我只是在用节点和链表来触及表面，所以我一直在尝试创建一个链表来打印出 1-10 的节点。但是，它充满了问题。该程序给了我运行时错误和分段错误，我在运行 valgrind 时也有错误。
评论更适合我，表明我(希望)知道每个命令在做什么
#include <stdio.h>
#include <stdlib.h>
int main(void) {
typedef struct node {
int value;
struct node* next;
}
node;
//creates nodes for head, tmp, content
node* head = NULL;
node* tmp = NULL;
node* content = NULL;
head->next = content; //head node points to content
for (int i = 1; i <= 10; i++) {
content = malloc(sizeof(node)); //creates new node
content->value = i; //node data becomes i
tmp->next = content; //tmp node points to content node
tmp = tmp->next; //tmp node becomes next content node
content->next = NULL; //content node points to null
printf("%i ", content->value); //see node value
}
while (head != NULL) {
node* temp = head;
head = head->next;
free(temp);
}
return 0;
}
好的，让我们分解一下
逐节
#include <stdio.h>
#include <stdlib.h>
int main(void) {
// Data section
typedef struct node {
int value;
struct node* next;
}
node;
node* head = NULL; // a pointer to what will be the head of the list
node* content; // a utility pointer
node* temp; // a utility pointer
int i; // iteration var
// Code Section
for (i = 1; i <= 10; i++) {
if (head == NULL) {
// The list is empty when this condition triigers
// we initialize the first node and assign head to point to it
head = (node*)malloc(sizeof(node)); // allocate memory to the pointer. Also make sure the memory pointed to is of a node type
// This is important because it will allow us use -> operator to assign values to the pointed
// at memory
head -> value = i; // assign the value i to the value field of the pointed to memory
head -> next = NULL; // assign the next pointer of the pointed to memory to NULL
} else {
// iterate over the list till we reach the end.
// Once we do, assign more memory at the end, assign the memory a value and make it's next pointer be NULL
content = head;
while (content -> next != NULL) {
content = content -> next;
}
content -> next = (node*)malloc(sizeof(node));
content -> next -> value = i;
}
}
while (head != NULL) {
temp = head;
printf("Node data: %dn", temp -> value);
head = head->next;
free(temp);
}
return 0;
}
数据部分
// Data section
typedef struct node {
int value;
struct node* next;
}
node;
node* head = NULL; // a pointer to what will be the head of the list
node* content; // a utility pointer
node* temp; // a utility pointer
int i; // iteration var
当我学会用 C 编程时，如果你没有在函数顶部预先声明你的函数使用的变量，编译器会抱怨。出于某种原因，它现在可以工作了。
当您将某物声明为指针时，您只是在声明它。在将内存分配给指针之前，无法为其赋值。
代码部分(我不知道还能怎么称呼它)
// Code Section
for (i = 1; i <= 10; i++) {
if (head == NULL) {
// The list is empty when this condition triigers
// we initialize the first node and assign head to point to it
head = (node*)malloc(sizeof(node)); // allocate memory to the pointer. Also make sure the memory pointed to is of a node type
// This is important because it will allow us use -> operator to assign values to the pointed
// at memory
head -> value = i; // assign the value i to the value field of the pointed to memory
head -> next = NULL; // assign the next pointer of the pointed to memory to NULL
} else {
// iterate over the list till we reach the end.
// Once we do, assign more memory at the end, assign the memory a value and make it's next pointer be NULL
content = head;
while (content -> next != NULL) {
content = content -> next;
}
content -> next = (node*)malloc(sizeof(node));
content -> next -> value = i;
}
}
分配内存并确保分配内存属于特定类型的正确方法是使用malloc并将其强制转换为适当的类型。正如您在某些行中看到的那样，例如content -> next = (node*)malloc(sizeof(node));.确保分配的内存类型是使用类型转换(node*)完成
的你做错了什么
head->next = content;错了。head在代码中出现此语句时NULL。它没有可以指向任何内容的next指针。tmp->next = content;同上content = malloc(sizeof(node));最适合我上面概述的类型转换- 正如其他人指出的那样，
malloc可能由于各种原因而失败。它返回一个您应该检查的NULL。 - 你从来没有真正让
head成为你名单的头
head->next = content; //head node points to content行没有任何意义。head不指向任何内容(您为其分配了 null)，内容也不指向，因此在这一点上说head->next = content毫无意义。您需要在开始循环之前为头节点分配内存，或者在循环中添加条件，例如在为内容分配内存后if(head == NULL) head = content。并检查以确保您的内存分配成功。