我可以用一点帮助。我正在尝试按年排序结构队列。
这是我的结构:
struct element{
int id;
int sign;
int year;
int month;
double amount;
struct element *next;
};
struct queue{
struct element *head;
struct element *tail;
struct element *heads;
struct element *temp;
struct element *temph;
int size;
};
这是我写的函数:
void sort(struct queue* queue){
if (queue->size == 0){
printf("Struct is empty\n");}
else {
struct element* head=queue->head;
struct element* heads=queue->heads;
struct element* temp=NULL;
struct element* temph=queue->head;
int i, size=queue->size;
for(i=0;i<size-1;i++){
heads=head->next;
if((head->year)>(heads->year)){
temp=head;
head=heads;
heads=temp;
}
head=head->next;
heads=NULL;
temp=NULL;
}
head=temph;
}
}
当我复制时,它会中断:if((head->year)>(heads->year))
。
我很确定我的问题是由于对head
旁边的结构的不正确引用造成的(我将其命名为heads
)。
答案 0 :(得分:1)
我省略了所有非重要的东西,并将链表冒泡排序减少到这个骨架。
void sort(struct queue* queue)
{
struct element **pp, *this;
if (!queue->head ){
fprintf(stderr, "OMG Struct is empty\n");
return;
}
for(pp = &queue->head; this = *pp; pp = &(*pp)->next){
struct element *other = this->next;
if (!this->next) break;
if (this->year < other->year) continue;
/*
** Now, Swap this (b) and other (c)
** old situation: @a -> (b) -> (c) -> (d)
** new situation: @a -> (c) -> (b) -> (d)
*/
*pp = other; /* @a -> (c) */
this->next = other->next; /* (b) -> (d) */
other->next = this; /* (c) -> (b) */
}
/* Note: when we get here, "this" will contain the last non-NULL node in the
** chain, and can be used to reset the tail-pointer
*/
return;
}