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  #1  
Old March 24th, 2007, 05:05 AM
ma740988
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Posts: n/a
Default composite types and std::swap

Consider the source snippet.


# include <iostream>
struct foo_struct {
int odx ;
int pdx ;
foo_struct ()
: odx ( 0 )
, pdx ( 0 )
{}
};

int main()
{

foo_struct fs1;
fs1.odx = 2 ;
fs1.pdx = 5 ;
foo_struct fs2;
std::swap ( fs1, fs2 ) ;
}

Ideally I could provide my own copy constructor and assignment
operator for foo_struct, nonetheless, I'm more interested in whether
the use of std::swap on the composite ( non-POD types ) makes the
program ill-formed?

Thanks in advance.

  #2  
Old March 24th, 2007, 06:05 AM
Alf P. Steinbach
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Posts: n/a
Default Re: composite types and std::swap

* ma740988:
Quote:
>
# include <iostream>
struct foo_struct {
int odx ;
int pdx ;
foo_struct ()
: odx ( 0 )
, pdx ( 0 )
{}
};
>
int main()
{
>
foo_struct fs1;
fs1.odx = 2 ;
fs1.pdx = 5 ;
foo_struct fs2;
std::swap ( fs1, fs2 ) ;
}
>
Ideally I could provide my own copy constructor and assignment
operator for foo_struct, nonetheless, I'm more interested in whether
the use of std::swap on the composite ( non-POD types ) makes the
program ill-formed?
No.

The effect is the same as

foo_struct temp;
temp = fs1; fs1 = fs2; fs2 = temp;

which just uses the compiler-generated copy assignment operator.

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