You could try consolidating the ethnicity and percentage into a single element, which makes adding and removing easier. You would start with a single row. Once selected, if the user needs to add another ethnicity to their profile, only then would they add another row with the + button. This way the choice of ethnicity and percentage is a single element, and you change both the ethnicity selection and amount close to each other, and just add or remove as needed. This element is sometimes known as a form stack.

I think you need to introduce a default value or recently used value already filled in the new form or some kind of checkbox for new form to auto populate the values from previous entry (mark it checked by default) This might speed up the adding entry process. Also, you can always have "Save and Add New" button to skip going to list view and adding the items quickly. Hope this helps.

It is why text columns in newspapers work. The reason is in gestalt laws, the law of proximity: We perceive items placed close together as belonging together. Columns are further apart than the items within them. Here is a nice (if old) page illustrating this: https://isle.hanover.edu/Ch05Object/Ch05ProxSim_evt.html

I don't think there is a maximum number, other than the difficulty of use. There are several solutions: Left side bar menu with main categories Scrolling window Horizontal slider Personally I would avoid the See more button, there are enough dynamic tools to show all items. Example https://www.staples.com/cleaning-supplies/cat_SC1683

Color is not the only visual contrast, try the others to use individually or in combination: Shape contrast Size contrast Weight contrast Figure/ground contrast Texture contrast Style contrast Position contrast

Many users will scroll horizontally if the design encourages them to do so. However, with 72 cards, they will get lost very quickly, whether horizontal or vertical. I would suggest you consider filter and sort controls to reduce the number of options visible and scrollable.

Can be checked classand at true perform the function. It's not possible to check. class♪ optionI don't see the difference. It'll be easier. https://stackoverflow.com/questions/10213620/how-to-check-if-an-option-is-selected http://jquery-docs.ru/traversing/is/

Function insert may, for example, be determined as shown in the demonstration programme below#include <iostream> struct ListMem { int listvalue; ListMem* next = 0; }; class List; class Iterator { public: void setValue(int a); void next(); int value(); bool equal(Iterator& other); private: ListMem* item = 0; friend class List; }; class List { public: void push(int a); Iterator begin(); Iterator end(); bool insert( int a, size_t n = 0 ); private: ListMem* cur = 0; ListMem* head = 0; }; void List::push( int a ) { head = new ListMem { a, head }; } bool List::insert( int a, size_t n ) { if ( n == 0 ) { push( a ); } else { ListMem **current = &head; while ( *current != nullptr &amp;&amp; --n ) current = &amp;( *current )-&gt;next; if ( n == 0 ) { ListMem *next = new ListMem { a, ( *current )-&gt;next }; ( *current )-&gt; next = next; } } return n == 0; } ;int main() { const int N = 10; List list; for ( int i = 0; i &lt; N; i++ ) list.insert( i, i ); return 0; }

If you need to change your first knot with the last one,you should have a reference to the last knot.The first knot should also be transmitted by reference.So first you have to write a function that's looking for an index on the last knot.She can look like she's going to be.newspapers ** getLastP(newspapers **head) { if (*head != NULL) { while ((*head)->next != NULL) head = &(*head)->next; } return head; } Then the function swap may look as follows:void swap(newspapers **head) { if (*head != NULL && (*head)->next != NULL) { newspapers **last = getLastP(head); newspapers *new_head = *last; if ((*head)-&gt;next == new_head) { new_head-&gt;next = *head; *last = NULL; } else { new_head-&gt;next = (*head)-&gt;next; *last = *head; (*last)-&gt;next = NULL; } *head = new_head; } } A demonstration programme is presented below.#include <stdio.h> #include <stdlib.h> typedef struct newspapers { int data; struct newspapers *next; } newspapers; int addNewspaper(newspapers **head, int data) { newspapers *tmp = malloc(sizeof(newspapers)); int success = tmp != NULL; if (success) { tmp-&gt;data = data; tmp-&gt;next = *head; *head = tmp; } return success; } newspapers ** getLastP(newspapers **head) { if (*head != NULL) { while ((*head)->next != NULL) head = &(*head)->next; } return head; } void swap(newspapers **head) { if (*head != NULL && (*head)->next != NULL) { newspapers **last = getLastP(head); newspapers *new_head = *last; if ((*head)-&gt;next == new_head) { new_head-&gt;next = *head; *last = NULL; } else { new_head-&gt;next = (*head)-&gt;next; *last = *head; (*last)-&gt;next = NULL; } *head = new_head; } } void printNewspapers(newspapers *head) { for (; head != NULL; head = head->next) printf("%d ", head->data); } int main() { const int N = 10; newspapers *head = NULL; for (int i = N; i != 0 &amp;&amp; addNewspaper(&amp;head, i); i--) { printNewspapers(head); printf("\n"); swap(&amp;head); printNewspapers(head); printf("\n\n"); } return 0; } Her withdrawal to the console followed10 10 9 10 10 9 8 10 9 9 10 8 7 9 10 8 8 9 10 7 6 8 9 10 7 7 8 9 10 6 5 7 8 9 10 6 6 7 8 9 10 5 4 6 7 8 9 10 5 5 6 7 8 9 10 4 3 5 6 7 8 9 10 4 4 5 6 7 8 9 10 3 2 4 5 6 7 8 9 10 3 3 4 5 6 7 8 9 10 2 1 3 4 5 6 7 8 9 10 2 2 3 4 5 6 7 8 9 10 1

Well, at least you don't initiate. pnew->firstYou have some kind of garbage in it when you try to add a list to it. Turn it off before the cycle.But it's definitely C? You have two functions with the same name.void add2list(Neighboor **nnhead, int val); void add2list(Node **pphead, int val, int neighoods[]); It works great at C++, but not C.

