[英]Reading array within a struct causes a segmentation fault
我創建了一個鏈表,每個節點包含一個結構作為元素和一個指向下一個節點的指針,如下所示
list.h
typedef struct node {
group data;
struct node *next;
} node;
typedef struct group {
unsigned int elements_count;
unsigned int closed;
unsigned int members[4];
} group;
list.c
node *add(node *head, group toadd) {
node *n_node = (node*) malloc(sizeof(node));
if(n_node != NULL) {
n_node->next = head;
group *n_group = &n_node->data;
/* Copy the values of the group into the created node */
n_group->elements_count = toadd.elements_count;
n_group->closed = toadd.closed;
for(int i = 0; i < 4; i++)
n_group->members[i] = toadd.members[i];
}
else {
throw_error("malloc returned a NULL pointer");
}
return n_node;
}
當我嘗試讀取數組的第一個元素( node->data.members[0]
)時出現問題。 Valgrind說,問題是大小為4的無效讀取,其中地址沒有被堆棧,未分配或(最近)被釋放。 即使使用malloc分配每個節點,為什么也會出現分段錯誤?
編輯:
main.c
node *group_list = NULL;
/* Other code here.. */
group *cur_group = is_present(group_list, msg_gest.mtype);
if(cur_group == NULL) {
// The group isn't still present in the group list, then add it
group new_group = {
.elements_count = 0,
.closed = 0,
.members = {-1, -1 , -1, -1}
};
new_group.members[new_group.elements_count++] = msg_gest.mtype;
new_group.members[new_group.elements_count++] = msg_gest.to_add;
new_group.closed = msg_gest.to_close;
group_list = add(group_list, new_group);
} else {
cur_group->members[cur_group->elements_count++] = msg_gest.to_add;
cur_group->closed = msg_gest.to_close;
}
存在
group* is_present(node *head, int matr) {
group *c_group;
node *c_node = head;
while(c_node != NULL) {
c_group = &c_node->data;
if(*(c_group->members) == matr) // !!Segmentation fault is caused by this read
return c_group;
printf("\n%d", *(c_group->members));
c_node = c_node->next;
}
return NULL;
}
我認為問題是由堆溢出引起的,為解決此問題,我對節點結構進行了如下修改
typedef struct node {
group* data;
struct node *next;
} node;
我在add
函數中分配了這樣的組
n_node->data = (group*) malloc(sizeof(group));
嘗試更換線
if(*(c_group->members) == matr)
用
if(c_group->members[0] == matr)
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.