Why a function returning by address can not be a lvalue?

Question

Why is it not possible to set a function returning an address while it is possible to set a function that returns a reference.

int* returnByAddress()
{
    int x = 20;
    return &x;
}

int& returnByReference()
{
    int x = 30;
    return x;
}

int main()
{
    returnByReference() = 23; // possible
    int x = 23;
    returnByAddress() = &x;   // not a lvalue error 
    *returnByAddress() = 245; // possible

    return 0;
}

Answer 1

To start off, your code is broken. Returning a reference or the address of a local variable is returning garbage. Never do this. Here I've rewritten your example to return something real.

int xByAddress = 20;
int xByReference = 30;

int* returnByAddress()
{
    return &xByAddress;
}

int& returnByReference()
{
    return xByReference;
}

int main()
{
    returnByReference() = 23; //possible
    int x = 23;
    returnByAddress() = &x; //not a lvalue error 
    *returnByAddress() = 245; //possible

    return 0;
}

In this example, xByReference is assigned the value 23. xByAddress is assigned the value 245. Of course, returnByAddress() = &x; is invalid.

The rule is: functions return rvalues. The only way around this is to return a reference.

returnByAddress() returns a pointer by rvalue. You can deference the pointer to get an lvalue of the thing pointed at.

Following this rule, you can return a reference to the pointer.

int xByAddress = 20;
int *xPointer = nullptr;

int*& returnByPointerReference()
{
    xPointer = &xByAddress;
    return xPointer;
}

int main()
{
    int x = 23;
    returnByPointerReference() = &x; //not a lvalue error     
    return 0;
}

This code changes xPointer to point at the address of x.

Answer 2

What you are trying in your examples is very dangerous.

Function results as lvalues can make sense, but you have to ensure that the objects exist after the function returns. Here an example that makes (kind of) sense:

#include <iostream>

/// get a R/W reference to a byte in a buffer
/// Warning: no range checking is performed
unsigned char& refByteOfBuffer(void* buf, const int offset)
{
    return reinterpret_cast<unsigned char*>(buf)[offset];
}

void testRefWrite()
{
    unsigned char buf[] = "Hello world!";
    refByteOfBuffer(buf, 1) = 'a';
    std::cout << buf << std::endl;
}

The output will be Hallo world!, after the e in buf at offset 1 was changed to a.