Deleting Node in Linked List C++ -

so i've been searching forums, im still new language , linked lists can barely decipher results.

basically made delete function linked list. can create list, traverse list, sort list, search list, , insert before node in linked list. recycled code insert locate point in list delete. main point of confusion how link previous points node after 1 deleting.

i won't write whole new linked list implementation can point out of problems code you.

the trick stay 1 node ahead of 1 want delete.

i have renamed entry current clarity

nodetype *current , *first, *next; int akey;  // line search start second element. // current =start->ptr; // should current = start;  // not needed. assuming last node has null value '->ptr' // last=start; next = current->ptr;  cout<<"input data delete"<<endl; cin>>akey;  // check if first node contains data // assuming list have @ least 1 element. i.e. current not null  while (current->adata == akey) {   // delete it.   delete current;   // update current while loop   current = next;   // update next too.   next = current->ptr; } // know first element doesn't contain data. // update pointer begging of new list if removed top. first = current;  // has unnecessary checks.  // ****specifically (akey!=current->adata)  // prevent entering loop if false. // while((akey!=current->adata)&&(current->ptr !=null)) while(next != null)  // should enough {      if(next->adata == akey)      {           // make current node (before 'deletion point')            // lined 1 after 'deletion point (next of next)           current->ptr = next->ptr;            // delete node.            delete next;           // make next pointer point new next.            next = current->ptr      }     // otherwise advance both current , next.      else {            current = next;            next = next->ptr;       } }  // use test it. current = first; while(current){     cout<<current->adata<<", ";     current = current->ptr; } 

this not cleanest way. similar implementation can see went wrong.