Your implementation is all wrong. For one thing, you are new‘ing temporary Node objects that you shouldn’t be allocating at all, and you are leaking them. And, you are not actually outputting the value of the Node that you delete. And, your whole function can be made much simpler, as you don’t need to enumerate … Read more
That’s why there are debug tools like gdb (just google for it 😉 ). This is the backtrace: #0 0x00000000004009ba in node::setnext (this=0x0, x=0x614cc0) at a.cpp:25 #1 0x0000000000400c18 in list::insertend (this=0x7fffffffdf10, x=45) at a.cpp:109 #2 0x00000000004008df in main () at a.cpp:137 that means in line 137 there is a function call (l.insertend(45)), then in line … Read more
I don’t understand how this change modifies the head too. I thought curr was just a variable. If this assignment were an assignment to curr you’d be right, and it wouldn’t mutate the list, but the assignment is not to curr, but to curr.next, and that is not a variable, but an attribute of a … Read more
You need both of those lines for the code to work correctly as far as I can tell. b.next = head.next makes b point at whatever head was pointing at (aka the first node in the list). And then head.next = b makes head point at b. Therefore these two lines insert b at the … Read more
Heading Hi, I have not looked thoroughly in the code . But , in a quick look I spotted the following basic problem nextPtr should be a pointer, you are creating an instance and copying the contents of the next structure. This is not efficient. I will declare it as below struct listNode *nextPtr; / … Read more
Think very carefully about the code you have for stepping through the list looking for where you can append your new value, especially the last line inside that loop body: while(nodeptr->next){ nodeptr=nodeptr->next; nodeptr->next=newNode; } Were you to actually single-step your code through a debugger, examining the list after each statement, you would see the problem … Read more
Yes – poll removes the first item in the list and returns it. This will then be evaluated in your if statement. On an unrelated note, it is worth mentioning that the size of edges will obviously change on every loop since polling edges reduces the element count by one. However, your for loop evaluates … Read more
This cannot work as void does not contain any member next. You must use your node structure struct tmpList * to access any members. Functions dealing with list manipulation should use the prope node type in the signature to get some type safety. If you really want to use void * in the signature, you … Read more
There are 2 problems: you are not saving the new head once it is returned (when you reach the end of the list), so the returned head is always going to be the one of the first stack of recursion, which is the original first node; you are not assigning node.next to null on the … Read more
This line while(temp->next != NULL) explicitly says “stop when you get to the element that points to end-of-list” (i.e stop when you get to the last element but before you use it). Instead use while(temp != NULL) which says “stop when you are no-longer in the list”. 4 solved Linked List in C not printing … Read more
Simply use std::is_sorted something like: if (std::is_sorted(std::begin(linked_list), std::end(linked_list)) { //… } solved How to check whether a linked list is sorted or not using c++?
I am not sure what del is, but assuming it is a node in your deleted words linked list, all your doing is assigning the deleted word to del in del = temp. If you want to add it onto the del linked list you have to create a new node and add it on … Read more
Because there is no free lunch. In general, if a data structure is more powerful than another, you will have to pay for that additional power somehow. Sometimes you will not be willing to pay that price. As for lists, you are correct, in terms of functionality, they are way better than arrays: You can … Read more
Consider the following Linked List: HEAD —> [“value2” | -]–> [“value3” | -]–> [“last_value” | null] Assuming current points to “value3”: HEAD —> [“value2” | -]–> [“value3” | -]–> [“last_value” | null] ↑ then current.next != null is true, since current.next is actually the “next” node. Now let’s assume current moved to “last_value”, now we’ll … Read more
In your Insert function, you never allocate newNode. Node* newNode; You need to allocate it like so: Node* newNode = new Node(); The program runs correctly after fixing this, and the output is: 24 http://ideone.com/16XL5W EDIT: Regarding your comment below, the following lines do not allocate anything: Node* newNode; struct Node* newNode; They are simply … Read more