C语言实现Flappy Bird小游戏
2019/7/10 22:31:24
本文主要是介绍C语言实现Flappy Bird小游戏,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!
本文实例为大家分享了C语言实现Flappy Bird小游戏的具体代码,供大家参考,具体内容如下
#include<stdio.h> #include<stdlib.h> #include<conio.h> #include<time.h> #include<Windows.h> /********函数变量声明********/ #define PR_Box printf("■") #define PR_Gold printf("★") #define PR_Ag printf("☆") #define PR_FBird printf("Ю") #define PR_DBird printf("Ф") #define PR_Land printf("┳┳┯") #define PR_Bg_TL printf("╔") #define PR_Bg_TR printf("╗") #define PR_Bg_DL printf("╚") #define PR_Bg_DR printf("╝") #define PR_Bg_X printf("═") #define PR_Bg_Y printf("║") #define PR_Blank printf(" "); int Grade = 1, C_Gold = 0, C_Ag = 0, Score = 0, Delay_time = 1000, Max_blank = 9, Distance = 18; typedef struct Birds { int x, y; int condition; }Birds; Birds * Bird; typedef struct Bg { int x, y; int l_blank; int reward[9]; struct Bg * pri; struct Bg * next; }Bg; Bg * Bg1; void Position(int x, int y) { COORD pos = { x - 1, y - 1 }; HANDLE Out = GetStdHandle(STD_OUTPUT_HANDLE); SetConsoleCursorPosition(Out, pos); } void CreatBird() { Bird -> x = 41; Bird -> y = 10; Bird -> condition = 0; } void CreatBg() { Bg * Bg2 = (Bg * ) malloc(sizeof(Bg)); Bg1 -> x = 90; Bg1 -> y = 8; Bg2 -> x = Bg1 -> x + Distance; Bg2 -> y = 9; Bg1 -> l_blank = Max_blank - Grade; Bg2 -> l_blank = Max_blank - Grade; Bg1 -> next = Bg2; Bg1 -> pri = Bg2; Bg2 -> next = Bg1; Bg2 -> pri = Bg1; } void InsertBg(Bg * p) { int temp; Bg * Bgs = (Bg * ) malloc(sizeof(Bg)); Bgs -> x = p -> pri -> x + Distance; Bgs -> l_blank = Max_blank - Grade; srand((int) time(0)); temp = rand(); if (temp % 2 == 0) //++ { if ((temp % 4 + p -> pri -> y + Max_blank - Grade) < 21) Bgs -> y = p -> pri -> y + temp % 4; else Bgs -> y = p -> pri -> y; } else { if ((p -> pri -> y - temp % 4) > 2) Bgs -> y = p -> pri -> y - temp % 4; else Bgs -> y = p -> pri -> y; } Bgs -> pri = p -> pri; Bgs -> next = p; p -> pri -> next = Bgs; p -> pri = Bgs; } void Check_Bg(Bg * q) { Bg * p = q; int i = 0, temp; while (++i <= 5) { if (p -> x > -4) p = p -> next; else { srand((int) time(0)); temp = rand(); if (temp % 2 == 0) //++ { if ((temp % 4 + p -> y + Max_blank - Grade) < 21) p -> y = p -> y + temp % 4; else p -> y = p -> y; p -> x = p -> pri -> x + Distance; p -> l_blank = Max_blank - Grade; } else { if ((p -> y - temp % 4) > 2) p -> y = p -> y - temp % 4; else p -> y = p -> y; p -> x = p -> pri -> x + Distance; p -> l_blank = Max_blank - Grade; } } } } void Loop_Bg(Bg * q) { Bg * p = q; int i = 0; while (++i <= 5) { p -> x = p -> x - 1; p = p -> next; if (Bird -> x == p -> x) { Score += 1; if (Score % 4 == 0 && Grade < 4) Grade++; } } } void Prt_Bg(Bg * q) { Bg * p = q; int i = 0, k, j; while (++i <= 5) { if (p -> x > 0 && p -> x <= 78) { for (k = 2; k < p -> y; k++) { Position(p -> x + 1, k); PR_Box; PR_Box; PR_Blank } Position(p -> x, p -> y); PR_Box; PR_Box; PR_Box; PR_Blank; Position(p -> x, p -> y + p -> l_blank); PR_Box; PR_Box; PR_Box; PR_Blank; k = k + p -> l_blank + 1; for (k; k <= 22; k++) { Position(p -> x + 1, k); PR_Box; PR_Box; PR_Blank; } Position(p -> x, 