mpuTest;
public:
    mpu getMpuTest() {
        return mpuTest;
    }
};

int main()
{
    Test3 test3Var;
    fs fsVar;
    fsVar.initialize(&test3Var.getMpuTest());
    fsVar.printSomething2();
    return 0;
}

>Solution :

The reason why you are getting this error is because test3Var.getMpuTest() is is returning by value, not by reference. When you return an object by value, a temporary – an rvalue – copy of the object is created. Hence the error message talks about rvalue:

main.cpp:38:42: error: taking address of rvalue [-fpermissive]
   38 |     fsVar.initialize(&test3Var.getMpuTest());

The & operator in C++ is only valid for lvalues – objects that have a memory location, not for rvalues (temporary objects), which is why the compiler is giving you this error.

Solution is to return by reference:

mpu& getMpuTest() {
    return mpuTest;
}