Assigning to a pointer returned by a function call

I'm trying to implement a simple LinkedList The following开发者_运维问答 code, producing this error:

error: lvalue required as left operand of assignment

The line is:

newNode->getNext() = _head;

Where getNext() returns a Node*, and i'm trying to make the new Node point at the current head before updating it. I'm probably thinking in Java and doing the transition wrong.

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Since you're implementing the class, you have several options:

1. implement a setNext(Node*) method (possibly private) and use newNode.setNext(_head);
2. assign to the node's _next member directly: newNode._next = _head;
3. make Node's constructors take the initial value of _next as an argument and pass _head when constructing newNode.

Since you didn't post your code, I took a guess at the name of the "next" pointer and called it _next. Also, the points 1-3 above are not mutually exclusive and are not the only possibilities.

P.S. Take a look at the C FAQ if you're unsure what "lvalue" and "rvalue" mean.

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Either, dereference:

Node* Node::getNext();
*(newNode->getNext()) = _head;

(assuming _head has type Node and the assignment operator was implemented with care in Node) or, what is probably better

Node& Node::getNext();
newNode->getNext() = _head;

But anyway, this is not the way you should be doing this. You may want to implement a setter.

Also, have you considered using std::list instead of re-inventing the wheel?

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You need to have the variable you are assigning to on the left hand side of the '=' sign.

_head = newNode->getNext();

After reading again, it doesn't really achieve what you want. I'd recommend implementing a "setNext" function that takes a node as an arguement, or just accessing the 'next' variable directly. newNode.next = _head; for example.

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You're placing a function call on the left side, which you really can't do. Are you sure you didn't mean...

_head = newNode->getNext();