CODEWITHSUNDEEP
c++classlvalue
Maulik Soneji
Maulik Soneji

Reputation: 177

C++ Error: lvalue required as unary '&' operand

I have created a class named node. It is defined as follows:

class node
{
    private:
        int key;
        node * next;

    public:
        void setkey(int key);
        void setnext(node * next);
        int getkey();
        node * getnext();
};

These functions are basic getter and setter functions. Now, if I try to do this:

MoveNode(&(bTail->getnext()),&start);

it shows error as :lvalue required as unary '&' operand

Now, if I update the class Node and keep (node * next) as public and access via this:

MoveNode(&(bTail->next),&start);

It shows no error.

Please help as to how can I get rid of the error.

Upvotes: 2

Views: 3871

Answers (3)

eerorika
eerorika

Reputation: 238411

MoveNode(&(bTail->getnext()),&start);

it shows error as :lvalue required as unary '&' operand

Of course it does. getnext returns a temporary value (a copy of the pointer) and temporaries are rvalues. Rvalues don't have an associated memory address, so you may not use them as an oprand for &.

MoveNode(&(bTail->next),&start);

It shows no error.

Well, bTail->next is an lvalue, so there is no syntax error there.

If you need to pass into MoveNode a pointer to a pointer, then that pointed to pointer must be stored somewhere. Since it's stored as a member in the node, you could return a reference instead of a copy of the pointer in which case you'd be returning an lvalue and your syntax would be correct.

Upvotes: 1

TartanLlama
TartanLlama

Reputation: 65720

MoveNode(&(bTail->getnext()),&start);

This doesn't work because you are trying to take the address of a temporary pointer returned by getnext. The fact that pointers can be temporary might sound strange at first, but essentially you aren't getting a pointer to where the node stores its next pointer, but to some meaningless memory location.

If you want really want &(bTail->getnext()) to return the address of next within that node, make getnext return a reference to that pointer:

node *& getnext();

Upvotes: 0

Petr
Petr

Reputation: 9997

When you return node* from getnext(), you create a temporary variable of node* type, so it has no sense to take its address by using &, this temporary variable will anyway be destroyed soon.

When you directly access next, you refer to a long-living variable, and you can take its address.

You might want to return a reference to node* from getnext():

node*& getnext() {return next;};

so that it will point to the same variable next and you will be able to take its address.

Upvotes: 2

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