4 ms·

&1[""] is the same as &(""[1])

• jessermeyer 4y ago
  Ah, so it's the nil address?
  • leni536 4y ago
    No, it's the address past the end of the array {'\0'}.
    • jessermeyer 4y ago
      Mind to unpack that for me? Here's how I unpack this: ""[1] is the 0 (termination) byte. The 0 byte is then interpreted as an address -- the nil address. But what's the use of asking for the address of the null terminator? Where is that stored exactly?
      • jussij 4y ago
        The "" will define a null terminated char array to represent the string. But as it string contains no text it's a char array that only requires one byte (i.e. it contains nothing but the null termination character). Now the first character of that char array is found here: ""[0] The second character is found here: ""[1] So the address of the second character is found here: &""[1] But as the string was represented by one byte char array that second address is past the end of the string. So it's actually an undefined address.
        • leni536 4y ago
          Nit: one past end of an array is actually fine as an address.
          • jussij 4y ago
            Not always. Consider this 'test.cpp' example: #include <stdio.h> #include <string.h> struct a_struct { char a[1]; char b[2]; a_struct() { strcpy(a, ""); strcpy(b, "b"); } } x; int main() { printf("before:\n"); printf("a : '%s'\n", x.a); printf("b : '%s'\n", x.b); char *c = &1[x.a]; *c = 'c'; printf("after:\n"); printf("a : '%s'\n", x.a); printf("b : '%s'\n", x.b); //b is now corrupt } Compile and running this code: C:\temp>g++ test.cpp C:\temp>a.exe before: a : '' b : 'b' after: a : '' b : 'c'
            • leni536 4y ago
              Yes, dereferencing past the end is not fine, but the address itself is.
      • leni536 4y ago
        ""[1] is not the termination byte, it's the byte after that. Its address is taken there with &, which yields a const char*. The (char *) cast is only there to cast away const.
  • bno1 4y ago
    The address of the nil byte + 1
    • jessermeyer 4y ago
      But isn't `""` 0 terminated? So the first offset past the nil byte is 0, interpreted as an address.
      • oldgradstudent 4y ago
        "" is empty, so ""[0] is the nil byte. The code uses "" as an arbitrary address, and ""[1] as that address + 1. This way this-"" gets you 1-based index into the array. It assumes the compiler dedfuplicates strings, making the behavior of this program undefined.
  • hoseja 4y ago
    The evilest way to write `nullptr`.
    • shultays 4y ago
      it is not nullptr, it is address past \0 in "", which is an invalid address
      • hoseja 4y ago
        Hence, evilest. (It got me.)
      • 3836293648 4y ago
        No, the address itself is not invalid, you're just not allowed to dereference it. Pointers have to point at a valid object or, in the case of arrays, one past the end of the array. It's what `std::end` returns