Are new std::map elements initialized?

[int2str@gmail.com]

Hi!

Given the following code:

#include <map>
#include <string>
#include <cassert>

int main()
{
std::map< std::string, unsigned > my_map;
++my_map["Test"];

assert( my_map["Test"] == 1 );
}

Can I rely on the fact that the newly created map element with key
"Test" will be created with a value of 0?

Josuttis's "The C++ Standard Library" seems to indicate so (pg. 207).

The Standard briefly states (23.3.1.2 para 1):

T& operator[] (const key_type& x);

Returns: (*((insert(make _pair(x, T()))).first)). second

So it looks like the default constructor for T is called. For PODs like
"unsigned", does the default constructor (?) guarantee
0-initialization?

Thanks for your insights!

Cheers,
Andre

[Gregory]

Re: Are new std::map elements initialized?

As I know there is no such thing as default constructor for C/C++ basic
types.

Try

void foo()
{
int i, j; // no default constructor
}

Neither i nor j are assigned 0. They both are uninitialized (junk).

Gregory

[red floyd]

Re: Are new std::map elements initialized?

Gregory wrote:[color=blue]
> As I know there is no such thing as default constructor for C/C++ basic
> types.
>
> Try
>
> void foo()
> {
> int i, j; // no default constructor
> }
>
> Neither i nor j are assigned 0. They both are uninitialized (junk).
>
> Gregory[/color]

but i = int() does initialize to 0.
So, If a map uses:
return (*((insert(make _pair(x, T()))).first)). second
as Josuttis claims, then if T is int, then the make_pair() call with
int() as the second parameter creates a pair with 0 as the second element.

[Gregory]

Re: Are new std::map elements initialized?

Yes, you are right. I wrote a test and my_map["Test"] returned 0.
I also checked g++ STL implementation we have and indeed
map::operator[] calls integer constructor - int() as you said.

_Tp& operator[](const key_type& __k) {
iterator __i = lower_bound(__k );
// __i->first is greater than or equivalent to __k.
if (__i == end() || key_comp()(__k, (*__i).first))
__i = insert(__i, value_type(__k, _Tp()));
return (*__i).second;
}


Gregory

[Jonathan Mcdougall]

Re: Are new std::map elements initialized?

int2str@gmail.c om wrote:[color=blue]
> Hi!
>
> Given the following code:
>
> #include <map>
> #include <string>
> #include <cassert>
>
> int main()
> {
> std::map< std::string, unsigned > my_map;
> ++my_map["Test"];
>
> assert( my_map["Test"] == 1 );
> }
>
> Can I rely on the fact that the newly created map element with key
> "Test" will be created with a value of 0?[/color]

Yes. std::map::opera tor[] returns (*((insert(make _pair(x,
T()))).first)). second and T() default-initializes an object. For
non-PODs, the default ctor is called; for arrays, each element is
default-initialized; other types are zero-initialized.


Jonathan


Jonathan