subscripted value is neither array nor pointer nor vector when assigning an array element value

Search for the error in Google. The error message seems pretty obvious. Or share your code. We dont have magic crystal balls to predict the reasons.

Search for the error in Google. The error message seems pretty obvious. Or share your code. We dont have magic crystal balls to predict the reasons.

#include<stdio.h> sort(int arr[],int n) { int i,a,b[10],c[10]; a = arr[-1]; for(i=0;i<n;i++) { if(a<=a[i]) b[i]=a[i]; else c[i]=a[i]; } for(i=0;i<n;i++) printf("\n%d&%d",b[i],c[i]); } void main() { int arr[25],n; printf("enter the input:"); scanf("%d",&n); for(int i=0;i<n;i++) { printf("enter the numbers:"); scanf("%d",&arr[i]); } sort(arr,n); }

What is arr[-1]?

I want to compare the values of 2 arrays with last no of the array

I just tried to split an array and tried to store them in different variables

If else Block

a is not an array. It is an integer. You should use arr[i]. If a function doesnt return a value specify that it returns void. Your sort function doesnt specify one. And main is supposed to return an int.

a is not an array. It is an integer. You should use arr[i]. If a function doesnt return a value specify that it returns void. Your sort function doesnt specify one. And main is supposed to return an int.

wilding, i see the c standard specification says the negative array index is semantically correct, however the result is memory value before the array.

wilding, i see the c standard specification says the negative array index is semantically correct, however the result is memory value before the array.

Hypothetically speaking, Is it possible to write window manager via ncurses? How complicated can it get?

Can it be TUI? To manage files and etc, but make Linux boot up to it

You could design a ncurses based TUI application and modify the kernel command line parameters to boot into text mode by default and set systemd to start this application after the booting process is completed. The user can then navigate through the interface to use the system. But the functionality would be very limited You would have to worry about how your application is going to display files and so on. How you are going to start processes? Your program would be taking on the role of a shell.

Exactly what I wanted to hear, I guessed the functionality going to be very very basic. I want to experiment some variaty of things

Hi guys How to create c++ libraries I was search it from google but I can not found I need full tutorial

Just source code libraries or compiled libraries? On Linux or on Windows?

Simply just write a collection of classes and functions in C++, and bundle it into one project. That will be a library. What exactly do you mean by C++ libraries anyway?

With source code For linux

Expose your header files and compile your C++ files into a shared object (.so).

With source code For linux

I am going to work with pc drivers

like this is am right ( n & ( (n-1) == 0 ) )

Yes

Yes

hi, can anyone explain the functions of the program? (write how they work) #include <iostream> using namespace std; // функция для сортировки методом вставки (по убыванию) void insertionSort_m(int arr[], int left, int right) { for (int i = left + 1; i <= right; i++) { int temp = arr[i]; int j = i - 1; while (j >= left && arr[j] < temp) { arr[j+1] = arr[j]; j--; } arr[j+1] = temp; } } // функция для сортировки методом вставки (по возрастанию) void insertionSort_b(int arr[], int left, int right) { for (int i = left + 1; i <= right; i++) { int temp = arr[i]; int j = i - 1; while (j >= left && arr[j] > temp) { arr[j+1] = arr[j]; j--; } arr[j+1] = temp; } } // Функция для слияния отсортированных кусков void merge(int arr[], int l, int m, int r) { int len1 = m - l + 1, len2 = r - m; int left[len1], right[len2]; for (int i = 0; i < len1; i++) left[i] = arr[l + i]; for (int i = 0; i < len2; i++) right[i] = arr[m + 1 + i]; int i = 0; int j = 0; int k = l; while (i < len1 && j < len2) { if (left[i] <= right[j]) { arr[k] = left[i]; i++; } else { arr[k] = right[j]; j++; } k++; } while (i < len1) { arr[k] = left[i]; k++; i++; } while (j < len2) { arr[k] = right[j]; k++; j++; } } void timSort(int arr[], int n, int choise) { for (int i = 0; i < n; i+=32) if(choise == 1) insertionSort_b(arr, i, min((i+32-1),(n-1))); else insertionSort_m(arr, i, min((i+32-1),(n-1))); for (int size = 32; size < n; size = 2*size) { for (int left = 0; left < n; left += 2*size) { int mid = left + size - 1; int right = min((left + 2*size - 1),(n-1)); if(mid < right) merge(arr, left, mid, right); } } } int main() { setlocale(LC_ALL, "Russian"); int size; cout << "Введите размер массива: "; cin >> size; int arr[size]; for(int i=0; i < size; i++){ cout << "arr[" << i << "] = "; cin >> arr[i]; } cout << endl << "Ваш массив: " << endl; for(int i=0; i < size; i++){ cout << arr[i] << " "; } int choise; cout << endl << "Каким способом хотите отсортировать массив ?" << endl; cout << "1 - по возрастанию, 2 - по убыванию" << endl; cout << "Ваш выбор: "; cin >> choise; if (choise ==1 || choise ==2) { timSort(arr, size, choise); cout << endl << "Массив после сортировки: " << endl; for(int i=0; i < size; i++){ cout << arr[i] << " "; } }else cout << "Вы выбрали неправильный пункт!"; return 0; }

