Why is the address-of operator ignored here?Code:#include <stdio.h> #include <string.h> int main(void) { char * ptr ; char string0[16] ; char string1[16] ; char string2[16] = "THIS IS DATA" ; memcpy(&string0, &string2, sizeof(string2)-1) ; printf("string0 = %s\n", string0 ) ; memcpy( string1, string2, sizeof(string2)-1) ; printf("string1 = %s\n", string1 ) ; return 0 ; }
Mainframe assembler programmer by trade. C coder when I can.
The address-of operator is not ignored at all. The name of an array denotes a pointer to its firest element. That makes that the address of the first element is the same as the address of the array. The type, however, is different (char* and char**, respectivly). It compiles since both types can be implicitly casted to a pointer to void that memcpy expects.
Why do you think it's being ignored?
A pointer to the whole array has the same value (but not the same type) as a pointer to the first element of the array.Code:$ cat foo.c #include <stdio.h> #include <string.h> int main(void) { char * ptr ; char string0[16] ; char string1[16] ; char string2[16] = "THIS IS DATA" ; memcpy(&string0, &string2, sizeof(string2)-1) ; printf("string0 = %s\n", string0 ) ; memcpy( string1, string2, sizeof(string2)-1) ; printf("string1 = %s\n", string1 ) ; return 0 ; } $ gcc foo.c $ ./a.out string0 = THIS IS DATA string1 = THIS IS DATA
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
> The type, however, is different (char* and char**, respectivly).
Not quite.
A pointer to a whole array isn't a char**, but something like char (*)[16] - in this example.
Code:#include <stdio.h> int main(void) { char string2[16] = "THIS IS DATA" ; char (*s)[16] = &string2; char *t = string2; char **w = &string2; return 0 ; } $ gcc foo.c foo.c: In function ‘main’: foo.c:6:15: warning: initialization of ‘char **’ from incompatible pointer type ‘char (*)[16]’ [-Wincompatible-pointer-types] 6 | char **w = &string2; | ^
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
I said it was "ignored" because the end result was the same. I wasn't aware that I could use &array. I've always just used array or &array[0]. Now I have a third option.
I was looking at the type differences (char * and char **) and expecting an error when using the &.
Thank you both for the lesson.
Mainframe assembler programmer by trade. C coder when I can.
FWIW For the sake of ease I've been not precise in both cases as I should have explained it as "array to pointer decay" when it gets passed to the parameter of memcpy.
You didn't get an error (or even a warning) in this case because memcpy takes void*—it doesn't care nor check what pointer types you're passing to it.Originally Posted by Dino
If you had a function that takes char* and you passed &string2 to it, then you could expect a warning or error.
