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Help me set up a double list.

zymir

Can't make a two-way list. It's a one-dimensional understanding of how to create, but there's a two-way poor.

What wrote:

#include <list>
using namespace std;
struct G{

G*k;

};

list<list<G>> *n;
void main() {

n = new list<list<G>>();// Это вроде верно

n.push_back(new G());// а тут как быть?

// n[0]. ... тоже не работает(

}

courtlanda

First, you don't have to create std::list by new♪ This type has a semantic value as it is. intand he should be used the same way.

Second, you need to set up at least one of the lists you made before you put it there. G()♪

list<list<G>> n;
n.push_back(list<G>()); // положили новый список в список верхнего уровня
n.back().push_back(G());

Either you can use the initial.

list<list<G>> n = {{G()}}; // список а в нем список с одним G

or emplace_back

list<list<G>> n;
n.emplace_back({G()}); // добавляем список из одного G

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int q1 = 0; .... a_i = (q1 * (1 - ((q2 + 1) / (q2 + 5)))); Here you are. a_i Just zero. Just because q1*♪ ♪ ♪if (fabs(a_i) >= eps) I see you're falling into the branch. else and get out of the cycle. ♪ ♪The amount of separation has already been mentioned:
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hasNext - That's it. return truethe sequence is infinite.You're taking the last revolving number into the private field.next In the cycle, it's found the next one behind the stored plain, updates the stored and returns it.That's it.If the upper limit is to be added, it may be:The safe number doesn't exceed the limit.Maintain the following simple number (number or limit + 1, if reached).You're keeping your current stored in the variable, and you're gonna keep your guarded pd. You return the one in the variable.If you need to filter the numbers out of the mass.Compareing a secure index with a length of mass.Instead of the number, you keep the index in the mass.Like in the past, but you're using the mass cycle.Now that's all:
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In this sentence mas = new int[n, n]; ^^^^ a comma operator is used. This proposal is equivalent to the following:mas = new int[n]; ^^ You probably need to distribute the masses to model the behaviour of the two-storey mass. For exampleint **mas, n; ^^^^^^^^ //... mas = new int *[n]; for ( int i = 0; i < n; i++ ) { mas[i] = new int[n]; } Once the masses have been used, the memory they occupy should be released in reverse.for ( int i = 0; i < n; i++ ) delete [] mas[i]; delete [] mas;
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A pointer is only a variable whose particularity is that it stores memory addresses. Thus, in the following example:int array[100]; int *ptr = array; By copying the pointer you are not creating a copy of 100 positions array, but you take the memory address where that variable starts and that's what it stores ptr.Well, what happens if we later work with ptr?ptr[0] = 10; std::cout << array[0]; We just changed the first position of array. Something logical and normal on the other hand. However, what happens if we change the memory direction ptr?int array2[5]; ptr = array2; ptr[0] = 11; std::cout << array[0]; Now array It hasn't been modified. Something that, on the other hand, is totally logical since before we have made it ptr Aim to array2 instead of a array.That is, the changes that are made in the memory region marked by the pointer will be visible changes from other variables, while making the pointer point to another region of memory will be an exclusively local change, only affects the pointer.Well, in the function you indicate:void remove_first_item(node_t **head) { node_t *new_node = 0; printf("%d\n", (*head)->next); new_node = (*head)->next; free(*head); *head = new_node; } It is easy to verify that using simple pointers the program will give problems:void remove_first_item(node_t *head) { node_t *new_node = 0; printf("%d\n", head->next); new_node = head->next; free(head); head = new_node; // <<--- Problema } Look at the highlighted line, in that line we are changing the memory address stored in head. That change, as we have commented a moment ago, is a local change, that is, only this variable head It will be the one affected by the change... the rest of the variables will not be affected. After leaving the program the function the change made in this last instruction is lost and the pointer head that you have created out of this function:typedef struct node { int data; struct node *next; } node_t; node_t *head = NULL; head = (node_t *)malloc(sizeof(node_t)); // <<--- ESTE He's gonna keep pointing to the element you just erased.

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