2016-02-14 77 views
0

所以我有一個方法,我試圖查看哪個鏈表更大。如果我發現大於我想設置一個指針的指針,那麼當我在鏈表上進行算術運算時,我將減去較大的較小值,無論它是鏈表A還是鏈表B。儘管有一些問題。首先,當我看到在我新分配的指針中是否有數據時,我得到一個錯誤,說「無效類型參數' - >'(有'結構節點')」這是我的代碼,任何幫助將不勝感激!分配指向鏈接列表的指針:不包含數據

void subtraction(struct node** headOne, struct node** currOne, struct node** tailOne, struct node** headTwo, struct node** currTwo, struct node** tailTwo, struct node** headThree, struct node** tailThree, int lengthOne, int lengthTwo){ 
int numberOne, numberTwo, diff; 
struct node* longest; 
struct node* shortest; 
printf("tailOne data = %d\n",(*tailOne)->data); 
    printf("tailTwo data = %d\n",(*tailTwo)->data); 
if(lengthTwo > lengthOne){ 
    longest = *currTwo; 
    shortest = *currOne; 
} 
else if (lengthOne > lengthTwo){ 
    longest = *currOne; 
    shortest = *currTwo; 
} 
else{ 
    if(((*tailOne)->data) > ((*tailTwo)->data)){ 
     longest = *currOne; 
     shortest = *currTwo; 
    } 
    else{ 
     longest = *currTwo; 
     shortest = *currOne; 
    } 
} 
while(longest){ 
    printf("longest = %d",(*longest)->data); 
} 

}

int main(){ 
//initials 
int i, number, lengthOne, lengthTwo; 
char ch = 1; 
//node pointers 
struct node* headOne = NULL; 
struct node* currOne = NULL; 
struct node* tailOne = NULL; 
struct node* headTwo = NULL; 
struct node* currTwo = NULL; 
struct node* tailTwo = NULL; 
struct node* headThree = NULL; 
struct node* currThree = NULL; 
//create linked list 
lengthOne = createLL(&headOne,&currOne, &tailOne, ch, number); 
lengthTwo = createLL(&headTwo,&currTwo, &tailTwo, ch, number); 
scanf("%c",&ch); 
if (ch == '+'){ 
addition(&headOne, &currOne, &headTwo, &currTwo, &headThree, &currThree, lengthOne, lengthTwo); 
} 
else if(ch == '-'){ 
subtraction(&headOne, &currOne, &tailOne, &headTwo, &currTwo, &tailTwo, &currThree, &headThree, lengthOne, lengthTwo); 
} 
+0

該功能原型是瘋了。 10個參數?這無疑是一個標誌,你將太多的東西組合成一個功能。 – xaxxon

+0

我同意一些更好的設計,我可以縮小參數很多,但我現在更關注指針和鏈表。 – user3260745

回答

1

應該longest->data(*longest).data,不(*longest)->data

+0

工作!我想我只是完全被指針弄糊塗了。不過謝謝。 – user3260745