C - 关于字符串的预处理器声明

Question

所以我得到了这个我从教科书上打出来的小程序。

#include <stdio.h>
#define string char* 


int main(void) {
    string a[] = { "I", "like", "to", "fight," },
    b[] = {"pinch,","and","bite."};

    printf("%s %s %s %s %s %s %s\n", a[0], a[1], a[2], a[3], b[0], b[1], b[2]);

    return 0;
}

就目前而言，它会抛出以下错误:

exit status 1
main.c: In function 'main':

main.c:7:18: error: excess elements in char array initializer
  b[] = {"pinch,","and","bite."};

main.c:7:18: note: (near initialization for 'b')

main.c:7:24: error: excess elements in char array initializer
  b[] = {"pinch,","and","bite."};

main.c:7:24: note: (near initialization for 'b') 

显然，我们的目标是让这项工作成功。该书暗示，在 #define 预处理器中，可以添加一个字符以使其工作。而且，错误都集中在 b 字符串周围，我认为它会是 b。但事实并非如此。要么，要么我将 b 放在了错误的位置。我对这个问题的理解是，虽然它为 a 字符串创建了足够的空间，但它不是为 b 字符串创建的。

如有任何意见，我们将不胜感激。

Answer 1

好吧，如果你有

int *a, b;

您可能很快就会发现您有一个 int 指针和一个 int。您的代码也发生了同样的情况

string a[] = { "I", "like", "to", "fight," },
b[] = {"pinch,","and","bite."};

变成了

char* a[] = { "I", "like", "to", "fight," },
b[] = {"pinch,","and","bite."};

所以你有一个 char 指针数组和一个 char 数组。您需要添加的单个字符是 * 以使 b 成为字符指针数组。

char* a[] = { "I", "like", "to", "fight," },
*b[] = {"pinch,","and","bite."};