How to multiply two number which are in a link list in C++?

Question

I mean multiply two numbers which every digits are in every node of the link list finally put the digits of the result in every node of link list like I explained before.
For example:
Note: suppose the list below is link list and the digits in them are the data of node in link list.

first Number:

1

2

0

second Number:

1

0

the result of multiply:

1

2

0

Vulpes · Answer

If you don't like using the end() and rend() methods then I've altered it so that it just counts the number of iterations rather than testing for the end of the list: #include #include #include using namespace std; int get_number(list digits) { int number = 0; int multiplier = 1; list ::reverse_iterator rpos = digits.rbegin(); for (int i = 0; i { number += (*rpos) * multiplier; multiplier *= 10; rpos++; } return number; } int main() { list digits1 digits2 digits3 digits4; list ::iterator pos; int count1 count2 count3 count4 digit; int number1 number2 number3 number4 result result2; ofstream file; cout cin >> count1; cout for(int i = 0; i { cout cin >> digit; digits1.push_back(digit % 10); // only add last digit if more than one } cout cin >> count2; cout for(int i = 0; i { cout cin >> digit; digits2.push_back(digit % 10); // only add last digit if more than one } number1 = get_number(digits1); number2 = get_number(digits2); number3 = number1 * number2; number4 = number1 + number2; result = number3; result2 = number4; while (number3 > 0) { digit = number3 % 10; digits3.insert(digits3.begin() digit); number3 /= 10; } // print the third linked list count3 = 1; cout pos = digits3.begin(); for (int i = 0; i { cout count3++; pos++; } while (number4 > 0) { digit = number4 % 10; digits4.insert(digits4.begin() digit); number4 /= 10; } // print the fourth linked list count4 = 1; cout pos = digits4.begin(); for (int i = 0; i { cout count4++; pos++; } // now append the results to a file file.open(joe.txt ofstream::out | ofstream::app); file file file.close(); return 0; }

Vulpes · Answer

I can only write the program with the methods which the list class actually has. What difference does it make if it gives the correct results?

Joe Wilson · Answer

Thank you for your answer but I have requested to rewrite the multiply and sum functions for doubly linked list which doesn't end at last node and it has back to first node of linked list. please respond my request.

Vulpes · Answer

This very simple program should illustrate what I said in my previous post: #include #include using namespace std; int main() { list numbers; list ::iterator pos; list ::reverse_iterator rpos; numbers.push_back(2); // add at back numbers.push_back(3); // add at back numbers.push_front(1); // add at front // list should now contain 1 2 3 // iterate through nodes to prove this for (pos = numbers.begin(); pos != numbers.end(); pos++) { cout } // move iterator back 3 nodes pos--; pos--; pos--; // print value of the current node cout // now iterate through nodes in reverse order for (rpos = numbers.rbegin(); rpos != numbers.rend(); rpos++) { cout } return 0; } The output is: 1 2 3 1 3 2 1 Notice that if end() and rend() pointed to the last node itself rather than the (theoretical) position just after that then the 'for' statement would not be able to deal with the last node. You'd have to deal with it separately after the 'for' statement had ended.

Joe Wilson · Answer

I didn't get your comments very well could you please describe your answer in codes?

Vulpes · Answer

std::list is a doubly linked list. An iterator is provided to traverse through its nodes. This has a begin() method which points to the first node and an end() method which points to a theoretical position just after the last node. The end() method is defined like this so that when you reach it you know there are no more nodes in the list. The '++' operator is overloaded to increment the iterator and you can also decrement it with the '--' operator. There is also a reverse iterator which you can use to iterate backwards through the list which has the analogous methods rbegin() and rend(). So it's all good clean object oriented stuff :) Check this link for full details of std::list: http://www.cplusplus.com/reference/list/list/

Joe Wilson · Answer

Thank you for your answers and I request you to rewrite these linked list functions in doubly linked list which last pointer of that isn't null and it is pointing the first node of doubly linked list.

Vulpes · Answer

Well the linked list contains the digits of a number in reverse order of significance - the earlier nodes contain the most significant digits. So when we re-assemble the number from the digits we need to start with the later - least significant - nodes and therefore need to iterate backwards through the nodes. Fortunately the List class provides a reverse iterator for this purpose. So all we need to do is to iterate in reverse through the nodes de-referencing the pointer as we do so and multiplying by successive multiples of 10 until we eventually rebuild the number.

