Question

这是练习：

编写一个名为monthName()的函数，该函数将类型enum month的值作为其参数（如本章中所定义），并返回指向包含月份名称的字符串的指针。通过这种方式，您可以使用以下语句显示enum month变量的值：

printf("%s\n", monthName(aMonth));

我写了我的版本并得到了我期望的输出。但是，我确信必须有一个更好的方法来实现它，而不必每个月使用一个case的switch语句。如何用更好的设计编写这个函数？

#include <stdio.h>

enum month { January = 1, February, March, April, May, June, 
        July, August, September, October, November, December };

char *monthName(enum month m){
    switch (m) {
        case January:
            return "January";
            break;
        case February:
            return "February";
            break;
        case March:
            return "March";
            break;
        case April:
            return "April";
            break;
        case May:
            return "May";
            break;
        case June:
            return "June";
            break;
        case July:
            return "July";
            break;
        case August:
            return "August";
            break;
        case September:
            return "September";
            break;
        case October:
            return "October";
            break;
        case November:
            return "November";
            break;
        case December:
            return "December";
            break;
        default:
            return "Not a valid month";
    }
}

int main(void)
{
    enum month aMonth = 1;

    printf("%s\n", monthName(aMonth));

    aMonth = 2;
    printf("%s\n", monthName(aMonth));

    aMonth = 6;
    printf("%s\n", monthName(aMonth));

    return 0;
}

Answer 1

您可以创建一个包含字符串的数组，并使用枚举对其进行索引。

const char *monthNames[] = { "Not a valid month", "January", "February", "March", 
                             "April", "May", "June", "July", "August", "September", 
                             "October", "November" "December" };

const char *monthName(enum month m)
{
    if (m < 1 || m > 12) {
        return monthNames[0];   // string for invalid month
    } else {
        return monthNames[m];
    }
}

Answer 2

您可以使用简单的查找表，因为枚举以1：

开头

#include <stdio.h>

enum month { January = 1, February, March, April, May, June, 
        July, August, September, October, November, December };

const char *months_str[] = {
    "January", "February", "March", "April", "May", "June", "July",
    "August", "September", "October", "November", "December", NULL
};

const char *monthName(enum month m) {
    if(m < January || m > December)
        return "Invalid month";

    return months_str[m-1];
}

int main(void)
{
    enum month aMonth = 1;

    printf("%s\n", monthName(aMonth));

    aMonth = 2;
    printf("%s\n", monthName(aMonth));

    aMonth = 6;
    printf("%s\n", monthName(aMonth));

    return 0;
}

Answer 3

开关很好（但在break之后删除了return，它们是无法到达的噪音），但阵列可以更简单。

在C99及以上版本中，您可以使用枚举符号使其超级清晰（理论上使得字符串的源代码顺序无关紧要）：

const char *monthName(month m)
{
    static const char * const names[] = {
      [January] = "January",
      [February] = "February",
      ... and so on ...
   };
   return names[m];
}

数组static只能初始化一次，当然const因为它没有变化。它在函数内部，以避免乱丢顶层空间，但名称仅在该函数内部使用。

int a[] = { [99] = 4 };仍然是100个整数的数组。

枚举和将月份名称作为字符串返回的函数

3 个答案: