我正在上 C 入门课，我的教授指派我们为当前作业编写一个名为“茶会”的游戏。我已经完成了游戏的编码，并且它在很大程度上可以工作，但是有一些我似乎无法解决的问题。
游戏规则很简单:两名玩家轮流旋转旋转器(通过输入“0”并按回车键来模拟)，并收集茶会的所有 7 个元素。第一个获得全部 7 件元素的玩家获胜。唯一的问题是，除非你先有一个盘子，否则你无法收集三明治、水果或甜点。如果您落在“失去一个棋子”方 block 上，则必须放弃一个棋子。这两个错误都来自游戏中丢失棋子的实例，所以我认为错误一定源自“get_lost_piece”函数。
其中之一是“玩家”数组中的棋子编号很奇怪，因为它们比应有的值高了 1 个值。另一个错误是，当玩家在有需要盘子的元素时试图拿走盘子时，它应该打印出“抱歉，不带盘子吃饭是不礼貌的。输入另一个选择:”，但相反，我得到一个无限循环，显示“你丢失了第 1 项。”
这是我的代码:
#include <stdio.h>
#include <time.h>
#define SLOW_MODE 1
#define NUMPLAYERS 2
#define NUMPIECES 7
#define MAXLEN 20
#define NO_WINNER -1
const char CHOICES[NUMPIECES+1][MAXLEN] = {"PLATE", "NAPKIN", "TEA CUP", "CREAM AND SUGAR", "SANDWICH", "FRUIT", "DESSERT", "LOSE A PIECE"};
void update_player(int player[], int square);
int get_lost_piece(int player[]);
int search(int piece_list[], int choice);
int get_spin();
void init_player(int player[]);
int get_winner(int players[][NUMPIECES]);
int get_next_player(int player_num);
int count_pieces(int player[]);
void print_player(int player[], int player_num);
int main() {
srand(time(0));
int players[NUMPLAYERS][NUMPIECES];
// Initialize each player in the game.
int i;
for (i=0; i<NUMPLAYERS; i++)
init_player(players[i]);
int player_number = 0;
// Play until we get a winner.
int status = get_winner(players);
while (status == NO_WINNER) {
int dummy;
// In slow mode, we stop before every spin.
if (SLOW_MODE) {
printf("Player %d, it is your turn. Type 0 and enter to spin.\n", player_number+1);
scanf("%d", &dummy);
}
// Get the current player's spin and print out her pieces.
int square = get_spin();
printf("Player %d, have landed on the square %s.\n", player_number+1, CHOICES[square]);
update_player(players[player_number], square);
print_player(players[player_number], player_number+1);
// Update the game status.
player_number = get_next_player(player_number);
status = get_winner(players);
printf("\n\n");
}
printf("Congrats player %d, you win!\n", status+1);
return 0;
}
// Pre-conditions: player stores the contents of one player and square is in between 0 and 7, inclusive.
// Post-conditions: The turn for player will be executed with the given square selected.
void update_player(int player[], int square) {
if (square == 7) {
if (count_pieces(player) == 0)
{
printf("There is no piece for you to lose. Lucky you, sort of.\n");
return;
}
else{
int q;
q = get_lost_piece(player);
player[q]= 0;
}
return;
}
player[square]=search(player, square);
// Restricted by having no plate!
if (player[0] == 0) {
if(square == 4 || square == 5 ||square == 6){
printf("Sorry, you can't obtain that item because you don't have a plate yet.\n");}
else{
if (player[square] == 0){
player[square]++;
}
}
}
// Process a regular case, where the player already has a plate.
else {
if (player[square] == 0){
player[square]++;
}
}
}
// Pre-conditions: player stores the contents of one player that has at least one piece.
