casting char * to char[]

Question

Example:

char str[10];

gets(str);
str = (char[10]) strtok(str, " "); // type error here

Since strtok() returns a char *, I get an type error without that casting. With it I get the following:

error: cast specifies array type

What is the best to fix this code?

Answer 1

You should be assigning the result of strtok into a separate char* variable. You can't assign it back into str.

Answer 2

You should not be assigning the reult of strtok() back to your str variable in the first place. Use a separate variable instead, eg:

char str[10]; 
gets(str); 
char *token = strtok(str, " ");
//use token as needed...

Answer 3

Oh man, be careful with that gets()! Related question

---

You can't assign to arrays (in other words, use them as lvalues).

char *p = "string";
char array[10];
array = p; /* invalid */

Apart from that, you're not using strtok() correctly. The pointer returned points to the subsequent token, so you might want to create a separate char pointer to store it.

Answer 4

You cannot assign anything to an array. Even this simplistic program will fail:

char *foo(void) { }

int main(int argc, char *argv[])
{
        char a[1];

        a = foo();

        return 0;
}

As indeed it does:

$ make fail
cc     fail.c   -o fail
fail.c: In function ‘main’:
fail.c:7:4: error: incompatible types when assigning to type ‘char[1]’ from type ‘char *’
make: *** [fail] Error 1

Either re-define str as char *str or figure out some other way to re-write your program to not attempt to assign to an array. (What does the surrounding code look like? The code you've pasted doesn't really make sense anyway...)

Answer 5

You can get parameter before calling your function:

char mystr[] = "192.168.0.2";
split_ip(myster[]);
char * split_ip( char  ip_address[]){

unsigned short counter = 0;
char *token;
token = strtok (ip_address,".");
while (token != '\0')
{

printf("%s\n",token);

token = strtok ('\0', ".");
}
}// end of function def