How do you make it possible that after filling the form for the first list, the verification automatically opens the form for "destructions"As an option, the URL to redirect to the form of a "violation" after the creation of a "check" can be submitted via the Source Parameter.And the records on the "violation" list should be linked to a certain verification (e.g. ID field)?On the form of a "check" after the page is downloaded, javascript's gonna find the last ID on the violation list and put it in the field.This option is fast (but not the best) and uses standard forms for listing elements (NewForm.aspx).It might be possible to create a form, a solution that would resolveThere is a need to make some violations for one verification, if necessary, easy to do so.

rotate :: Int -> [a] -> [a] rotate _ [] = [] rotate _ xs@[x] = xs rotate 0 xs = xs rotate n xs'@(x:xs) | n > 0 = rotate (n-1) (xs ++ [x]) | n < 0 = rotate (n+1) (last xs' : init xs') With regard to endless lists: there is no possibility of checking the list indefinitely, so the endless or no list should be determined by the programmer itself, the function of the endless list rotation is therefore limited to the removal of the left element and the transition to the following, i.e., no negative rotation, and the task of rotation of the endless list is not correct, in my view.

The mistake is that you work with a minimum element or a copy of it. On your previous question, as a result of one of the rights, I've all ruled the prototype that made a mistake in coping. In theory, your component transport code should look like:Single_List* max(Find_Max_Elem(Head)); // Обратите внимание, здесь тоже должен быть указатель, чтобы работать именно с элементом списка, а не с его копией Single_List* preMin(Find_PreMin_Elem(Head)); Single_List* min(preMin->Next); preMin->Next = min->Next; min->Next = max->Next; max->Next = min;

public DoubleNode(int value) { this.value = value; } public static DoubleNode doubleNodeGen(int length) { return doubleNodeGenRec(length, null); } private static DoubleNode doubleNodeGenRec(int length, DoubleNode prev) { if (length == 0) { return null; } int value = (int) (Math.random() * 10); DoubleNode node = new DoubleNode(value); node.prev = prev; node.next = doubleNodeGenRec(length - 1, node); return node; }

In general, the index list may be equal. NULL♪ Therefore, the starting code of the function is incorrect and may cause the programme &apos; s uncertain behaviourstruct Node *delete_all(struct Node *list, const char *str){ struct Node *temp,*a; if((strcmp(list->word,str))!=0){ //^^^^^^^^^^^^^^^^^^^^ list можкт быть равен NULL a=list; while(a->next->next!=NULL){ //^^^^^^^^^^^^^^^^^ поле a->next может быть равно NULL //.... Besides, if the first knot has a line that matches a given line, you only remove this knot. Other nodes having a concomitant line shall not be removed as the output occurs.And the condition in this cycle while(a->next->next!=NULL){ It doesn't make sense, because if the list has only two knots and the second knot has a concomitant line, because of the conditions in the cycle, this second knot will not be processed because a->next->next It's early. NULL♪The function may be written as follows:struct Node * delete_all( struct Node *list, const char *str ) { struct Node *curr = list; struct Node *prev = NULL; while ( curr ) { if ( strcmp( curr-&gt;word, str ) == 0 ) { struct Node *tmp = curr; if ( prev ) { prev-&gt;next = curr-&gt;next; } else { list = curr-&gt;next; } curr = curr-&gt;next; free( tmp-&gt;word ); free( tmp ); } else { prev = curr; curr = curr-&gt;next; } } return list; }

I don't understand what your problem is.So I'm gonna propose a code that allows the list to be filled with elements of the body. I hope that if it's not what you need, you will inform me in your commentary to the answer.Here's the demonstration program.#include <iostream> template <typename T> struct Node { T data; Node *next; }; template <typename T> Node<T> * arrayToList( const T a[], size_t n ) { Node<T> *head = nullptr; Node<T> **current = &head; for ( size_t i = 0; i &lt; n; i++, current = &amp;( *current )-&gt;next ) { *current = new Node&lt;T&gt; { a[i], nullptr }; } return head; } template <typename T> void displayList( Node<T> *head ) { for ( ; head; head = head->next ) std::cout << head->data << ' '; } int main() { int a[] = { 2, 1, 4, 3, 6, 5, 7, 8 }; const size_t N = sizeof( a ) / sizeof( *a ); for ( int x : a ) std::cout &lt;&lt; x &lt;&lt; ' '; std::cout &lt;&lt; std::endl; Node&lt;int&gt; *list = arrayToList( a, N ); displayList( list ); std::cout &lt;&lt; std::endl; } Its conclusion to the console:2 1 4 3 6 5 7 8 2 1 4 3 6 5 7 8 If your compiler doesn't support the initialization list for the operator. newthe proposal*current = new Node<T> { a[i], nullptr }; you can replace the following proposals*current = new Node<T>; ( *current )->data = a[i]; ( *current )->next = nullptr; Either you can write for class. Node The designer, so that we can rewrite everything in one line like, for example,*current = new Node<T>( a[i], nullptr );

