How to gauge the pointer in the link list in C++?

Question

First ask user to enter some numbers then put them in a link list next ask user to choose the location of the pointer which point a node in link list then gauge (move) the pointer to the beginning of the link list or ask user to choose the number that pointer will be moved in link list

for example:

if user choose number 3 the pointer move left 3 nodes from the location which have been chosen before by user, at last show the data of the node which pointer point after movement .

Joe Wilson · Answer

Thank you for your answer by the way how can i move pointer forward . for example: if user choose number 3 the pointer move right 3 nodes from the location which have been chosen before by user at last show the data of the node which pointer point after movement .

Vulpes · Answer

Try this: #include #include #include using namespace std; int main() { list numbers; list ::iterator pos; int number count start i move; cout cin >> count; cout for(int i = 0; i { cout cin >> number; numbers.push_back(number); } cout do { cout cin >> start; } while(start count); pos = numbers.begin(); if (start > 1) { for(i = 2; i } cout do { cout cin >> move; } while (move start - 1); if (move > 0) { for(i = 1; i } cout return 0; }

Vulpes · Answer

I've altered the code so that the user can now move the pointer to the left or to the right of the initial position: #include #include using namespace std; int main() { list numbers; list ::iterator pos; int number count start i move; char lr; cout cin >> count; cout for(int i = 0; i { cout cin >> number; numbers.push_back(number); } cout do { cout cin >> start; } while(start count); pos = numbers.begin(); if (start > 1) { for(i = 2; i } cout do { cout cin >> lr; if (lr == 'L') lr = 'l'; if (lr == 'R') lr = 'r'; } while (lr != 'l' && lr != 'r'); cout if (lr == 'l') { do { cout cin >> move; } while (move start - 1); if (move > 0) { for(i = 1; i } cout } else { do { cout cin >> move; } while (move count - start); if (move > 0) { for(i = 1; i } cout } return 0; }