[wizzairflyer]

My (admittedly naive) understanding of the ordeal leads me to believe that it is that the 1st would not segfault but the second will since declaring it as a const pointer will create additional memory constraints.

Testing it on my machine with the following code seems to validate this hypothesis.

  //file: test.c
  #include <stdio.h>
  const void * const_pointer = &const_pointer;
  void * const const_value   = &const_value;
  
  int main()
  {
      printf("%p\n", const_pointer);
      *(int*)const_pointer = 0;
      printf("%p\n", const_pointer);
  
      printf("---------------------------\n");
  
      printf("%p\n", const_value);
      *(int*)const_value = 0;
      printf("%p\n", const_value);
      return 0;
  }
  
  Result:
  
  $ gcc test.c
  test.c:4:30: warning: initialization discards ‘const’ qualifier from pointer target type [-Wdiscarded-qualifiers]
  void * const const_value   = &const_value;
                              ^

  $ ./a.out
  0x55b29ebfc010
  0x55b200000000
  ---------------------------
  0x55b29ebfbdb8

  Command terminated

As to why the first one doesn't also result in a segfault, I don't know.

[nybble41]

In the first version, const_pointer is a non-const variable (so located in .data) holding a pointer to potentially constant data—you can't modify the data through that pointer without a typecast, but the actual location in memory may be mutable. That's why you don't get a segfault when you cast away the const and modify the data—the destination (const_pointer) is not const even though the const-qualified pointer would allow it to be.

In the second case the const_value variable itself is const-qualified and thus located in .rodata, but the pointer itself is not const-qualified so nothing prevents you from attempting to modify the data through that pointer. This is why you get a compiler warning about discarding the 'const' qualifier in the initialization. Since const_value is in .rodata, writing to it through the pointer causes a segfault.

As Sean1708 pointed out, it's more obvious what is going on if you place the 'const' qualifier immediately before the thing it's modifying, which is either the pointer operator or the variable name, never the type itself:

    void const *const_pointer = &const_pointer;
    void *const const_value   = &const_value;

What would something like "const int" even mean on its own, anyway? There is no such thing as a mutable integer. It's the memory location holding the integer which may be either mutable or immutable.

[Sean1708]

My assembly knowledge is limited, but it looks like both const_pointer and const_value get put into .rodata (read-only data). In both cases you're trying to change what is at the memory location that the pointer points to, in the first case it's the pointer that's in .rodata so you can change what it points to, but in the second case it's the value that's in .rodata so you can't change it.

Edit: Actually I don't think the pointer is put anywhere, rather its value is stored in .data (non-read-only data) so it can be mutated without issue. Again though, my assembly isn't amazing.