feat: 好几章
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3.编程思维体系构建/3.4.6.8.移动方向.md
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3.编程思维体系构建/3.4.6.8.移动方向.md
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# 8.移动方向
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<em>传统的文本冒险使用</em><em>指南针方向</em><em>进行导航。</em>
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例如，我们在第 6 章中绘制的地图上，玩家可能想<strong>向东</strong>移动，从田野移动到洞穴。我们可以通过给连接<strong>Cave</strong>的通道标上“east”来实现这一点。但是，我们首先需要解决两个问题。
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1. 我们可能仍然想把这段通道称为“entrance”和“east”。但现在，一个对象只能有一个标签。
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2. 在更大的地图上，具有更多的位置和道路，标签“east”将被定义多次。到目前为止，标签在我们的游戏中是全球独一无二的，没有重复项。
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思考题：你能否想出解决办法？
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这些问题同样适用于其他对象，而不仅仅是通道。
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在我们的冒险中，我们有一枚银币和一枚金币。一方面，如果不接受在只有一枚硬币存在的地点得到硬币，那将是愚蠢的。
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另一方面，在两种硬币都在同一位置存在的情况下，玩家应该可以有两个拾取选择。
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这立即将我们带到了解析器的第三个问题：
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1. 一个标签只能是一个单词;“ sliver coin”这是俩单词，他不接受啊
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所有三个问题都将在本章中解决，从问题#1 开始。它通过为每个对象提供一个标签列表来解决，而不仅仅是一个标签。
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# object.h
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```c
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typedef struct object {
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const char *description;
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const char **tags;
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struct object *location;
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struct object *destination;
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} OBJECT;
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extern OBJECT objs[];
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#define field (objs + 0)
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#define cave (objs + 1)
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#define silver (objs + 2)
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#define gold (objs + 3)
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#define guard (objs + 4)
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#define player (objs + 5)
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#define intoCave (objs + 6)
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#define exitCave (objs + 7)
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#define wallField (objs + 8)
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#define wallCave (objs + 9)
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#define endOfObjs (objs + 10)
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```
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# object.c
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```c
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#include <stdio.h>
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#include "object.h"
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static const char *tags0[] = {"field", NULL};
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static const char *tags1[] = {"cave", NULL};
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static const char *tags2[] = {"silver", "coin", "silver coin", NULL};
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static const char *tags3[] = {"gold", "coin", "gold coin", NULL};
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static const char *tags4[] = {"guard", "burly guard", NULL};
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static const char *tags5[] = {"yourself", NULL};
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static const char *tags6[] = {"east", "entrance", NULL};
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static const char *tags7[] = {"west", "exit", NULL};
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static const char *tags8[] = {"west", "north", "south", "forest", NULL};
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static const char *tags9[] = {"east", "north", "south", "rock", NULL};
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//我们不固定标签长度，在结束的时候用NULL来标记
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OBJECT objs[] = {
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{"an open field" , tags0, NULL , NULL },
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{"a little cave" , tags1, NULL , NULL },
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{"a silver coin" , tags2, field, NULL },
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{"a gold coin" , tags3, cave , NULL },
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{"a burly guard" , tags4, field, NULL },
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{"yourself" , tags5, field, NULL },
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{"a cave entrance to the east", tags6, field, cave },
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{"an exit to the west" , tags7, cave , field },
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{"dense forest all around" , tags8, field, NULL },
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{"solid rock all around" , tags9, cave , NULL }
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//我们用NULL来阻绝进入一个你不知道的地方
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};
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```
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当然，要让这个改动生效，我们还需要调整<em>noun.c</em>中的<em>objectHasTag</em>函数。
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<em>同时，我们将让函数 getVisible</em>和<em>getPossession</em> 告知玩家他必须更具体的选择你到底是银币还是金币 ，而不是随机选择任何一个对象。
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# noun.h
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```c
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extern OBJECT *getVisible(const char *intention, const char *noun);
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extern OBJECT *getPossession(OBJECT *from, const char *verb, const char *noun);
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```
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# <strong>noun.c</strong>
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```c
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#include <stdbool.h>
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#include <stdio.h>
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#include <string.h>
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#include "object.h"
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#include "misc.h"
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static bool objectHasTag(OBJECT *obj, const char *noun)
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{
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if (noun != NULL && *noun != '\0')
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{
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const char **tag;
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for (tag = obj->tags; *tag != NULL; tag++)
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{
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if (strcmp(*tag, noun) == 0) return true;
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}//扫描对象的tag列表
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}
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return false;
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}
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static OBJECT ambiguousNoun;//我们需要这玩意的地址帮助我们把它当成一个返回值
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static OBJECT *getObject(const char *noun, OBJECT *from, DISTANCE maxDistance)
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{
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OBJECT *obj, *res = NULL;
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for (obj = objs; obj < endOfObjs; obj++)
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{
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if (objectHasTag(obj, noun) && getDistance(from, obj) <= maxDistance)
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{
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res = res == NULL ? obj : &ambiguousNoun;//标签不明确的解决方案哦
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}
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}
