c - 程序不输出存储在数组中的字符串

Question

我正在编写一个程序来显示姓名和公寓数量，但是，我存储姓名的数组无法显示姓名，表示他们身份不明。无论如何要显示数组中包含的字符串？另外，我似乎在显示的公寓数量下方显示了 n 的值，是否有办法摆脱这个？这是我的代码:

#include <stdio.h>

int main(void)
{
    int i;
    char name[] = {North, West, South, East};
    int apt[] = {24, 30, 14, 18};
    const int n = 5;

    printf("Name    No. of Apartments\n");
    for (i = 0; i < n; i++)
            printf("%c        %d\n", name[i],  apt[i]);

    return 0;

}

Answer 1

这是您的代码，已更正:

#include <stdio.h>

int main(void)
{
    int i;
    char *name[] = {"North", "West", "South", "East"}; /* You're declaring an array of characters, you need an array of strings/pointers */
    int apt[] = {24, 30, 14, 18};
    const int n = 4; /* you have 4 elements in your array, not 5 */

    printf("Name    No. of Apartments\n");
    for (i = 0; i < n; i++)
        printf("%s %d\n", name[i],  apt[i]); /* %c is for characters, you need %s for strings */

    return 0;
}