Names of the functions say it by yourself, here you have 2 variations of sorting algorithm by insertions. Just google it, there are a lot articles which could explain you the principle of that working. You have no need to get deeper with this info, just discover principles how these algorithms are built

hi, can anyone explain the functions of the program? (write how they work) #include <iostream> using namespace std; // функция для сортировки методом вставки (по убыванию) void insertionSort_m(int arr[], int left, int right) { for (int i = left + 1; i <= right; i++) { int temp = arr[i]; int j = i - 1; while (j >= left && arr[j] < temp) { arr[j+1] = arr[j]; j--; } arr[j+1] = temp; } } // функция для сортировки методом вставки (по возрастанию) void insertionSort_b(int arr[], int left, int right) { for (int i = left + 1; i <= right; i++) { int temp = arr[i]; int j = i - 1; while (j >= left && arr[j] > temp) { arr[j+1] = arr[j]; j--; } arr[j+1] = temp; } } // Функция для слияния отсортированных кусков void merge(int arr[], int l, int m, int r) { int len1 = m - l + 1, len2 = r - m; int left[len1], right[len2]; for (int i = 0; i < len1; i++) left[i] = arr[l + i]; for (int i = 0; i < len2; i++) right[i] = arr[m + 1 + i]; int i = 0; int j = 0; int k = l; while (i < len1 && j < len2) { if (left[i] <= right[j]) { arr[k] = left[i]; i++; } else { arr[k] = right[j]; j++; } k++; } while (i < len1) { arr[k] = left[i]; k++; i++; } while (j < len2) { arr[k] = right[j]; k++; j++; } } void timSort(int arr[], int n, int choise) { for (int i = 0; i < n; i+=32) if(choise == 1) insertionSort_b(arr, i, min((i+32-1),(n-1))); else insertionSort_m(arr, i, min((i+32-1),(n-1))); for (int size = 32; size < n; size = 2*size) { for (int left = 0; left < n; left += 2*size) { int mid = left + size - 1; int right = min((left + 2*size - 1),(n-1)); if(mid < right) merge(arr, left, mid, right); } } } int main() { setlocale(LC_ALL, "Russian"); int size; cout << "Введите размер массива: "; cin >> size; int arr[size]; for(int i=0; i < size; i++){ cout << "arr[" << i << "] = "; cin >> arr[i]; } cout << endl << "Ваш массив: " << endl; for(int i=0; i < size; i++){ cout << arr[i] << " "; } int choise; cout << endl << "Каким способом хотите отсортировать массив ?" << endl; cout << "1 - по возрастанию, 2 - по убыванию" << endl; cout << "Ваш выбор: "; cin >> choise; if (choise ==1 || choise ==2) { timSort(arr, size, choise); cout << endl << "Массив после сортировки: " << endl; for(int i=0; i < size; i++){ cout << arr[i] << " "; } }else cout << "Вы выбрали неправильный пункт!"; return 0; }

#Solution The code shown in this puzzle would be a bit weird if you have not heard of function-try-block. It is used to associate a try-catch block with the entire function. If used with a constructor, it covers the member initializer list and base class constructor calls as well. So if an exception is thrown during member initialization or from a base class constructor, it can be caught here. As to why this code throws an exception again even after it is handled already, read Herb Sutter's excellent explanation here: http://www.gotw.ca/gotw/066.htm

class Singleton { private: static Singleton *m_singleton; public: static Singleton &Get() { return *m_singleton; } void Print() { cout << "we are inside singleclass " << endl; } }; Singleton* Singleton::m_singleton=nullptr; int main() { Singleton::Get().Print(); Singleton::Get().Print(); Singleton::Get().Print(); Singleton::Get().Print(); }

yes that is null ...u can check by printing it ...check this for more clarity https://stackoverflow.com/questions/28482809/c-access-static-members-using-null-pointer