Why do I get a warning when I do this:
char (*arr_ptr)[];
char str[] = "string";
arr_ptr = str;
The warning is:
assignment from incompatible pointer type
But when I change the assignment to:
arr_ptr = &str;
The warning disappears. I thought str, &str, and &str[0] were all the same. So how would str or &str[0] be incompatible?
I think my code is being compiled by GCC since I'm running it on TextMate.
In C, str, &str and &str[0] refer to the same address - an array's label is simply a way of referring to the address of the start of a contiguous block of memory. str and &str are obviously of different types, however, &str being of type char (*)[].
If you want arr_ptr to point to "string", you merely have to do:
char *arr_ptr = str;
arr_ptr is simply of type char*, a single level pointer to char.
EDIT: In case I accidentally avoided the question, the reason why assigning &str to char (*arr_ptr)[] removes the error is because of the fact that &str is of type char (*)[], and str is only of type char[] (which can decay to char*).
str or &str[0] is incompatible. You just changed the definition of arr_ptr which avoided the question.
arr_ptr is a pointer to array of chars. Wouldn't an array of pointers to char would simply be *foo[]?
str != &str. In the context, str is an array of characters, and &str is a pointer to an array of characters, and even if the numerical value of the address is the same, the types are different so the equality doesn't apply.I thought
str,&strand&str[0]were all the same.
No. str is &str[0] - both denote a pointer to the first array element.
&str points to the same place, but is a pointer to the complete array.
The diffrence is the type if you dereference the pointer.
Try
printf("%d %d %d",
(int)sizeof(*str),
(int)sizeof(*(&str)),
(int)sizeof(*(&str[0])));
.
sizeof yields a size_t. Wouldn't it be better to use %z, or is that less portable?
%z doesn't even exist. %zu is the correct format for size_t, but it's new in C99, and some C implementations probably don't support it; Microsoft in particular has been less than enthusiastic about supporting C99. Converting to int and using %d is portable as long as the size doesn't exceed INT_MAX (which is at least 32767). You can also cast to unsigned long and use %lu as long as the size doesn't exceed ULONG_MAX (which is at least 2147483647).I guess what you were trying was this:
#include <stdio.h>
int main()
{
char *arr_ptr;
char str[] = "string";
arr_ptr = str;
printf("%s \n", arr_ptr);
printf("%s \n", str);
return 0;
}
The following program adds one more dimension for your understanding.
int main()
{
char **arr_ptr;
char *str[] = {"string1", "string2", "string3"};
int i;
arr_ptr = str;
for (i=0 ; i<3 ; i++) {
printf("%s \n", arr_ptr[i]);
printf("%s \n", str[i]);
}
return 0;
}
Hope this helps!
