链接列表存储从1到1000的素数

正如您将在以下程序的注释中看到的那样，我应该创建一个列表，存储从1到1000的所有素数并释放节点。

只有两个function是我的工作。 但是，我还没弄清楚为什么这个程序不能编译。 你们看到了这个错误吗？ 这是一个已经交完的作业，所以这仅供我个人参考。

#include  #include  #include  /* given data structure declaration */ struct record { int data; struct record * next; }; typedef struct record RecordType; /* DO NOT MODIFY */ /* print a list */ void print_list(RecordType * list) { RecordType * visitor = list; int count = 0; while (visitor != NULL) { printf("%d ", visitor->data); visitor = visitor->next; count++; } printf("\n"); printf("There are %d items in the list.\n", count); } /* MY WORK HERE */ /* free every node in the list */ void free_list(RecordType * list) { while (list->data != 2){ free(list->next); list->next = list; } } /* MY WORK HERE */ /* this function may call other functions created by students */ /* create a list storing all prime numbers in [1, 1000] in ascending order */ /* return a pointer to the starting point of the list */ RecordType * create_list_prime_in_1_to_1000() { RecordType * begin, *tail, *temp; int i = 0; begin = malloc(sizeof(RecordType)); begin->data = 0; begin->next = NULL; tail = begin; while(i data = ++i; tail -> next = temp; tail -> temp; tail -> next = NULL; } } int isPrime(int n){ int d; for (d = 2; d data = 2; start->next = malloc(sizeof(RecordType)); start->next->data = 3; start->next->next = malloc(sizeof(RecordType)); start->next->next->data = 5; start->next->next->next = malloc(sizeof(RecordType)); start->next->next->next->data = 7; start->next->next->next->next = NULL; print_list(start); free_list(start); /* i am expected to expected to build a list iteratively rather than hard-code */ start = create_list_prime_in_1_to_1000(); print_list(start); free_list(start); return 0; } 

Answer 1

您将tail声明为：

 RecordType * begin, *tail, *temp; 

和RecordType为：

 struct record { int data; struct record * next; }; typedef struct record RecordType; 

接下来你有：

 tail -> temp; 

这不起作用，因为RecordType没有名为temp的成员。

我认为它应该是：

 tail = temp; 

运行时错误的原因似乎是因为：

 void free_list(RecordType * list) { while (list->data != 2){ free(list->next); list->next = list; } } 

哪个不对。 你需要这样的东西：

 void free_list(RecordType * list) { // keep going till there are nodes. while (list){ // save the link to the rest of the nodes. RecordType *temp = list->next; // free the current node. free(list); // repeat the process starting at the next node. list = temp; } }