我有2个结构:wires和wireLLN(导线的链表节点)。我们的想法是findWire()函数在链表中查找具有与给定连线(wireName)同名的连线的节点。当添加第一条线时,这不会产生任何问题,因为头节点为空,因此创建了一个新节点,其中线作为节点的线路'属性。 addNode中的printf调用显示了这一点。但是,在第二轮findWire()上,尝试访问“线路”。头节点的属性导致分段错误。 (我已经评论过段错误出现在哪里)
typedef struct wires {
bool value;
char* name;
} Wire;
typedef struct wireLLN {
Wire wire;
struct wireLLN * nextNode;
} WireLLN;
//head of the linked list
WireLLN * headWire = NULL;
// adds a node to the linked list
void addNode(WireLLN* head, WireLLN* node){
if (headWire == NULL){
headWire = node;
printf("Head node was null. Adding node with wire name: %s\n", headWire->wire.name); //this prints no problem
}
else if (head->nextNode == NULL)
head->nextNode = node;
else
addNode(head->nextNode, node);
}
//finds if a wire with a given name already exists, if not returns null
Wire * findWire(char wireName[], WireLLN * head){
if (headWire != NULL){
puts("head wasnt null");
printf("HEAD NODE ADDRESS: %s\n", head);
printf("HEAD WIRE: %s\n", head->wire); //SEG FAULT HERE
if (strcmp(head->wire.name, wireName) == 0){
puts("1");
return &head->wire;
} else if (head->nextNode == NULL){
puts("2");
return NULL;
} else {
puts("3");
return findWire(wireName, head->nextNode);
}
} else return NULL;
}
// assigns a wire to a gate if it exists, otherwise creates a new one then assigns it
Wire assignWire(char wireName[]){
Wire * result = findWire(wireName, headWire);
if (result == NULL){
Wire wire = makeWire(wireName);
WireLLN node;
node.wire = wire;
addNode(headWire, &node);
return wire;
} else {
return *result;
}
}
感谢您的时间。
答案 0 :(得分:0)
您有内存释放问题。 Yoiu忘记了一旦功能停止运行,你的记忆将被删除。 这发生在assignWireFunction。
您可能希望将其更改为:
Wire assignWire(char wireName[]){
Wire * result = findWire(wireName, headWire);
if (result == NULL){
//to prevent the data being lost after the function returns
WireLLN* node = (WireLLN*)malloc(sizeof(WireLLN));
node->wire = makeWire(wireName);
addNode(headWire, node);
return node->wire;
} else {
return *result;
}
}
你的NULL检查没有警告你的原因是,因为你给addNode的指针在函数返回后停止被分配内存。 然后你访问那个内存(地址是相同的),但它不是你被允许写任何东西的内存。