c-使用主参数的链表

Question

我设法构建了一个链表，其中显示了我作为参数提供给它的int，但我无法实现一个显示命令行参数的链表。

例如/aout 20 40 60 80将作为输出[20] [40] [60] [80]

我认为列表的头应该变成argv[1]，head->下一个应该是argv[2]等

这是我现在的代码：

#include <stdlib.h>
#include <stdio.h>
typedef struct s_list t_list;
struct s_list
{
int     value;
t_list  *next;
};
void    add_node_end(t_list **head, int value);
void    add_node_start(t_list **head, int value);
int     remove_first(t_list **head);
int     remove_last(t_list *head);
int     remove_by_index(t_list **head, int n);
void    print_list(t_list *head);
void    add_node_end(t_list **head, int value)
{
t_list  *current;
current = *head;
while (current->next != NULL)
current = current->next;

current->next = malloc(sizeof(t_list));
current->next->value = value;
current->next->next = NULL;
}
void    add_node_start(t_list **head, int value)
{
t_list  *new;
new = malloc(sizeof(t_list));

new->value = value;
new->next = *head;
*head = new;
}
int remove_first(t_list **head)
{
t_list  *next_node;
int     retvalue;

retvalue = -1;
next_node = NULL;

if (*head == NULL)
return (-1);

next_node = (*head)->next;
retvalue = (*head)->value;
free(*head);
*head = next_node;
return (retvalue);
}
int remove_last(t_list *head)
{
t_list  *current;
int     retvalue;
retvalue = 0;
if (head->next == NULL)
{
retvalue = head->value;
free(head);
return (retvalue);
}

current = head;
while (current->next->next != NULL)
current = current->next;
retvalue = current->next->value;
free(current->next);
current->next = NULL;

return(retvalue);
} 
int remove_by_index(t_list **head, int n)
{
t_list  *current;
t_list  *temp_node;
int retvalue;
int i;

i = 0;
retvalue = -1;

if (n == 0)
return (remove_first(head));

while (i < n - 1)
{
if (current->next == NULL)
return (-1);
current = current->next;
i++;
}
temp_node = current->next;
retvalue = temp_node->value;
current->next = temp_node->next;
free(temp_node);
return(retvalue);
}
void    print_list(t_list *head)
{
t_list  *current;
current = head;

while (current != NULL)
{
printf("%dn", current->value);
current = current->next;
}
}
int main(void)
{
t_list  *head;
head = NULL;

add_node_start(&head, 10);
add_node_end(&head, 20);
add_node_end(&head, 30);
add_node_end(&head, 40);
add_node_end(&head, 50);
print_list(head);
return (0);
}

你有主意吗？感谢

Answer 1

您需要更改main()的定义，并且需要一种将字符串转换为整数(atoi()、strtol()等(的方法：

int main(int argc, char **argv) {
t_list  *head = NULL;
if(argc > 1)
add_node_start(&head, atoi(argv[1]));
for(int i = 2; i < argc; i++)
add_node_end(&head, atoi(argv[i]));
print_list(head);
return 0
}

将add_node_start()和add_node_end()的实现组合成一个函数会很方便，这样调用代码更容易使用。作为奖励，add_node_end()在head指向NULL指针的情况下调用时会做一些明智的事情(而不是崩溃(：

void add_node_end(t_list **head, int value) {
if(!*head) {
add_node_start(head, value);
return;
}
t_list  *current = *head;
while (current->next != NULL)
current = current->next;
current->next = malloc(sizeof(t_list));
current->next->value = value;
current->next->next = NULL;
}

int main(int argc, char **argv) {
t_list  *head = NULL;
for(int i = 1; i < argc; i++)
add_node_end(&head, atoi(argv[i]));
print_list(head);
return 0
}

Answer 2

有点偏离主题。Allan Wind的回答很好。

当涉及到简短的例程时，变量名可能会使环境变得混乱(可能会掩盖问题(

void print_list( t_list *p ) {
while( p ) {
printf( "%dn", p->value );
p = p->next;
}
}

如果您希望在链表中灵活地附加和预附加节点，可以通过传递第三个参数将这两个函数合并为一个函数。在本例中，当"atEnd"为0(false(时，代码将将新节点前置到列表的开头。(相反，如果"atEnd"为非零，则将"appending"添加到链表。(.

void add_node( int atEnd, t_list **head, int value ) {
t_list *new = malloc( sizeof *new );
/* omitting test for NULL */
new->value = value;
new->next = NULL;
if( atEnd ) {
for( t_list *p = *head; p->next; p = p->next )
;// just marching
p->next = new;
}
else {
new->next = *head;
*head = new;
}
}

更少的代码和更少的"；特殊用途"；函数减少了错误可能隐藏的行数。