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Remove return value from Lst_Concat. None of the callers ever checked

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it. Remove stuff that was needed for circular lists.
This commit is contained in:
harti 2004-12-08 16:47:19 +00:00
parent debc547816
commit bcfafac95e
2 changed files with 20 additions and 36 deletions
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2
usr.bin/make/lst.h
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@ -125,7 +125,7 @@ ReturnStatus Lst_Remove(Lst *, LstNode *);
#define Lst_Replace(NODE, D) (((NODE) == NULL) ? FAILURE : \
(((NODE)->datum = (D)), SUCCESS))
/* Concatenate two lists */
ReturnStatus Lst_Concat(Lst *, Lst *, int);
void Lst_Concat(Lst *, Lst *, int);
/*
* Node-specific functions

54
usr.bin/make/lst.lib/lstConcat.c
View File

@ -70,7 +70,7 @@ __FBSDID("$FreeBSD$");
* New elements are created and appended the the first list.
*-----------------------------------------------------------------------
*/
ReturnStatus
void
Lst_Concat(Lst *list1, Lst *list2, int flags)
{
LstNode *ln; /* original LstNode */
@ -78,40 +78,27 @@ Lst_Concat(Lst *list1, Lst *list2, int flags)
LstNode *last; /* the last element in the list. Keeps
* bookkeeping until the end */
if (!Lst_Valid(list1) || !Lst_Valid(list2)) {
return (FAILURE);
}
if (list2->firstPtr == NULL)
return;
if (flags == LST_CONCLINK) {
if (list2->firstPtr != NULL) {
/*
* We set the nextPtr of the last element of list two to be NULL
* to make the loop easier and so we don't need an extra case --
* the final element will already point to NULL space and the first
* element will be untouched if it existed before and will also
* point to NULL space if it didn't.
*/
list2->lastPtr->nextPtr = NULL;
/*
* So long as the second list isn't empty, we just link the
* first element of the second list to the last element of the
* first list. If the first list isn't empty, we then link the
* last element of the list to the first element of the second list
* The last element of the second list, if it exists, then becomes
* the last element of the first list.
*/
list2->firstPtr->prevPtr = list1->lastPtr;
if (list1->lastPtr != NULL) {
list1->lastPtr->nextPtr = list2->firstPtr;
} else {
list1->firstPtr = list2->firstPtr;
}
list1->lastPtr = list2->lastPtr;
}
} else if (list2->firstPtr != NULL) {
/*
* We set the nextPtr of the last element of list 2 to be NULL to make
* the loop less difficult. The loop simply goes through the entire
* Link the first element of the second list to the last element of the
* first list. If the first list isn't empty, we then link the
* last element of the list to the first element of the second list
* The last element of the second list, if it exists, then becomes
* the last element of the first list.
*/
list2->firstPtr->prevPtr = list1->lastPtr;
if (list1->lastPtr != NULL)
list1->lastPtr->nextPtr = list2->firstPtr;
else
list1->firstPtr = list2->firstPtr;
list1->lastPtr = list2->lastPtr;
} else {
/*
* The loop simply goes through the entire
* second list creating new LstNodes and filling in the nextPtr, and
* prevPtr to fit into list1 and its datum field from the
* datum field of the corresponding element in list2. The 'last' node
@ -121,7 +108,6 @@ Lst_Concat(Lst *list1, Lst *list2, int flags)
* the first list must have been empty so the newly-created node is
* made the first node of the list.
*/
list2->lastPtr->nextPtr = NULL;
for (last = list1->lastPtr, ln = list2->firstPtr;
ln != NULL;
ln = ln->nextPtr)
@ -145,6 +131,4 @@ Lst_Concat(Lst *list1, Lst *list2, int flags)
list1->lastPtr = last;
last->nextPtr = NULL;
}
return (SUCCESS);
}