Question

I ran into a compiler error that didn’t make much sense to me:

#include <memory> using namespace std;  auto_ptr<Table> table = db->query('select * from t'); 

error: conversion from ‘Table*’ to non-scalar type ‘std::auto_ptr< Table>’ requested

However, the following line does work:

auto_ptr<Table> table(db->query('select * from t')); 

What is it about this definiton of the constructor that prevents it from working as I expect? I thought that initialized declarations used the constructors.

Here’s my auto_ptr‘s constructor (from the SGI STL):

explicit auto_ptr(element_type* __p = 0) throw() : _M_ptr(__p) { } 

Answer 1

It’s the ‘explicit’ keyword.

template <typename T> struct foo {   explicit foo(T const *)   {   } };   template <typename T> struct bar {   bar(T const *)   {   } };   int main(int argc, char **argv) {   int a;   foo<int> f = &a; // doesn't work   bar<int> 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<int> const &); void quux(bar<int> 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>);   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. }