Joe Wilson · Answer

I have several problems in get_number function could you please explain more about this function?(Bold parts) int get_number(list digits) { int number = 0; int multiplier = 1; list ::reverse_iterator rpos; for (rpos = digits.rbegin(); rpos != digits.rend(); rpos++) { number += (*rpos) * multiplier; multiplier *= 10; } return number; }

Vulpes · Answer

OK I've amended the code to deal with that: #include #include #include using namespace std; int get_number(list digits) { int number = 0; int multiplier = 1; list ::reverse_iterator rpos; for (rpos = digits.rbegin(); rpos != digits.rend(); rpos++) { number += (*rpos) * multiplier; multiplier *= 10; } return number; } int main() { list digits1 digits2 digits3 digits4; list ::iterator pos; int count1 count2 count3 count4 digit; int number1 number2 number3 number4 result result2; ofstream file; cout cin >> count1; cout for(int i = 0; i { cout cin >> digit; digits1.push_back(digit % 10); // only add last digit if more than one } cout cin >> count2; cout for(int i = 0; i { cout cin >> digit; digits2.push_back(digit % 10); // only add last digit if more than one } number1 = get_number(digits1); number2 = get_number(digits2); number3 = number1 * number2; number4 = number1 + number2; result = number3; result2 = number4; while (number3 > 0) { digit = number3 % 10; digits3.insert(digits3.begin() digit); number3 /= 10; } // print the third linked list count3 = 1; cout for (pos = digits3.begin(); pos != digits3.end(); pos++) { cout count3++; } while (number4 > 0) { digit = number4 % 10; digits4.insert(digits4.begin() digit); number4 /= 10; } // print the fourth linked list count4 = 1; cout for (pos = digits4.begin(); pos != digits4.end(); pos++) { cout count4++; } // now append the results to a file file.open(joe.txt ofstream::out | ofstream::app); file file file.close(); return 0; }

Joe Wilson · Answer

Than you for your answers and by the way i want to add other function that sum my numbers and after creation of these sum and multiply function save the results of sum and multiply in text file. For example: For example: Note: suppose the list below is link list and the digits in them are the data of node in link list. first Number: 1 2 0 second Number: 1 0 0 0 the result of multiply: 1 2 0 0 0 0 the result of sum: 1 1 2 0

Vulpes · Answer

I've altered the code (alterations highlighted) to do that: #include #include #include using namespace std; int get_number(list digits) { int number = 0; int multiplier = 1; list ::reverse_iterator rpos; for (rpos = digits.rbegin(); rpos != digits.rend(); rpos++) { number += (*rpos) * multiplier; multiplier *= 10; } return number; } int main() { list digits1 digits2 digits3; list ::iterator pos; int count1 count2 count3 digit; int number1 number2 number3 result ; ofstream file; cout cin >> count1; cout for(int i = 0; i { cout cin >> digit; digits1.push_back(digit % 10); // only add last digit if more than one } cout cin >> count2; cout for(int i = 0; i { cout cin >> digit; digits2.push_back(digit % 10); // only add last digit if more than one } number1 = get_number(digits1); number2 = get_number(digits2); number3 = number1 * number2; result = number3; while (number3 > 0) { digit = number3 % 10; digits3.insert(digits3.begin() digit); number3 /= 10; } // print the third linked list count3 = 1; cout for (pos = digits3.begin(); pos != digits3.end(); pos++) { cout count3++; } // now append the result to a file file.open(joe.txt ofstream::out | ofstream::app); file file.close(); return 0; }

Joe Wilson · Answer

Thank you for your nice answer by the way how can I finally send the answer of multiply to text file. Please guide me.

Vulpes · Answer

Try this: #include #include using namespace std; int get_number(list digits) { int number = 0; int multiplier = 1; list ::reverse_iterator rpos; for (rpos = digits.rbegin(); rpos != digits.rend(); rpos++) { number += (*rpos) * multiplier; multiplier *= 10; } return number; } int main() { list digits1 digits2 digits3; list ::iterator pos; int count1 count2 count3 digit; int number1 number2 number3; cout cin >> count1; cout for(int i = 0; i { cout cin >> digit; digits1.push_back(digit % 10); // only add last digit if more than one } cout cin >> count2; cout for(int i = 0; i { cout cin >> digit; digits2.push_back(digit % 10); // only add last digit if more than one } number1 = get_number(digits1); number2 = get_number(digits2); number3 = number1 * number2; while (number3 > 0) { digit = number3 % 10; digits3.insert(digits3.begin() digit); number3 /= 10; } // print the third linked list count3 = 1; cout for (pos = digits3.begin(); pos != digits3.end(); pos++) { cout count3++; } return 0; }