118 lines
6.3 KiB
JavaScript
118 lines
6.3 KiB
JavaScript
// 客户端入参校验：重复牌id / 越界id / 非数组 / 非整数（check_cards_valid + check_cards_inhand）
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// 回归两个真实缺陷：① 重复id 被 can_playcard 当成对子/拖拉机；② 埋牌传重复id 只埋下 1 张
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require('./_shim');
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const { mod, setup, make108, id } = require('./_rpc.js');
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const P = require('../class.paiju.js');
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const t = require('./_assert')();
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// 造一个最小可用牌局：108 张牌全发给指定座位，主花色=1
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function newPaiju(step, banker, owner) {
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const paiju = {
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cards: make108(), step, banker, call: 65, flower: 1,
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seatlist: [[[0, 0], [0, 0], [0, 0], [0, 0], [-1, -1]], [[0, 0], [0, 0], [0, 0], [0, 0], [-1, -1]], [[0, 0], [0, 0], [0, 0], [0, 0], [-1, -1]]],
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playproc: {}, playhistory: [], callproc: [], idx: 1
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};
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for (let i = 0; i < 108; i++) paiju.cards[i].dealowner = owner + 1;
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paiju.method = {
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check_cards_valid: c => P.check_cards_valid(paiju, c),
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check_cards_inhand: (c, s) => P.check_cards_inhand(paiju, c, s),
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do_burycard: c => P.do_burycard(paiju, c),
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do_playcard: c => P.do_playcard(paiju, c),
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get_seat_cards: s => P.get_seat_cards(paiju, s),
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get_seat_zhucards: s => P.get_seat_zhucards(paiju, s),
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have_baofu: () => (paiju.seatlist[0][4][0] == 0 || paiju.seatlist[1][4][0] == 0 || paiju.seatlist[2][4][0] == 0),
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get_liangpai: () => null,
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get_curmultiple: () => P.get_curmultiple(paiju),
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get_mustcard: s => P.get_mustcard(paiju, s)
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};
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return paiju;
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}
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// ============ L1 check_cards_valid：正 / 反 / 边界 ============
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const pj = newPaiju(5, 0, 0);
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const V = c => P.check_cards_valid(pj, c);
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t.eq('valid 正 单张', V([0]), true);
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t.eq('valid 正 多张不重复', V([0, 1, 2, 53]), true);
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t.eq('valid 边界 最大牌id 107', V([107]), true);
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t.eq('valid 边界 最小牌id 0', V([0]), true);
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t.eq('valid 反 undefined', V(undefined), false);
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t.eq('valid 反 null', V(null), false);
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t.eq('valid 反 字符串', V('012'), false);
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t.eq('valid 反 对象', V({ 0: 1, length: 1 }), false);
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t.eq('valid 反 空数组', V([]), false);
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t.eq('valid 反 重复id', V([5, 5]), false);
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t.eq('valid 反 多张中含重复', V([1, 2, 3, 2]), false);
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t.eq('valid 反 越界上界108', V([108]), false);
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t.eq('valid 反 负数id', V([-1]), false);
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t.eq('valid 反 小数id', V([1.5]), false);
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t.eq('valid 反 数字字符串', V(['5']), false);
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// ============ L1 check_cards_inhand ============
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t.eq('inhand 正 在手上', P.check_cards_inhand(pj, [0, 1], 0), true);
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t.eq('inhand 反 不在手上(别人的)', P.check_cards_inhand(pj, [0], 1), false);
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t.eq('inhand 反 重复id(经 valid 拦截)', P.check_cards_inhand(pj, [0, 0], 0), false);
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t.eq('inhand 反 越界id(经 valid 拦截)', P.check_cards_inhand(pj, [999], 0), false);
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pj.cards[3].playround = 1; // 已打出
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t.eq('inhand 反 已打出的牌', P.check_cards_inhand(pj, [3], 0), false);
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pj.cards[4].playround = 0; // 已埋
