This seems like a silly question. I have an array of chars and want to store the address of the array in another variable, but can't seem to declare the correct type for the array address (I'm using gcc):

IN:

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

OUT:

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;
      |                         ^

This is expected, I have read elsewhere that the type of cha should be char(*)[6]. But when I try to declare arrayAddress with this type, my program fails:

IN:

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

OUT:

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
       ^

How do I define arrayAddress correctly?

CodePudding user response：

It is written:

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

Notice that the name of variable gets tucked in middle of the expression.

CodePudding user response：

Arrays decay to pointers.

  char cha[] = "abcde";

  char *p1 = cha;

  char (*arrayptr)[sizeof(cha)] = &cha;

cha, &cha[0] and &cha reference the same first element of the array cha, the only difference is the type.

1. cha and &cha[0] has type: pointer to char
2. &cha has type: pointer to array of 6 char elements.

CodePudding user response：

If your compiler supports typeof extension (gcc does) then you can define the pointer as:

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

Or even cleaner as:

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