void myFun(Node* head) {Node* current = head;while (current != nullptr) {if((current->data)%2==0)std::cout << current->data << " "; current = current->next;}std::cout << std::endl; }Output of above code in linked list
Question
void myFun(Node* head) {Node* current = head;while (current != nullptr) {if((current->data)%2==0)std::cout << current->data << " "; current = current->next;}std::cout << std::endl; }Output of above code in linked list
Solution
The provided code is a function in C++ that traverses a linked list and prints out the data of each node that contains an even number.
Here's a step-by-step explanation:
-
The function
myFunis defined with a single parameterheadwhich is a pointer to the first node of a linked list. -
A new pointer
currentis initialized to point to the head of the linked list. -
A
whileloop is started which continues untilcurrentbecomesnullptr, i.e., until we have traversed all nodes in the linked list. -
Inside the loop, an
ifcondition checks whether the data in the current node is even. This is done by taking the modulo of the data with 2 (current->data % 2) and checking if it equals 0. -
If the data is even, it is printed to the standard output followed by a space.
-
The
currentpointer is then moved to the next node in the list (current = current->next). -
Once the loop has traversed all nodes, a newline character is printed to the standard output
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