C如何获得球员及其名字的数量

时间:2017-06-05 01:35:29

标签: c

您好我正在尝试学习如何编程,我希望用户输入有多少玩家以及他们的名字。在看了一些youtube视频并在网上寻找其他地方之后,我认为我的以下代码是正确的,但我看不出我哪里出错了。如果有人能帮忙那就太棒了。

int main() {
  int i, player_num; 
  char names[8][25];

  printf("\n\n");  

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


  for(i = 0; i < player_num; i++) {
    printf("Enter the player's name: ");
    scanf("%s", names[i]);  //enters name and creates a newline <enter key>
    getchar();                   //removes the newline from the keyboard buffer
  }
  printf("\n\n");
  for(i = 0; i < player_num; i++)
    printf("\n%s", names[i]);

  printf("\n\n\t\t\t     press enter when ready");

  getchar();   //holds the console window open until you press enter
  return 0;
}

1 个答案:

答案 0 :(得分:0)

您想输入播放器的数量。您似乎已经编写了输入玩家姓名的代码。因此,要输入播放器的数量,您需要使用一个scanf()

int main() 
{
   int i, player_num; 
   char names[8][25];
   printf("\n\n");  
//user inputs value of player_num, here, as you have now
   printf("Enter the number of player(from 1 to 8)\n");
   scanf_s("%d",&player_num,sizeof(int));
   for(i = 0; i < player_num; i++) {
     printf("Enter the player's name: ");
     scanf_s("%s", names[i],25);  //enters name and creates a newline <enter key>
     getchar();               //removes the newline from the keyboard buffer
   }
   printf("\n\n");
   for(i = 0; i < player_num; i++)
      printf("\n%s", names[i]);
   printf("\n\n\t\t\t     press enter when ready");
   getchar();   //holds the console window open until you press enter
   return 0;
}

希望这有助于:)