Question

我正在尝试创建一个游戏，并且当用户进入游戏时，有时会为该玩家创建一个结构。如果玩家输入名称，游戏会提示玩家以下内容：

Type player name:> George
Choose one of the following Otions:
    [+]1) to remove George from player List.
    [+]0) to keep George.

玩家需要输入3个名字，并决定是否应删除或保留其中的一个或全部。

我的问题是，如果玩家决定从列表中删除一个或多个列表，我将无法保留已创建（保留）的列表。

这是程序的一部分，可以编译：

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

#define PERCENT 5
#define LEFT "left"
#define RIGHT "Right"

struct node {
    int player_ID;
    struct node* prev;
    struct node* next;
    char move_player[256];
    char player_name[ 31 ];
} node;

struct node *root;
struct node *current;
struct node *player_node( int player_ID, const char *player_name );
void delete_player ( void *player_list );

int main( void )
{
    char name[256] = { 0 };
    int i = 0, clean;

    root = player_node( PERCENT, RIGHT );
    current =  root;

    do{
        struct node *player = player_node( PERCENT, LEFT );
        player->prev = current;
        current->next = player;
        current = player;

        printf( "Type player name:> ");
        if ( fgets( name, 256, stdin ) == NULL )
        {
            printf("Error, fgets()\n");
            exit( EXIT_FAILURE );
        }
        name[ strcspn( name, "\n" ) ] = '\0';
        strncpy( player->player_name, name, strlen( name ) );
        printf( "Choose one of the following Otions:\n" );
        printf( "\t[+]1) to remove %s from player List.\n", current->player_name );
        printf( "\t[+]0) to keep %s.\n", current->player_name );


        int opt = 0;
        if ( scanf( "%d", &opt) != 1 )
        {
            printf("Error, scanf()\n");
        }else if ( opt == 1 )
        {
            delete_player ( player );
        }
        while ( ( clean = getchar() ) != '\n' && clean != EOF );
        i++;
    }while( i < 3 );

    struct node *tmp = root->next;
    while ( tmp != NULL )
    {
        printf( "Name = %s", tmp->player_name );
        tmp = tmp->next;
    }
    free( root );
}

struct node *player_node( int player_ID, const char *const movement )
{
    struct node *player = malloc( sizeof( struct node ) );
    player->prev = NULL;
    player->next = NULL;
    player->player_ID = player_ID;
    strncpy( player->move_player, movement, strlen( movement) );
    strncpy( player->player_name, "NULL", 5 );
    return player;
}

void delete_player ( void *player_list )
{
    struct node *local_client = (struct node* )player_list;
    if ( local_client == current )
    {
        current = local_client->prev;
        current->next = NULL;
    } else
    {
        local_client->prev->next = local_client->next;
        local_client->next->prev = local_client->prev;
    }
    free( local_client );
}

如果代码到达此地址：

struct node *tmp = root->next;
while ( tmp != NULL )
{
    printf( "Name = %s", tmp->player_name );
}

列表中没有剩余的玩家，因为没有打印列表。

为什么列表为空？

Answer 1

使用strcpy()代替strncpy()。后者比前者not safer。

更改：

strncpy(player->player_name, name, strlen( name ));

收件人：

strcpy(player->player_name, name);

对此通话执行相同操作：

strncpy( player->move_player, movement, strlen( movement) );

因为在两种情况下，您都不复制NULL终止符。

然后，当printf()或任何标准字符串函数处理您的字符串时，它不知道何时停止...

问题的核心是[strlen()] 3在不考虑NULL终止符的情况下计算字符串的长度：

C字符串的长度由终止的空字符确定：AC字符串的长度与字符串开头和终止的空字符之间的字符数一样（不包括终止的空字符本身）。

我也将strncpy( player->player_name, "NULL", 5 );更改为同质。

如果必须使用strncpy()，则只需在第三个参数中加1，如下所示：

strncpy(player->player_name, name, strlen( name ) + 1);

但是，请记住，在此代码中使用strncpy()是毫无意义的，只会增加复杂性并降低代码的简洁度，从而只会降低可读性。

创建动态分配的链接列表

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