Question

到目前为止，我能使代码工作的唯一方法是使用switch语句。有没有办法可以带走这个switch语句并创建一个数组。我听说我可以这样做：char name[11]= {"name 1", "name 2"};等等但是我不知道我将如何在程序中打印出来。因为我的数字我只是将它分配给一个非数组变量并使用printf来打印它。

我的代码：

#include <stdio.h>
int main(){
int i;
int player [11] = {1,    2,   10,   13,   21,  22,  24,   25, 31,32,   33};
int points [11] = {60, 297, 11, 373, 154, 52, 555, 218, 29, 242, 257};
int games [11] = {33, 35,  12,  35,   35,  35,  35,   35, 22,35,   35};
int bestplayer = 0;
float bestppg = 0.0;
float ppg [11] ;
for (i=0; i<11; i++){
    ppg[i] = (float)points [i] / (float)games [i] ;
    printf("%d \t %d \t %d \t %.1f ppg\n", player[i], games[i], points[i],ppg[i]);
    if (ppg[i]>bestplayer){
        bestplayer = player[i];
        bestppg = ppg[i];
    }
}
printf("\nThe player with the most points per game is #%d ", bestplayer);
switch(bestplayer){
    case 1:
    printf("Player 1");
    break;
    case 2:
    printf("Player 2");
    break;
    case 10:
    printf("Player 3");
    break;
    case 13:
    printf("Player 4");
    break;
    case 21:
    printf("Player 5");
    break;
    case 22:
    printf("Player 6");
    break;
    case 24:
    printf("Player 7");
    break;
    case 25:
    printf("Player 8");
    break;
    case 31:
    printf("Player 9");
    break;
    case 32:
    printf("Player 10");
    break;
    case 33:
    printf("Player 11");
    break;
    default:
    printf("Invalid Player");
    break;
}
printf(" with %.1f ppg.\n",bestppg);
return 0;
}

Answer 1

在当前结构中使用char*数组的主要问题是您没有跟踪最佳播放器的索引。如果你这样做，那么你可以创建一个数组并将其编入索引。

#include <stdio.h>

int main(){
    int i;
    int player [11] = {1,    2,   10,   13,   21,  22,  24,   25, 31,32,   33};
    int points [11] = {60, 297, 11, 373, 154, 52, 555, 218, 29, 242, 257};
    int games [11] = {33, 35,  12,  35,   35,  35,  35,   35, 22,35,   35};
    const char* names[11] = {
        "Jaylon Tate","Joseph Bertrand","Jaylon Tate","Tracy Abrams","Malcolm Hill","Maverick Morgan","Rayvonte Rice","Kendrick Nunn","Austin Colbert","Nnanna Egwu","Jon Ekey"
    };
    int bestplayer = 0;
    float bestppg = 0.0;
    float ppg [11] ;
    int bestIndex = 0;
    for (i=0; i<11; i++){
        ppg[i] = (float)points [i] / (float)games [i] ;
        printf("%d \t %d \t %d \t %.1f ppg\n", player[i], games[i], points[i],ppg[i]);
        if (ppg[i]>bestplayer){
            bestplayer = player[i];
            bestppg = ppg[i];
            bestIndex = i;
        }
    }

    printf("\nThe player with the most points per game is #%d %s with %.1f ppg.\n", bestplayer, names[bestIndex], bestppg);
    return 0;
}

Answer 2

像这样：

int i;
char *names[5] = { "name1", "name2", "name3", "name4", "name5" };

for (i = 0; i<5; i++)
{
    printf("%s\n", names[i]);
}

还有一件事。声明

char name[11];

为单个名称声明空格，长度为10个字符，末尾为零终结符，而不是您想要的数组。

Answer 3

通过记住最佳播放器的索引来避免你的switch，将其用作const字符串数组中的索引：

#include <stdio.h>

   int main(){
   int i;
   int player [11] = {1,    2,   10,   13,   21,  22,  24,   25, 31,32,   33};
   int points [11] = {60, 297, 11, 373, 154, 52, 555, 218, 29, 242, 257};
   int games [11] = {33, 35,  12,  35,   35,  35,  35,   35, 22,35,   35};
   int bestplayer = 0;
   float bestppg = 0.0;
   float ppg [11] ;

   int best = 0;
   for (i=0; i<11; i++){
    ppg[i] = (float)points [i] / (float)games [i] ;
    printf("%d \t %d \t %d \t %.1f ppg\n", player[i], games[i], points[i],ppg[i]);
    if (ppg[i]>bestplayer){
        bestplayer = player[i];
        bestppg = ppg[i];
        best = i;
    }
   }

   const char* const players[] = { "Jaylon Tate", "Joseph Bertrand", "Jaylon Tate", "Tracy Abrams", "Malcolm Hill", "Maverick Morgan", "Rayvonte Rice", "Kendrick Nunn", "Austin Colbert", "Nnanna Egwu", "Jon Ekey"  };

   printf("\nThe player with the most points per game is %s with %.1f ppg.\n", players[best], bestppg);
}

你如何在C中创建一个字符串数组？

3 个答案: