How to create an endless loop

Question

How the task sounds:

Write a program "directory". Create two one-dimensional arrays. One array stores ICQ numbers, the second - phone numbers. Implement user menu:

1. Sort by ICQ numbers

2. Sort by phone numbers

3. List users

4. Exit

Question:

Everything that I wanted to realize was done, but I don’t like goto and I would like to understand how to do without it.

The code itself:

#include <iostream> #include <ctime> using namespace std; int main() { setlocale(LC_ALL, "Russian"); srand(time(0)); const int ICQ = 10; const int N = 10; int arr1[ICQ] = {}; int arr2[N] = {}; int num = 0; tryAgain2: for (int i = 0; i < ICQ; i++) { arr1[i] = rand() % 101; } for (int i = 0; i < N; i++) { arr2[i] = rand() % 90000000 + 9999999; } tryAgain: cout << "№\t Номер телефона \t Номер ICQ" << endl; cout << "1.\t " << arr2[0] << "\t\t " << arr1[0] << endl; cout << "2.\t " << arr2[1] << "\t\t " << arr1[1] << endl; cout << "3.\t " << arr2[2] << "\t\t " << arr1[2] << endl; cout << "4.\t " << arr2[3] << "\t\t " << arr1[3] << endl; cout << "5.\t " << arr2[4] << "\t\t " << arr1[4] << endl; cout << "6.\t " << arr2[5] << "\t\t " << arr1[5] << endl; cout << "7.\t " << arr2[6] << "\t\t " << arr1[6] << endl; cout << "8.\t " << arr2[7] << "\t\t " << arr1[7] << endl; cout << "9.\t " << arr2[8] << "\t\t " << arr1[8] << endl; cout << "10.\t " << arr2[9] << "\t\t " << arr1[9] << endl; cout << "\n1.\t Отсортировать по номерам ICQ" << endl; cout << "2.\t Отсортировать по номерам телефона" << endl; cout << "3.\t Вывести список пользователей" << endl; cout << "4.\t Выход" << endl << endl; cin >> num; switch (num) { case 1: system("cls"); for (int i = 1; i < ICQ; ++i) { int k = i; while (k > 0 && arr1[k - 1] > arr1[k]) { int tmp = arr1[k - 1]; arr1[k - 1] = arr1[k]; arr1[k] = tmp; tmp = arr2[k - 1]; arr2[k - 1] = arr2[k]; arr2[k] = tmp; k -= 1; } } goto tryAgain; case 2: system("cls"); for (int i = 1; i < N; ++i) { int k = i; while (k > 0 && arr2[k - 1] > arr2[k]) { int tmp = arr2[k - 1]; arr2[k - 1] = arr2[k]; arr2[k] = tmp; tmp = arr1[k - 1]; arr1[k - 1] = arr1[k]; arr1[k] = tmp; k -= 1; } } goto tryAgain; case 3: system("cls"); goto tryAgain2; break; case 4: return 0; } }

If your task is to arr1[ICQ] arrays (for ICQ numbers) and arr2[N] (for phone numbers) to be of the same size (otherwise the table printing will be nonsense), then specify the size of these arrays by one constant (for example , N_USERS ), and not 2 as you have now.

freim freim 4,390 3 ten 27 · Accepted Answer · 2019-02-07T11:41:50

goto alone is easy to clean. For a more elegant program, you need to somehow rework the algorithm, and just an equivalent conversion to for will look like this:

 #include <iostream> #include <ctime> using namespace std; constexpr int ICQ = 10; constexpr int N = 10; void Reinit(int arr1[], int arr2[]) { for (int i = 0; i < ICQ; i++) { arr1[i] = rand() % 101; } for (int i = 0; i < N; i++) { arr2[i] = rand() % 90000000 + 9999999; } } int main() { setlocale(LC_ALL, "Russian"); srand((unsigned)time(0)); int arr1[ICQ]; int arr2[N]; int num = 0; Reinit(arr1, arr2); for (;;) { cout << "№\t Номер телефона \t Номер ICQ" << endl; cout << "1.\t " << arr2[0] << "\t\t " << arr1[0] << endl; cout << "2.\t " << arr2[1] << "\t\t " << arr1[1] << endl; cout << "3.\t " << arr2[2] << "\t\t " << arr1[2] << endl; cout << "4.\t " << arr2[3] << "\t\t " << arr1[3] << endl; cout << "5.\t " << arr2[4] << "\t\t " << arr1[4] << endl; cout << "6.\t " << arr2[5] << "\t\t " << arr1[5] << endl; cout << "7.\t " << arr2[6] << "\t\t " << arr1[6] << endl; cout << "8.\t " << arr2[7] << "\t\t " << arr1[7] << endl; cout << "9.\t " << arr2[8] << "\t\t " << arr1[8] << endl; cout << "10.\t " << arr2[9] << "\t\t " << arr1[9] << endl; cout << "\n1.\t Отсортировать по номерам ICQ" << endl; cout << "2.\t Отсортировать по номерам телефона" << endl; cout << "3.\t Вывести список пользователей" << endl; cout << "4.\t Выход" << endl << endl; cin >> num; switch (num) { case 1: system("cls"); for (int i = 1; i < ICQ; ++i) { int k = i; while (k > 0 && arr1[k - 1] > arr1[k]) { int tmp = arr1[k - 1]; arr1[k - 1] = arr1[k]; arr1[k] = tmp; tmp = arr2[k - 1]; arr2[k - 1] = arr2[k]; arr2[k] = tmp; k -= 1; } } break; case 2: system("cls"); for (int i = 1; i < N; ++i) { int k = i; while (k > 0 && arr2[k - 1] > arr2[k]) { int tmp = arr2[k - 1]; arr2[k - 1] = arr2[k]; arr2[k] = tmp; tmp = arr1[k - 1]; arr1[k - 1] = arr1[k]; arr1[k] = tmp; k -= 1; } } break; case 3: system("cls"); Reinit(arr1, arr2); break; case 4: return 0; } } }

Minor notes:

1) For srand you need to do time_t castes.

2) It is not necessary to initialize the arrays during the declaration, they will still be initialized later.

3) After goto no need to put a break .

Thanks to you, I learned about the existence of "constexpr", I read about it in more detail today.
Be so kind as to explain this: void Reinit (int arr1 [], int arr2 []) Reinit (arr1, arr2);
Also did not quite understand about the "time_t" how and where to use?
@Blacit, - 1) Reinit - I just put a piece of code into a function to remove the second label.
Otherwise, two intersecting blocks were obtained, and this did not fit into a simple cycle.
2) time_t - the time function returns a result of the time_t type, and srand requires an unsigned int argument.
You can not tell how to add more usernames moved with phones and icq.
And how can you save the original information on the arrays, so that by pressing the 3 button - we would have the original table, which was created from the very beginning.
@Blacit, well, I can’t rewrite your program, I just don’t have time, but I can tell you the direction.
First, define a struct that will describe the user — name, phone, everything.
Then create one single vector that will hold these structures.

αλεχολυτ 21.1k 9 39 92 · Answer 2 · 2019-02-07T11:44:55

for(;;) - this is an infinite loop. Banal answer, but still.

Thank you very much :) Above provided the entire code. - Blacit

How to create an endless loop

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