"expression must be an l-value or function designator" error when taking the address of this

Question

I'm trying to do this in C++:

class Abc
{
   int callFunction1()
};

void function1(Abc** c1) {//do something}

int Abc::callFunction1()
{
   function1(&this);
   return 0;
}

And I get "expression must be an l-value or function designator" error in visual studio 2015. So I don't understand where I go wrong. To my knowledge, &this should have the type Abc** right?

The function definition isn't mine to change. So I can't just change the parameter type.

In what case do you need to use a pointer to a pointer? I'd argue that it's not the proper approach most of the time — UnholySheep, Sep 20 '17 at 20:00
`"&this" should have type Abc** right?` It would have such type, but `this` is not an lvalue, as the error says. You can't apply `&` to it. — HolyBlackCat, Sep 20 '17 at 20:00
Change the function signature to: `void function1(Abc* c1)` and in then: `function1(this);` — Raindrop7, Sep 20 '17 at 20:01
@UnholySheep this is defined by others, I'm working on a project based on their defination... — disccip, Sep 20 '17 at 20:01
Thanks for everyone's help, as I said I can't change the definition, so are there any chance for me to achieve that? — disccip, Sep 20 '17 at 20:04

Arnav Borborah · Accepted Answer · 2017-09-21T14:46:09.297

The error is clear enough. Since this is not an lvalue, you cannot take its address. If you just want the address of the object, then just pass this, not &this, and change the function declaration to:

void function1(Abc* c1) //To just pass a pointer

However, since you mentioned you cannot change the definition of the function, you can create a temporary variable and pass its address:

auto temp = this;
function1(&temp);

How this works:

Since this is a prvalue and cannot have its address taken, you need something to point to it to turn it into an lvalue, here temp.
Now that temp points to this, taking temp's address will effectively take this's address, albeit indirectly.
Therefore, since you are passing the address of an lvalue to function1, the code compiles and works as expected.

"The error is clear enough" I have to disagree because "lvalue" is not a word, and without knowing more about it, the error is opaque. It's a little rude to imply beginners should just know what the problem is by reading that, I think... — Adam Barnes, Jul 09 '21 at 12:47

score 5 · Answer 2 · answered Sep 20 '17 at 20:18

From the C++ Standard (9.2.2.1 The this pointer)

1 In the body of a non-static (9.2.1) member function, the keyword this is a prvalue expression whose value is the address of the object for which the function is called.

and (5.3.1 Unary operators)

3 The result of the unary & operator is a pointer to its operand. The operand shall be an lvalue or a qualified-id....

To make it more clear consider the following code snippet.

If for example you have a declaration

int x = 10;

then you may not write

int **p = &&x;

In the right expression &x is a prvalue and according to the second quote from the Standard you may not apply the unary operator & to the prvalue.

You could write

int *q = &x;
int **p = &q;

because q is lvalue.

score 4 · Answer 3 · answered Sep 20 '17 at 20:12

The expression this is an rvalue, the same way that the expressions 137 or 'a' are, and so you can't take its address.

If you want to get a pointer to a pointer to this, you'll need to create a new variable of the right type:

auto* ptr = this;
doSomething(&ptr);

"expression must be an l-value or function designator" error when taking the address of this

3 Answers3