Why doesn't auto_ptr construction work using = syntax

Answer 1

It's the "explicit" keyword.

template 
struct foo
{
  explicit foo(T const *)
  {
  }
};


template 
struct bar
{
  bar(T const *)
  {
  }
};


int main(int argc, char **argv)
{
  int a;
  foo f = &a; // doesn't work
  bar b = &a; // works
}

The "explicit" keyword prevents the constructor from being used for implicit type conversions. Consider the following two function prototypes:

void baz(foo const &);
void quux(bar const &);

With those definitions, try calling both functions with an int pointer:

baz(&a);  // fails
quux(&a); // succeeds

In the case of quux, your int pointer was implicitly converted to a bar.

EDIT: To expand on what other people commented, consider the following (rather silly) code:

void bar(std::auto_ptr);


int main(int argc, char **argv)
{
  bar(new int()); // probably what you want.

  int a;
  bar(&a); // ouch. auto_ptr would try to delete a at the end of the
           // parameter's scope

  int * b = new int();
  bar(b);
  *b = 42; // more subtle version of the above.
}