c - Segmentation fault deleting nodes from singly linked list -

c - Segmentation fault deleting nodes from singly linked list -

this case working on

[11] -> [12] -> [13] -> null

i trying delete elements liked list above(example) maintain getting segfault , on running gdb doesnot help much. not looking reply , explanation on going wrong logically.

here code

int list:: remove( int val ) { listnode *headnode = _head; listnode *tempnode = null; if(headnode->_value == val){ tempnode = headnode->_next; delete headnode; _head = tempnode; } else { while(headnode->_value != val){ tempnode = headnode; headnode = headnode->_next; } tempnode->_next = headnode->_next; delete headnode; } }

you're not accounting next conditions:

the list may empty; i.e. _head null; the value may not in list @ all. your function declared homecoming int, makes no such return

assuming rest of code right (and big assumption), i'm all-but-certain you're trying do:

void list::remove( int val ) { listnode *headnode = _head; listnode *tempnode = null; while (headnode && headnode->_value != val) { tempnode = headnode; headnode = headnode->next; } if (headnode) { if (tempnode) tempnode->next = headnode->next; else _head = headnode->next; delete headnode; } }

alternatively, if inclined can (arguably) simpler utilizing pointer-to-pointer traverse pointers in list, not values. worth investigating how next works, still covers bases described previously, using actual pointers in list nodes themselves, including _head, by-address rather by-value, thereby eliminating need walk-behind temporary pointer:

void list::remove( int val ) { listnode **pp = &_head; while (*pp && (*pp)->_value != val) pp = &(*pp)->next; if (*pp) { listnode *p = *pp; *pp = p->next; delete p; } }

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