map question

[michael]

Hi, it happen to me this thing: I create lightmap[4] element and dimensione map became 4; is I do lightmap[6] = Light(); then 6 element...why this?

map<int,Light> lightmap;
lightmap[4] = Light();

map<int,Light> lightmap;
lightmap[4] = Light();

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I don't understand what you are asking.

[mickey]

lightmap[4] = light();
int i = lightmap.size();
printf("ADDING LIGHT dimensione <map> %i\n" ,i); //this print 4...

lightmap[4] = light();
int i = lightmap.size();
printf("ADDING LIGHT dimensione <map> %i\n" ,i); //this print 4...

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I thought lightmap[4] = light(); create an element with index <int=4, > ...

[michael]

lightmap[4] = light();
int i = lightmap.size();
printf("ADDING LIGHT dimensione <map> %i\n" ,i); //this print 4...

lightmap[4] = light();
int i = lightmap.size();
printf("ADDING LIGHT dimensione <map> %i\n" ,i); //this print 4...

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I thought lightmap[4] = light(); create an element with index <int=4, > ...

I am still not completely understanding you. If you mean that it is creating them element lightmap[4] then yes it does do that. If you are mean that it should create elements lightmap[0] through lightmap[4] then no it does not do that.

lightmap.size() only returns how many elements the map has, not the key of any one particular element. So, if you did the following:

map<int, Light> lightmap;
lightmap[3] = light();
lightmap[4] = light();
lightmap[5] = light();
int i = lightmap.size();
printf("ADDING LIGHT dimensione <map> %i\n" ,i);

map<int, Light> lightmap;
lightmap[3] = light();
lightmap[4] = light();
lightmap[5] = light();
int i = lightmap.size();
printf("ADDING LIGHT dimensione <map> %i\n" ,i);

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The output would be 3.

A map is declared as "map <key type, value type, ...>" where the key type is the type of key you will be using. You can then use the brackets [] to get or set the value at map element [key]. So when you call lightmap[4] you are NOT telling it to create lightmap[0] through lightmap[4], you are simply saying I am holding this value at element location named 4 in the map named lightmap. You could have declared your map as:
map<string, Light> lightmap;

And changed
lightmap[4] = light();
to
lightmap["Four"] = light();

Does this help?

[bood]

lightmap[4] = light();
int i = lightmap.size();
printf("ADDING LIGHT dimensione <map> %i\n" ,i); //this print 4...

lightmap[4] = light();
int i = lightmap.size();
printf("ADDING LIGHT dimensione <map> %i\n" ,i); //this print 4...

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I thought lightmap[4] = light(); create an element with index <int=4, > ...

The `i` should be 1, and I cann't reproduce your problem...
here's my code

class A
{
};
int main()
{
  map<int,A> m;
  m[4]=A();
  cout<<m.size();
  return 0;
}

class A
{
};
int main()
{
  map<int,A> m;
  m[4]=A();
  cout<<m.size();
  return 0;
}

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and it prints `1`

[mickey]

my problem probabily was another:

map<int, Light> lightmap;
lightmap[3] = light();
lightmap[4] = light();.
lightmap[5] = light();
int i = lightmap.size();
printf("ADDING LIGHT dimensione <map> %i\n" ,i); "ADDING LIGHT dimensione <map> %i\n" ,i); //this should print 3

map<int, Light> lightmap;
lightmap[3] = light();
lightmap[4] = light();.
lightmap[5] = light();
int i = lightmap.size();
printf("ADDING LIGHT dimensione <map> %i\n" ,i); "ADDING LIGHT dimensione <map> %i\n" ,i); //this should print 3

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now I need to to a thing (but I dont know how):

int size = lightmap.size();
for (int i=0; i<size; i++) {
 lightmap[i].DOSomething();
}

int size = lightmap.size();
for (int i=0; i<size; i++) {
 lightmap[i].DOSomething();
}

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here above I need to put in a for a my Light class function; but the for is working on elements 0,1,2; I need to use map with KEY 3,4,5 (they're iserted before).
Are u understand? thanks

[wysota]

for(QMap<int, Light>::iterator iter = lightmap.begin(); iter!=lightmap.end();++iter){
  iter.data().doSomething();
}

for(QMap<int, Light>::iterator iter = lightmap.begin(); iter!=lightmap.end();++iter){
  iter.data().doSomething();
}

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The same goes with std::map, just the iterator may behave different (just look at its docs to see what member you need to access). It'll probably be "iter->doSomething()".

[bood]

wysota's right, just use the`iterator`

And seems your English's poor

[Michiel]

And seems your English's poor

And it seems your English's poor.

[bood]

And it seems your English's poor.

I thought it can be ignored in oral English...can't it?

[mickey]

Are we speaking?

[wysota]

Stop the flame, please.

[Michiel]

I thought it can be ignored in oral English...can't it?

Not if you prepend it with 'And'. Anyway, what relevance does it have? As long as it's not 1337-speak and we can understand each other...