Re: Abusing const/mutable with std::set
"terminator" <email@example.com> wrote in message
On Oct 4, 6:48 pm, "Victor Bazarov" <v.Abaza...@comAcast.net> wrote:
On 4 Oct, 15:13, "Victor Bazarov" <v.Abaza...@comAcast.net> wrote:
I wish to modify some values of items contained in a set after they
have been added. [..] How acceptable is this?
And are there other reasons the items should be treated as const,
besides the ordering?
No, there are no other reasons. However, to be entirely correct,
you should perhaps consider storing the iterator following the item
you're trying to change, remove the item, change it, and then insert
it again using the stored iterator as the hint. It will be constant
AFA the complexity is concerned. And you'll have no mutable members
in your structs... Try it.
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Ok, thanks for the quick reply. I will look into removing and
reinserting the object - my only concern is that other iterators may
be pointing at it (in my triangle and mesh classes). If I remove and
reinsert it then it will be at the same point in the sequence, but I
doubt if this guarantees existing iterators will still be valid.
That's a valid concern. Most likely they will become invalid.
Though I can probably change my code so I don't create other iterators
until the vertex object is finished being set up. I'll have a look...
mutable my_data data;
typedef set<const_remover,my_comp,my_alloc> my_set;
Is VC++ .net 2003 non standard compliant as far as set goes then? This
compiled and runs, and according to this discussion it shouldn't. I was
planning on seeing what chaning i to mutable would do, but it runs with or
without it being mutable.
double x, y, z;
bool operator<( const Foo& lhs, const Foo& rhs )
if ( lhs.x < rhs.x )
else if ( lhs.x == rhs.x )
if ( lhs.y < rhs.y )
else if ( lhs.y == rhs.y )
if ( lhs.z < rhs.z )
Inst.x = 1.0;
Inst.y = 2.0;
Inst.z = 3.0;
Inst.i = 1234.0;
Bar.insert( Inst );
std::set<Foo>::iterator it = Bar.find( Inst );
it->i = 5.0;
it->x = 7.0;
for ( std::set<Foo>::iterator it = Bar.begin(); it != Bar.end(); ++it )
std::cout << it->x << " " << it->y << " " << it->z << " " << it->i
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