arrays - 如何将数组[]的地址存储在变量中

Question

这似乎是一个愚蠢的问题。我有一个字符数组，想要将数组的地址存储在另一个变量中，但似乎无法声明数组地址的正确类型(我正在使用 gcc):

---

输入:

int main(void){
  char cha[] = "abcde";
  char **arrayAddress = &cha;
}

输出:

arrayaddress.c: In function ‘main’:
arrayaddress.c:3:25: warning: initialization of ‘char **’ from incompatible pointer type ‘char (*)[6]’ [-Wincompatible-pointer-types]
    3 |   char **arrayAddress = &cha;
      |                         ^

---

这是预料之中的，我已阅读 elsewhere cha 的类型应为 char(*)[6]。但是当我尝试用这种类型声明 arrayAddress 时，我的程序失败了:

---

输入:

int main(void){
  char cha[] = "abcde";
  char (*)[6]arrayAddress = &cha;
}

输出:

arrayaddress.c: In function ‘main’:
arrayaddress.c:3:10: error: expected identifier or ‘(’ before ‘)’ token
    3 |   char (*)[6]arrayAddress = &cha;
      |          ^
make: *** [<builtin>: arrayaddress] Error 1
       ^

---

如何正确定义arrayAddress？

Answer 1

是written :

char (*arrayAddress)[6] = &cha;

请注意，变量名称隐藏在表达式中间。