could not register reservation events on device "unregistered device", status: failure. event registration will be retried on next path eval.
亲，您这看起来像是一段设备相关的错误提示信息呀。您是想让我根据这个为您编写一篇解释它的文章，还是用代码来处理类似的情况呢？ [2024-11-27 13:34:15 | AI写代码神器 | 42点数解答]
相关提问
- [字节豆包] could not register reservation events on device "unregistered device", status: failure. event registration will be retried on next path eval.（42点数解答 | 2024-11-27 13:34:15）101
- [字节豆包] #include <stdio.h> #include <stdlib.h> #include <string.h> typedef struct sqlist //单链表结构体 { int data; struct sqlist *next; //指针域 }sqlist; void initlist (sqlist *l, sqlist *m, sqlist *n, sqlist *h) { //todo list yes l->next = null; m->next = null; n->next = null; h->next = null; } void emp(sqlist *l) { if (l->next==null) //todo list)yes printf("链表为空表。\n"); else printf("链表为非空表。\n"); } void length(sqlist *l) { sqlist *p; int length = 0; p=l->next; while(p!=null) { length++; p=p->next; } //todo list（345点数解答 | 2024-10-23 23:11:22）250
- [字节豆包] skill={ trigger:{ global:"useCard1", }, audio:2, forced:true, firstDo:true, filter:function(event,player,card){ if(get.color(event.card)!='black') return false; return event.card.name=='nanman'&&player!=event.player||event.card.name=='wanjian'&&player!=event.player||event.card.name=='taoyuan'&&player.hp<player.maxHp||event.card.name=='wugu'; }, content:function(){}, mod:{ targetEnabled:function(card){ if((get.type(ca（211点数解答 | 2025-02-01 13:23:26）279
- [字节豆包] #include <stdio.h> #include <stdlib.h> #include <string.h> typedef struct sqlist //单链表结构体 { int data; struct sqlist *next; //指针域 }sqlist; void initlist (sqlist *l, sqlist *m, sqlist *n, sqlist *h) { //todo list yes l->next = null; m->next = null; n->next = null; h->next = null; } void emp(sqlist *l) { if (l->next==null) //todo list)yes printf("链表为空表。\n"); else printf("链表为非空表。\n"); } void length(sqlist *l) { sqlist *p; int length = 0; p=l->next; while(p!=null)（428点数解答 | 2024-10-23 23:09:51）255
- [字节豆包] content: async function(event, trigger, player) { const [target] = event.targets; const [card] = event.cards; trigger.cancel(); await player.discard(event.cards); const { result } = await player.chooseControlList( true, function(event, player) { const target = _status.event.target; let att = get.attitude(player, target); if (target.hasSkillTag("maihp")) att = -att; return att > 0 ? 0 : 1; }, ["令" （179点数解答 | 2025-05-17 20:49:48）194
- [字节豆包] trigger: { player: "useCardToPlayer", }, filter: filter(event, player) { if (!event.isFirstTarget || !["basic", "trick"].includes(get.type(event.card)) || _status.currentPhase !== player) { return false; } if (player.getHistory("useCard", evt => evt?.targets?.length).indexOf(event.getParent()) != 0) { return false; } return game.hasPlayer(current => lib.filter.targetEnabled2(event.card, player, current) && !event.ta（1422点数解答 | 2025-11-11 22:44:00）43
- [字节豆包] skill={ audio: 2, trigger: { global: "useCardToTargeted", }, forced: true, locked: false, filter: filter(event, player) { if (event.player == player || event.target != player) { return false; } return event.player.getHistory("useCard", evt => evt?.targets.includes(player)).indexOf(event.getParent()) > 0; }, content: async content(event, trigger, player) { player .when({ global: "useCardAfter" }) （951点数解答 | 2025-11-11 22:46:53）89
- [字节豆包] "createa算法分析及结果 结点的数据结构 typedef struct lnode { int data; // 数据域 struct lnode *next; // 指针域 } linknode; void createa(linknode *&l, int a[],int n) { linknode *p; l=(linknode *) malloc(sizeof (linknode)); l->next=null; for (int i=0; i< n; i++) { p = (linknode *) malloc(sizeof(linknode)); p->data=a[i] ; p->next= l->next ; l->next = p; } } ① 算法时间复杂度 ： ② 算法空间复杂度：_________。 调用createf 时对应形参a[]的实参为int b[]={8，9，5，7，3}，执行完createf后： ③ 单链表中第一（139点数解答 | 2024-11-14 00:33:32）265
- [阿里通义] "createa算法分析及结果 结点的数据结构 typedef struct lnode { int data; // 数据域 struct lnode *next; // 指针域 } linknode; void createa(linknode *&l, int a[],int n) { linknode *p; l=(linknode *) malloc(sizeof (linknode)); l->next=null; for (int i=0; i< n; i++) { p = (linknode *) malloc(sizeof(linknode)); p->data=a[i] ; p->next= l->next ; l->next = p; } } ① 算法时间复杂度 ： ② 算法空间复杂度：_________。 调用createf 时对应形参a[]的实参为int b[]={8，9，5，7，3}，执行完createf后： ③ 单链表中第一（502点数解答 | 2024-11-14 00:35:46）224
- [百度文心] "createa算法分析及结果 结点的数据结构 typedef struct lnode { int data; // 数据域 struct lnode *next; // 指针域 } linknode; void createa(linknode *&l, int a[],int n) { linknode *p; l=(linknode *) malloc(sizeof (linknode)); l->next=null; for (int i=0; i< n; i++) { p = (linknode *) malloc(sizeof(linknode)); p->data=a[i] ; p->next= l->next ; l->next = p; } } ① 算法时间复杂度 ： ② 算法空间复杂度：_________。 调用createf 时对应形参a[]的实参为int b[]={8，9，5，7，3}，执行完createf后： ③ 单链表中第一（487点数解答 | 2024-11-14 00:36:23）172
- [百度文心] "createa算法分析及结果 结点的数据结构 typedef struct lnode { int data; // 数据域 struct lnode *next; // 指针域 } linknode; void createa(linknode *&l, int a[],int n) { linknode *p; l=(linknode *) malloc(sizeof (linknode)); l->next=null; for (int i=0; i< n; i++) { p = (linknode *) malloc(sizeof(linknode)); p->data=a[i] ; p->next= l->next ; l->next = p; } } ④ 单链表中最后一个元素是：_________。 ⑤ 链表中总共结点数是： 。"（539点数解答 | 2024-11-14 00:36:52）195
- [DeepSeek] 这是无名杀的一个技能，如何让他去除不可声明限定技、觉醒技、隐匿技、使命技、主公技等特殊技能的限制skill={ unique: true, audio: 2, trigger: { global: "phaseBefore", player: ["enterGame","phaseBegin","phaseEnd"], }, filter(event, player, name) { if (event.name != "phase") return true; if (name == "phaseBefore") return game.phaseNumber == 0; return player.storage.rehuashen?.character?.length > 0; }, async cost(event, trigger, player) { if (trigger.name !== "phase" || event.triggername === "phaseBefore") { event.result = { bool: true, cost_data: ["（556点数解答 | 2025-06-29 08:59:58）202