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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T14:26:06+00:00 2026-05-14T14:26:06+00:00

Consider the following list: [LinkNode * head — LinkNode * node1 — LinkNode *

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Consider the following list:

[LinkNode * head — LinkNode * node1 — LinkNode * node2]

I’m creating a stack of FIFO.
I Call pop() which I want to pop node1.

LinkNode::LinkNode(int numIn) {
    this->numIn = numIn;
    next = null;
}
.
.
.
int LinkNode::pop() {
    Link * temp = head->next;
    head = temp->next;
    int popped = head->nodeNum;
    delete temp;
    Return numOut;

Question:
1) head should be a pointer or a LinkNode *?
2) Link * temp is created on the call stack and when pop finishes doesn’t temp delete automatically?
3) My major confusion is on what is the value of temp->next? Does this point to node1.next which equals node2?

Appreciate your help?

My reference is C++ for Java Programmers by Weiss.

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  1. Editorial Team
    Editorial Team
    2026-05-14T14:26:07+00:00Added an answer on May 14, 2026 at 2:26 pm
    1. LinkNode * is a pointer. So I’m not sure what you are asking.
    2. The variable goes out of scope but this does not automatically remove the dynamically allocated data. In C++, if you dynamically allocate data (call new) you need to free it (call delete)
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