5 ms·

As someone who avoids C++ these days ... Is this a compile time FizzBuzz? `(char*)&1[""]` is the empty string here returning its data segment address as the o

As someone who avoids C++ these days ... Is this a compile time FizzBuzz? `(char*)&1[""]` is the empty string here returning its data segment address as the offset? Not sure about the purpose of `1`. Is the iteration achieved by the constructor basically re-invoking itself?

vardump 4y ago
It's a static initialization time FizzBuzz. In this case, I think it's executed before main() is called, and not at compile time. Unless, of course, you have a very clever compiler that determines memory allocation is not actually allocating anything and that the output is a static string, and there are no side effects. Such a clever compiler could optimize it all into just one "puts" call.
- jessermeyer 4y ago
  Neither GCC nor Clang bake the final string into the data segment. If I had to guess, printf is the one preventing the more fancy optimizations to take place.
  - vardump 4y ago
    I think so too. Also, depending on stdlib output buffering, the external I/O behavior of 100 puts calls is potentially different from just one call. In other words, there might be a different number of stdout write-calls.
bno1 4y ago
&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'

vardump 4y ago

It's a static initialization time FizzBuzz. In this case, I think it's executed before main() is called, and not at compile time. Unless, of course, you have a very clever compiler that determines memory allocation is not actually allocating anything and that the output is a static string, and there are no side effects. Such a clever compiler could optimize it all into just one "puts" call.

jessermeyer 4y ago
Neither GCC nor Clang bake the final string into the data segment. If I had to guess, printf is the one preventing the more fancy optimizations to take place.
- vardump 4y ago
  I think so too. Also, depending on stdlib output buffering, the external I/O behavior of 100 puts calls is potentially different from just one call. In other words, there might be a different number of stdout write-calls.

bno1 4y ago

&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'