// Post-conditions: Executes asking a player which item they want to lose, and reprompts them
// until they give a valid answer.
int get_lost_piece(int player[]) {
int choice = -1;
// Loop until a valid piece choice is made.
while (1) {
if (choice == -1){
printf("Which piece would you like to lose?\n");
print_player(player,choice);
scanf("%d", &choice);}
if (player[choice] == 0 && choice < 7 && choice >= 0){
printf("Sorry, that was not one of the choices");
scanf("%d", &choice);
}
if (player[0] == 1 && choice == 4 || choice == 5 ||choice == 6){
printf("Sorry, it is bad manners to eat without a plate. Enter another choice:\n");
scanf("%d", &choice);
}
else{
printf("You lost piece %d\n", choice);
}
}
return choice;
}
// Pre-conditions: piece_list stores the contents of one player
// Post-conditions: Returns 1 if choice is in between 0 and 6, inclusive and corresponds to
// an item in the piece_list. Returns 0 if choice is not valid or if the
// piece_list doesn't contain it.
int search(int piece_list[], int choice) {
int i;
for (i=0; i<NUMPIECES; i++){
if(piece_list[i]==choice){
return 1;}
else {return 0;}
}
}
// Pre-condition: None
// Post-condition: Returns a random value in between 0 and 7, inclusive.
int get_spin() {
return rand() % 8;
}
// Pre-condition: None
// Post-condition: Initializes a player to have no pieces.
void init_player(int player[]) {
int j;
for (j=0; j< NUMPIECES; j++)
player[j]=0;
}
// Pre-condition: players stores the current states of each player in the tea party game.
// Post-condition: If a player has won the game, their 0-based player number is returned.
// In the case of no winners, -1 is returned.
int get_winner(int players[][NUMPIECES]) {
int i =0;
for (i=0; i<NUMPLAYERS; i++){
if(count_pieces(players[i]) == NUMPIECES) {
return i;}
}
return -1;
}
// Pre-condition: 0 <= player_num < NUMPLAYERS
// Post-condition: Returns the number of the next player, in numerical order, with
// a wrap-around to the beginning after the last player's turn.
int get_next_player(int player_num) {
player_num++;
if (player_num == NUMPLAYERS){
player_num = 0;
}
return player_num;
}
// Pre-conditions: player stores the contents of one player
// Post-conditions: Returns the number of pieces that player has.
int count_pieces(int player[]) {
int y, counter;
counter = 0;
for ( y = 0; y < 7; y++){
if(player[y] == 1){
counter++;
}
}
return counter;
}
// Pre-conditions: player stores the contents of one player and player_num is that
// player's 1-based player number.
// Post-conditions: Prints out each item the player has, numbered with the numerical
// "codes" for each item.
void print_player(int player[], int player_num) {
int i;
printf("\n");
printf("Player %d\n", player_num);
for (i=0; i < 7; i++){
if (player[i] == 1){
printf("%d. %s\n", i + 1, CHOICES[i]);
}
}
}
感谢您提前提供的帮助。我感觉解决方案就在我面前，但花了几天时间研究这个问题后，我很难发现问题。
最佳答案
- 需要
#include <stdlib.h>对于rand(),srand() - 添加
player_number参数get_lost_piece()并将其传递给print_player.
以下只是重构的一种思路。可以通过多种不同的方式来完成。
- 在循环开始时获取输入一次。
- 使用
continue每个if中的关键字重做循环的语句。
当我改变 get_lost_piece() 的逻辑时，它工作得很好。至:
while...
get choice
if choice < 1 || choice > 7 "1-7 please..." continue
if player[choice - 1] == 0 "sorry..." continue
if choice == 1 (and player[] contains foods) "bad manners..." continue
return choice - 1;
有用的提示
- 循环应该受到限制，并且需要为玩家提供退出选项。
- 查看FAQ entry :
scanf()返回错误，将内容留在输入流中。 - 尽早测试，经常测试。
- 打开编译器警告(按照 catcall 的建议)
关于C:创建一款游戏，但在某些情况下我遇到了一个错误和无限循环错误，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/10150612/