If you're creating a bi-linked list and you're going to add elements to the list and remove them at the end of the list, you need to support two indexes, one at the beginning of the list and the other at the end of the list. Otherwise the code will be ineffective.Therefore, you should declare two structures. One structure will define the list &apos; s node and the other structure will define the list itself.For example,struct node { int value; struct node *next; struct node *prev; }; struct list { struct node *head; struct node *tail; }; Then, main You can determine the object of the list, for example:struct list list = { NULL, NULL }; The addition of the element at the beginning of the list can be determined as follows:void add_node( struct list *list, int value ) { struct node *tmp = ( struct node * )malloc( sizeof( struct node ) ); if ( tmp ) { tmp-&gt;value = value; tmp-&gt;next = list-&gt;head; tmp-&gt;prev = NULL; if ( list-&gt;head ) list-&gt;head-&gt;prev = tmp; list-&gt;head = tmp; if ( !list-&gt;tail ) list-&gt;tail = tmp; } } The removal of elements at the end of the list may be as follows:void delete_node( struct list *list ) { if ( list->tail ) { struct node *tmp = list->tail; list-&gt;tail = list-&gt;tail-&gt;prev; if ( list-&gt;tail ) list-&gt;tail-&gt;next = NULL; else list-&gt;head = NULL; free( tmp ); } } Here's the demonstration program.#include <stdio.h> #include <stdlib.h> struct node { int value; struct node *next; struct node *prev; }; struct list { struct node *head; struct node *tail; }; void add_node( struct list *list, int value ) { struct node *tmp = ( struct node * )malloc( sizeof( struct node ) ); if ( tmp ) { tmp-&gt;value = value; tmp-&gt;next = list-&gt;head; tmp-&gt;prev = NULL; if ( list-&gt;head ) list-&gt;head-&gt;prev = tmp; list-&gt;head = tmp; if ( !list-&gt;tail ) list-&gt;tail = tmp; } } void delete_node( struct list *list ) { if ( list->tail ) { struct node *tmp = list->tail; list-&gt;tail = list-&gt;tail-&gt;prev; if ( list-&gt;tail ) list-&gt;tail-&gt;next = NULL; else list-&gt;head = NULL; free( tmp ); } } void show_list( struct list *list ) { for ( struct node *current = list->head; current; current = current->next ) { printf( "%d ", current->value ); } } void show_reversed_list( struct list *list ) { for ( struct node *current = list->tail; current; current = current->prev ) { printf( "%d ", current->value ); } } int main( void ) { const int N = 10; struct list list = { NULL, NULL }; for ( int i = 0; i &lt; N; i++ ) { add_node( &amp;list, i ); show_list( &amp;list ); printf( "\n" ); } for ( int i = 0; i &lt; N; i++ ) { show_reversed_list( &amp;list ); delete_node( &amp;list ); printf( "\n" ); } } Its conclusion to the console:0 1 0 2 1 0 3 2 1 0 4 3 2 1 0 5 4 3 2 1 0 6 5 4 3 2 1 0 7 6 5 4 3 2 1 0 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 2 3 4 5 6 7 8 9 3 4 5 6 7 8 9 4 5 6 7 8 9 5 6 7 8 9 6 7 8 9 7 8 9 8 9 9

You've got a wrong list. Instead of pointing at first location, you identified the local object of the knot.int main() { struct dirlist list, *curr; ^^^^^^^^^^^^^^^^^^^ It's not usually done. Either an additional structure is in place which contains an index to the first part of the list, or an index to the first line of the list as the list itself.In addition, the field of structure has not been initiated. For example list->next not assigned to an equal nil.I mean, the list's not real.In your case, the knots need to be removed as follows:struct dirlist * freeList( struct dirlist *dlist ) { struct dirlist *curr = dlist->next; while ( curr ) { struct dirlist *tmp = curr; curr = curr-&gt;next; free( tmp-&gt;name ); free( tmp ); } free( dlist-&gt;name ); dlist-&gt;next = nullptr; return dlist; } You should have shown the full code how the nodes are added to the list. Also if it's really C++ code, you should use new operators and delete, not malloc and free functions.

The two-way line can also be implemented through a dynamic or bi-linked list.Arbitrary access to the element. Two sides of the line have no such access. If she's released through the mass, you can add that possibility, but will it be just a line?The element in the middle allows all the data structures you have listed.I think the difficulty of getting on the list and the two-ways is the same. In the dynamics, there will be a copy of the elements after the midway to the forward position.

For start-up add() Add the bracket. } after exit (and adjust the rest of the code accordingly) and then we look further.In general, the idea of using global variables is flawed, try to rewrite the entire relationship between functions through the transfer of parameters and the return of the result.