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Home/ Questions/Q 7511153
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Editorial Team
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Editorial Team
Asked: May 29, 20262026-05-29T23:27:36+00:00 2026-05-29T23:27:36+00:00

Assume i have struct node{ int val; node* left; }; Now, I have a

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Assume i have

struct node{
  int val;
  node* left;
};

Now, I have a priority_queue<node> sth;
What does the following do:

node temp = sth.top();// does it perform copy
sth.pop();
temp.left = sth.top(); // is this allowed?

How do I pop an element from the queue and store it in temp.left?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-29T23:27:37+00:00Added an answer on May 29, 2026 at 11:27 pm
    node temp = sth.top();// does it perform copy
    

    Yes, this makes a copy of the top element and stores it in temp. temp.left points to the same place as top node’s left pointer. Note that top() actually returns a reference but this line asks for a copy. node &temp = sth.top(); would declare a reference to the top element.

    sth.pop();
    

    Removes the top node from the queue. You may have just broken your data structure here. Any pointers to that node object (e.g., left pointers of other nodes in the queue) are now invalid.

    temp.left = sth.top(); // is this allowed?
    

    No, this won’t compile. temp.left is a pointer and top() returns a reference to an object. At a minimum, you need to get the address of that object: temp.left = &sth.top();

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