Question

Why does

#include <iostream>
using namespace std;

int main() {
  cout << (char*)0x10 << endl; 
}

segfault, but

#include <iostream>
using namespace std;

int main() {
  cout << (void*)0x10 << endl; 
}

seems to work just fine?

Answer 1

Because

cout::operator <<(void*) 

prints a memory address, and

cout::operator <<(char*)

prints a null-terminated character array, and you run into undefined behaviour when you attempt to read the char array from 0x10.