3 ms·

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

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.

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.