23); for (k = 1; k < Distance / 3 - 2; k++) PR_Land; } p = p -> next; if (p -> x == 0) { for (j = 2; j < p -> y; j++) { Position(p -> x + 1, j); PR_Blank; PR_Blank; } Position(p -> x + 1, p -> y); PR_Blank; PR_Blank; PR_Blank; Position(p -> x + 1, p -> y + Max_blank - Grade); PR_Blank; PR_Blank; PR_Blank; j = j + Max_blank - Grade + 1; for (j; j <= 22; j++) { Position(p -> x + 1, j); PR_Blank; PR_Blank; } } } } void PrtBg() { int i; Position(1, 1); PR_Bg_TL; Position(79, 1); PR_Bg_TR; Position(1, 24); PR_Bg_DL; Position(79, 24); PR_Bg_DR; for (i = 3; i <= 78; i += 2) { Position(i, 1); PR_Bg_X; Position(i, 24); PR_Bg_X; } } void PrtBird() { Position(Bird -> x, Bird -> y - 1); PR_Blank; Position(Bird -> x, Bird -> y); PR_FBird; Position(38, 2); printf("Score:%d", Score); } int CheckYN(Bg * q) { Bg * p = q; int i = 0; while (++i <= 5) { if (Bird -> y > 23) return 0; if (Bird -> x == p -> x && Bird -> y <= p -> y) return 0; if ((Bird -> x == p -> x || Bird -> x == p -> x + 1 || Bird -> x == p -> x + 2) && Bird -> y == p -> y) return 0; if (Bird -> x == p -> x && Bird -> y > p -> y + p -> l_blank) return 0; if ((Bird -> x == p -> x || Bird -> x == p -> x + 1 || Bird -> x == p -> x + 2) && Bird -> y == p -> y + p -> l_blank) return 0; p = p -> next; } return 1; } void Prtfirst() { printf("══════════════════════════════════════\n"); printf(" ■■ ■■\n"); printf(" ■■ ■■\n"); printf(" ■■ ■■ C语言版 Flappy Bird\n"); printf(" ■■ ■■ 瞎搞人:yyposs\n"); printf(" ■■ ■■ 瞎搞日期:2014.2\n"); printf(" ■■ ■■ 耗时:4小时\n"); printf(" ■■■ ■■ 游戏说明:\n"); printf(" ■■ 1-按上箭头使鸟起飞\n"); printf(" ■■ 2-等级越高,难度越大!\n"); printf(" Ю ■■■\n"); printf("\n"); printf(" ■■■ 欢迎各路大神一起探讨\n"); printf(" ■■\n"); printf(" ■■\n"); printf(" ■■ ■■■ 【无版权,随意修改】\n"); printf(" ■■ ■■\n"); printf(" ■■ Ф ■■\n"); printf(" ■■ ■■\n"); printf(" ■■ ■■\n"); printf(" ■■ ■■\n"); printf(" ■■ ■■\n"); printf(" ┳┳┯┳┳┯┳┳┯┳┳┯┳┳┯┳┳┯┳┳┯┳┳┯┳┳┯┳┳┯┳┳┯┳┳┯┳\n"); system("pause"); Position(1, 1); int i = 0; while (i++ < 40 * 25) PR_Blank; } void main() { int i = 0; Bird = (Birds * ) malloc(sizeof(Birds)); Bg1 = (Bg * ) malloc(sizeof(Bg)); Prtfirst(); PrtBg(); CreatBg(); InsertBg(Bg1); InsertBg(Bg1); InsertBg(Bg1); CreatBird(); while (1) { if (!CheckYN(Bg1)) break; Check_Bg(Bg1); Prt_Bg(Bg1); PrtBird(); Loop_Bg(Bg1); Bird -> y = Bird -> y + 1; if (GetAsyncKeyState(VK_UP)) { Position(Bird -> x, Bird -> y - 1); PR_Blank; Bird -> y = Bird -> y - 4; } while (i++ < 500); { Sleep(100); } i = 0; } Position(38, 10); printf("You Lost!"); Position(1, 25); system("pause"); }
以上就是本文的全部内容,希望对大家的学习有所帮助,也希望大家多多支持找一找教程网。
这篇关于C语言实现Flappy Bird小游戏的文章就介绍到这儿,希望我们推荐的文章对大家有所帮助,也希望大家多多支持为之网!
- 2024-11-23增量更新怎么做?-icode9专业技术文章分享
- 2024-11-23压缩包加密方案有哪些?-icode9专业技术文章分享
- 2024-11-23用shell怎么写一个开机时自动同步远程仓库的代码?-icode9专业技术文章分享
- 2024-11-23webman可以同步自己的仓库吗?-icode9专业技术文章分享
- 2024-11-23在 Webman 中怎么判断是否有某命令进程正在运行?-icode9专业技术文章分享
- 2024-11-23如何重置new Swiper?-icode9专业技术文章分享
- 2024-11-23oss直传有什么好处?-icode9专业技术文章分享
- 2024-11-23如何将oss直传封装成一个组件在其他页面调用时都可以使用?-icode9专业技术文章分享
- 2024-11-23怎么使用laravel 11在代码里获取路由列表?-icode9专业技术文章分享
- 2024-11-22怎么实现ansible playbook 备份代码中命名包含时间戳功能?-icode9专业技术文章分享