段錯誤當試圖打印鏈表

Question

有問題的問題代碼：

currnode = ((currnode)->next); 

在我的printList功能

。每次嘗試將當前節點的值更改爲下一個節點時，都會收到分段錯誤。爲什麼是這樣？我玩過這些指針，並且在網上做了一些搜索無濟於事。

#include<stdio.h>  
#include<stdlib.h> 
#define RANGE 1000 

typedef struct lnode { 
    int value; struct lnode *next; 
     } lnode; 

void printlist(lnode *list); 
void search(); 

int main(int argc, char *argv){ 

    time_t t; 
    srand((unsigned) time(&t)); 
    int i, times, num, a, b; 
    lnode **leven, **lodd;  

    printf("Please state the amount of numbers to be printed\n"); 
    scanf("%d", &times); 


    for(i = 0; i < times; i++){ 

     lnode **crnt, **pred; 
     a = nextnum();  
     b = (a & 1); 

     if (b == 0){ 


      printf("even\n"); 
      printf("The value of the even node is: %d\n", a); 
     /* search(*leven,**crnt,**pred, a); */ 
      insertatfront(&leven, a); 


     }else{ 

      printf("odd\n"); 
      printf("The value of the odd node is: %d\n", a); 
      /* search(&leven,crnt, pred, a);*/ 
      insertatfront(&lodd, a); 

     } 

    } 

    printlist(*lodd); 
    printlist(*leven); 
} 

int nextnum(){ 

    int i, rand1, num; 

    rand1 = rand()%RANGE; 
    return rand1; 
} 

void getnode(lnode **ptr) 
{ 

*ptr = malloc(sizeof(lnode)); 

} 



int insertatfront(lnode **list, int x){ 

    lnode *new_node, **pred, **crnt; 

    getnode(&new_node); 

    if(!new_node) return 0; 

    new_node->value = x; 
    new_node->next = *list; 

    printf("The node at the next is %d\n", new_node->next->value); 


    *list = new_node; 

    /*search(*list,&crnt,&pred, x);*/ 

} 

void printlist(lnode *list){ 

    int *val; 

    int print; 
    struct lnode *tmp; 
    lnode *currnode; 


    currnode = list; 

    while(currnode != NULL){ 

    val = &((currnode)->value); 

    printf("%d\n", val); 


    currnode = ((currnode)->next); 
    printf("printed successfully\n");  

    } 
    printf("done \n"); 

    } 



void search(lnode *list, lnode **crnt, lnode **pred, int x) 
    { 
     *crnt = list; 
     *pred = NULL; 

      while(*crnt) { 
       if((*crnt)->value == x) return; 
        *pred = *crnt; *crnt = (*crnt)->next; 
      } 
     } 

Answer 1

您可能需要提高警戒級別。有一些明顯的問題。例如：

insertatfront(&leven, a); 

insertatfront()需要lnode**，但&leven是lnode***。

Answer 2

最後一個節點的下一個不是null，它是一個未定義的值，所以當程序執行undefinedValue.next時，會發生分段錯誤。