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return res;
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}
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OBJECT *getVisible(const char *intention, const char *noun)
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{
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OBJECT *obj = getObject(noun, player, distOverthere);
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if (obj == NULL)
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{
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if (getObject(noun, player, distNotHere) == NULL)
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{
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printf("I don't understand %s.\n", intention);
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}
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else
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{
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printf("You don't see any %s here.\n", noun);
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}
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}
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else if (obj == &ambiguousNoun)//模糊匹配
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{
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printf("Please be specific about which %s you mean.\n", noun);
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obj = NULL;
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}
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return obj;
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}
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OBJECT *getPossession(OBJECT *from, const char *verb, const char *noun)
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{
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OBJECT *obj = NULL;
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if (from == NULL)
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{
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printf("I don't understand who you want to %s.\n", verb);
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}
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else if ((obj = getObject(noun, from, distHeldContained)) == NULL)
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{
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if (getObject(noun, player, distNotHere) == NULL)
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{
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printf("I don't understand what you want to %s.\n", verb);
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}
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else if (from == player)
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{
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printf("You are not holding any %s.\n", noun);
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}
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else
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{
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printf("There appears to be no %s you can get from %s.\n",
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noun, from->description);
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}
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}
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else if (obj == &ambiguousNoun)//模糊匹配
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{
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printf("Please be specific about which %s you want to %s.\n",
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noun, verb);
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obj = NULL;
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}
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else if (obj == from)
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{
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printf("You should not be doing that to %s.\n", obj->description);
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obj = NULL;
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}
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return obj;
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}
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```
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问题 #3 可以通过从函数<em>parseAndExecute</em>中删除一个 [空格](http://en.wikipedia.org/wiki/Space_(punctuation))字符来解决（下面的第 10 行）。这个解决方案远非完美（'silver' 和 'coin' 之间的双空格是大咩的），但直到我们在第 13 章中让自己成为一个更好的解析器之前。
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# parsexec.c
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```c
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#include <stdbool.h>
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#include <stdio.h>
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#include <string.h>
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#include "location.h"
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#include "inventory.h"
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bool parseAndExecute(char *input)
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{
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char *verb = strtok(input, " \n");
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char *noun = strtok(NULL, "\n");
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if (verb != NULL)
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{
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if (strcmp(verb, "quit") == 0)
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{
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return false;
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}
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else if (strcmp(verb, "look") == 0)
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{
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executeLook(noun);
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}
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else if (strcmp(verb, "go") == 0)
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{
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executeGo(noun);
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}
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else if (strcmp(verb, "get") == 0)
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{
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executeGet(noun);
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}
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else if (strcmp(verb, "drop") == 0)
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{
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executeDrop(noun);
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}
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else if (strcmp(verb, "give") == 0)
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{
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executeGive(noun);
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}
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else if (strcmp(verb, "ask") == 0)
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{
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executeAsk(noun);
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}
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else if (strcmp(verb, "inventory") == 0)
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{
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executeInventory();
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}
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else
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{
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printf("I don't know how to '%s'.\n", verb);
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}
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}
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return true;
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}
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```
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模块<em>main.c</em>、<em>inventory.c</em>、<em>location.c</em>、<em>move.c</em> 和<em>misc.c</em>保持不变
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现在对象数组 ( <em>object.c</em> ) 开始在多个维度上增长（特别是在引入多个标签的情况下），我们需要一种使其更易于维护的方法。
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猜猜看该怎么办？
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