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t.eq('inhand 反 已埋下的牌', P.check_cards_inhand(pj, [4], 0), false);
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// ============ L2 出牌：重复id 不得被当成一对 ============
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const pj2 = newPaiju(5, 0, 0);
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P.new_playround(pj2, 1, 0);
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const cid = id(1, 1, 13); // 方块K
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t.eq('出牌 反 重复id 被 mod 层拦截(valid)', P.check_cards_valid(pj2, [cid, cid]), false);
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// ============ L3 mod.maipai：8 个重复id 必须被拒、不得进入出牌阶段 ============
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const pj3 = newPaiju(3, 0, 0);
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const s3 = setup('00000', pj3);
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const dup = id(1, 1, 13);
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mod.maipai({ data: { seat: 0, cards: [dup, dup, dup, dup, dup, dup, dup, dup] }, conmode: 0, fromid: 0 });
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t.eq('埋牌 反 8个重复id 被拒(step仍为3)', pj3.step, 3);
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t.eq('埋牌 反 8个重复id 未埋下任何牌', pj3.cards.filter(c => c.playround === 0).length, 0);
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// 对照：8 张不同的牌可以正常埋下
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const pj4 = newPaiju(3, 0, 0);
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setup('00000', pj4);
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mod.maipai({ data: { seat: 0, cards: [0, 1, 2, 3, 4, 5, 6, 7] }, conmode: 0, fromid: 0 });
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t.eq('埋牌 正 8张不同牌 埋牌成功(step进5)', pj4.step, 5);
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t.eq('埋牌 正 实际埋下8张', pj4.cards.filter(c => c.playround === 0).length, 8);
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// ============ L3 mod.chupai：越界/非数组入参不得抛异常 ============
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const pj5 = newPaiju(5, 0, 0);
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P.new_playround(pj5, 1, 0);
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const s5 = setup('00000', pj5);
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let threw = null;
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try {
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mod.chupai({ data: { seat: 0, cards: [999] }, conmode: 0, fromid: 0 });
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mod.chupai({ data: { seat: 0, cards: undefined }, conmode: 0, fromid: 0 });
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mod.chupai({ data: { seat: 0, cards: 'x' }, conmode: 0, fromid: 0 });
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mod.chupai({ data: { seat: 0, cards: [dup, dup] }, conmode: 0, fromid: 0 });
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} catch (e) { threw = String(e); }
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t.eq('出牌 反 非法入参不抛异常', threw, null);
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t.eq('出牌 反 非法入参不产生任何出牌', pj5.cards.filter(c => c.playround > 0).length, 0);
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// ============ design §9：报无主后立即为全体三人刷新主牌统计 ============
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// 造 step5 牌局：seat0 手上只剩 1 张主牌（打掉即报无主），seat1/seat2 各持若干主牌
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const pj6 = newPaiju(5, 0, 0);
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for (let i = 0; i < 108; i++) pj6.cards[i].dealowner = -1; // 先全部清空
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const give = (cid, seat) => { pj6.cards[cid].dealowner = seat + 1; pj6.cards[cid].playround = -1; };
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give(id(1, 1, 5), 0); // seat0：主♦5（唯一主牌）
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give(id(1, 3, 9), 0); // seat0：副♥9（保证还有牌）
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give(id(1, 1, 13), 1); give(id(2, 1, 13), 1); give(id(1, 1, 12), 1); // seat1：主♦K对 + 主♦Q
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give(id(1, 1, 7), 2); // seat2：正7（固定主牌）
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P.new_playround(pj6, 1, 0);
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t.eq('报无主前 三家统计均为初始 [-1,-1]', pj6.seatlist.map(s => s[4]), [[-1, -1], [-1, -1], [-1, -1]]);
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P.do_playcard(pj6, [id(1, 1, 5)]); // seat0 打出唯一主牌 → 报无主
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t.eq('报无主 seat0 主牌清空 [0,0]', pj6.seatlist[0][4], [0, 0]);
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t.eq('报无主 立即刷新 seat1 [3张,1对]', pj6.seatlist[1][4], [3, 1]);
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t.eq('报无主 立即刷新 seat2 [1张,0对]', pj6.seatlist[2][4], [1, 0]);
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t.eq('报无主 seat0 主花色标志置1', pj6.seatlist[0][pj6.flower - 1], [1, 1]);
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t.eq('报无主 seat2 无主对→无对标志置1', pj6.seatlist[2][pj6.flower - 1][1], 1);
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process.exit(t.done('input') ? 0 : 1);
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