C語言實現(xiàn)Flappy Bird小游戲
更新時間:2018年12月24日 08:37:07 作者:一個全棧游戲開發(fā)者
這篇文章主要為大家詳細(xì)介紹了C語言實現(xiàn)Flappy Bird小游戲,具有一定的參考價值,感興趣的小伙伴們可以參考一下
本文實例為大家分享了C語言實現(xiàn)Flappy Bird小游戲的具體代碼,供大家參考,具體內(nèi)容如下
#include<stdio.h> #include<stdlib.h> #include<conio.h> #include<time.h> #include<Windows.h> /********函數(shù)變量聲明********/ #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語言版 Flappy Bird\n"); printf(" ■■ ■■ 瞎搞人:yyposs\n"); printf(" ■■ ■■ 瞎搞日期:2014.2\n"); printf(" ■■ ■■ 耗時:4小時\n"); printf(" ■■■ ■■ 游戲說明:\n"); printf(" ■■ 1-按上箭頭使鳥起飛\n"); printf(" ■■ 2-等級越高,難度越大!\n"); printf(" Ю ■■■\n"); printf("\n"); printf(" ■■■ 歡迎各路大神一起探討\n"); printf(" ■■\n"); printf(" ■■\n"); printf(" ■■ ■■■ 【無版權(quán),隨意修改】\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內(nèi)容,希望對大家的學(xué)習(xí)有所幫助,也希望大家多多支持腳本之家。
相關(guān)文章
C++11?成員函數(shù)作為回調(diào)函數(shù)的使用方式
這篇文章主要介紹了C++11?成員函數(shù)作為回調(diào)函數(shù)的使用方式,具有很好的參考價值,希望對大家有所幫助。如有錯誤或未考慮完全的地方,望不吝賜教2022-11-11C++數(shù)據(jù)精度問題的解決方案(對浮點數(shù)保存指定位小數(shù))
對浮點數(shù)保存指定位小數(shù),怎么解決這個問題呢?如果有小伙伴對C++數(shù)據(jù)精度問題的解決方案感興趣的朋友一起看看吧2017-08-08離線安裝visual?studio2022+QT5.12的實現(xiàn)步驟
近期有需求離線配置C++與QT環(huán)境,本文主要介紹了離線安裝visualstudio2022+QT5.12的實現(xiàn)步驟,具有一定的參考價值,感興趣的可以了解一下2024-06-06C語言實現(xiàn)打印數(shù)組以及打印注意事項說明
這篇文章主要介紹了C語言實現(xiàn)打印數(shù)組以及打印注意事項說明,具有很好的參考價值,希望對大家有所幫助,如有錯誤或未考慮完全的地方,望不吝賜教2024-01-01關(guān)于STL中l(wèi)ist容器的一些總結(jié)
list就是數(shù)據(jù)結(jié)構(gòu)中的雙向鏈表(根據(jù)sgi stl源代碼),因此它的內(nèi)存空間是不連續(xù)的,通過指針來進行數(shù)據(jù)的訪問,這個特點使得它的隨即存取變的非常沒有效率,因此它沒有提供[]操作符的重載2013-09-09