請問: 現在有兩個stucture,如下 typedef struct STRUCT_1 { struct STRUCT_1 *next; }CHAIN; typedef struct STRUCT_2 { CHAIN *CHAIN_1; CHAIN *CHAIN_2; }CHAIN_SET; void getchain_1end(CHAIN *chain) { while(chain->next != NULL) chain = chain->next; } void getchain_2end(CHAIN **chain) { CHAIN *pChain = (*chain); while(pChain->next != NULL) { pChain = pChain->next; } } void main() { CHAIN_SET Chain_Set; CHAIN *chain1->next=NULL, *chain2->next=NULL; ...... Chain_Set->CHAIN_1 = chain1; Chain_Set->CHAIN_2 = chain2; ...... getchain_1end(Chain_Set->CHAIN_1); getchain_2end(&(Chain_Set->CHAIN_2)); } Chain_Set連結兩個chain(chain_1,chain_2) 我想分別將兩個chain丟到同一個函數裡,找出chain的最尾端做新增 我原本觀念是用getchain_1end( )方式,但執行到chain = chain->next; 就錯誤,所以我改成getchain_2end( )方式,就可以執行... 我是認為這兩個傳值的方法應該是異曲同工,怎麼會有一個可行,一個不可行 Chain_Set->CHAIN_1和&(Chain_Set->CHAIN_2)差別在哪?

 
CHAIN *getchain_1end(CHAIN *chain)
{
while(chain->next != NULL)
chain = chain->next;    return chain;   //要把值傳回去
}    void main()
{
CHAIN_SET Chain_Set;    CHAIN *chain1 = new CHAIN;
CHAIN *chain11 = new CHAIN;
CHAIN *chain111 = new CHAIN;
CHAIN *chain2 = new CHAIN;
CHAIN *p;    chain1->next=chain11;
chain11->next=chain111;
chain111->next=NULL;
chain2->next=NULL;    Chain_Set.CHAIN_1 = chain1;
Chain_Set.CHAIN_2 = chain2;    p = getchain_1end(Chain_Set.CHAIN_1);    delete chain1;
delete chain11;
delete chain111;
delete chain2;
}