c - 用户想玩多少次游戏 c 语言

Question

我正在学习如何编写游戏，我已获得以下游戏规则。

• 规则 1:玩家必须定义停止回合的最小数字，该数字必须大于 10
• 规则 2:本轮所有得分均计入总分
• 规则 3:如果骰子落在 1，则该轮的所有分数均被没收
• 规则 4:第一个获得 101 分的玩家获胜!

我试图在用户输入有多少玩家后获取要输入的游戏数量。

我的问题是每次我输入玩家姓名时都会询问用户，具体取决于我输入的玩家数量

示例

How many Players: 3

How many games: 15

Enter Players name: John

How many games: 15 (I input this number again) I dont need this line here

Enter Players name: Mary

How many games: 15 (I input this number again) I dont need this line here

Enter Players name: Barry

How many games: 15 (I input this number again) I dont need this line here

这是我解决问题的代码:

//user inputs value of player_num, here, as you have now//

printf_s("Please type in the number of players: ");
scanf_s("%d", &player_num, sizeof(int));
for (i = 0; i < player_num; i++) {
    //user inputs value of num_game, here, as you have now//
    printf_s("Please type in the number of games: ");
    scanf_s("%d", &num_games, sizeof(int));
    num_games = (j = 1, num_games);

    printf_s("Enter the player's first name: ");
    scanf_s("%s", names[i], 25);  
    getchar();               
}
printf_s("\n");
for (i = 0; i < player_num; i++)
    printf_s("\n%s", names[i]);

Answer 1

之所以一直要求你“输入游戏次数”，是因为你把printf和scanf语句放在了循环中。

for (i = 0; i < player_num; i++) {
printf_s("Please type in the number of games: ");//this
scanf_s("%d", &num_games, sizeof(int));//this
num_games = (j = 1, num_games); //including this

printf_s("Enter the player's first name: ");// also this
scanf_s("%s", names[i], 25);  // and this
getchar();
}

因此，每次执行循环时，它都会要求您输入值。

相反，如果您希望它询问您一次，并且您希望语句只执行一次，则必须将其放在循环之外。

如果您愿意，您可以这样做:

get the number of games to be inputted in after the user enters how many players there are.

printf_s("Please type in the number of players: ");
scanf_s("%d", &player_num, sizeof(int));
printf_s("Please type in the number of games: ");
scanf_s("%d", &num_games, sizeof(int));
num_games = (j = 1, num_games);

它是

asking the user every time you enter the players names depending on how many players you have inputted because you've entered those lines "inside the loop";