Can't figure out this error...

AndreFratelli

Hi I'm having the following error message when compiling a project with VS2005

e:\projects\netsim\netsim\grafo.cpp(66) : error C2679: binary '=' : no operator found which takes a right-hand operand of type 'std::_Vector_const_iterator<_Ty,_Alloc>' (or there is no acceptable conversion)
with
[
_Ty=Ligacao *,
_Alloc=std::allocator<Ligacao *>
]
c:\program files (x86)\microsoft visual studio 8\vc\include\vector(392): could be 'std::_Vector_iterator<_Ty,_Alloc> &std::_Vector_iterator<_Ty,_Alloc>::operator =(const std::_Vector_iterator<_Ty,_Alloc> &)'
with
[
_Ty=Ligacao *,
_Alloc=std::allocator<Ligacao *>
]
while trying to match the argument list '(std::_Vector_iterator<_Ty,_Alloc>, std::_Vector_const_iterator<_Ty,_Alloc> )'
with
[
_Ty=Ligacao *,
_Alloc=std::allocator<Ligacao *>
]

The code is:

std::vector<Ligacao \*>::iterator it;

for (it=this->\_m\_graf\_map\[\_onde\].begin() ; it != this->\_m\_graf\_map\[\_onde\].end() ; it++)
{
	if (
		((\*it)->esquerda()->id() == \_esq && (\*it)->direita()->id() == \_dir) ||
		((\*it)->esquerda()->id() == \_dir && (\*it)->direita()->id() == \_esq)
	) return true;
}

Where _m_graf_map is defined as:

std::vector<std::vector<Ligacao *> > _m_graf_map;

Does anybody know what am I doing wrong? Fratelli

Naveen

if _m_graf_map is like

std::vector<std::vector<Ligacao *> > _m_graf_map;

Then the iterator it should be defined like

std::vector<std::vector<Ligacao *> >::iterator it;

isnt it?

nave [OpenedFileFinder]

AndreFratelli

At first I tried it too, but I'm not actually iterating through _m_graf_map but _m_graf_map[_ond], which has type std::vector<Ligacao *>. best regards Fratelli

Nibu babu thomas

AndreFratelli wrote:

std::vector >

Just a suggestion! Use a typedef to create type for this vector and will make things easier for you...

typedef std::vector<std::vector<Ligacao *> > LigacaoVector2d;

Then create iterators like LigacaoVector2d::iterator it;

Nibu babu thomas Microsoft MVP for VC++ Code must be written to be read, not by the compiler, but by another human being. Programming Blog: http://nibuthomas.wordpress.com

BadKarma

Hi, I don't see it either, but you can try to split the code up to make it easier top spot it. According to the error messages the iterator believes it should use the const iterator instead of the non-const version.

std::vector lstLigacao = this->_m_graf_map[_onde];
std::vector::iterator it;

for(it = lstLigacao.begin() ; it != lstLigacao.end() ; it++)
{
	if (
		((*it)->esquerda()->id() == _esq && (*it)->direita()->id() == _dir) ||
		((*it)->esquerda()->id() == _dir && (*it)->direita()->id() == _esq)
	) return true;
}

Learn from the mistakes of others, you may not live long enough to make them all yourself.

AndreFratelli

That works! But I can't afford to copy an entire array like that =/ If that's the case, I rather use vector<>::size_type to iterate through the elements But why does this happen? I mean, lstLigacao and _m_graf_map[_ond] have exactly the same type, right? So why does the iterator work with one, but not the other? :confused: Best regards Fratelli

AndreFratelli

Yes, I usually do that.. But when dealing with arrays I also define an "iterator" type, like: typedef std::vector<std::vector<Ligacao *> > LigacaoVector2d; typeded LigacaoVector2d::iterator iterator; That way, I can even do: iterator it; Thx anyway ;) =) best regards Fratelli

Nibu babu thomas

AndreFratelli wrote:

typeded LigacaoVector2d::iterator iterator;

It's better to use it directly so that readers of your code can understand from where this iterator is being used/to which vector this iterator belongs. Isn't it?

Nibu babu thomas Microsoft MVP for VC++ Code must be written to be read, not by the compiler, but by another human being. Programming Blog: http://nibuthomas.wordpress.com

BadKarma

Hi, I have tried your code, and it works. I'm using VS2005. But you don't need to copy the array, you could use a reference to it;

std::vector<Ligacao *>& refList = _m_graf_map[_ond];

and use the reference

Learn from the mistakes of others, you may not live long enough to make them all yourself.

AndreFratelli

True =) but, on the other hand, if you have a class with the following definitions:

template <typename T> class MyClass
{
typedef vector<T> my_vector;
typedef my_vector::iterator iterator;
};

Then you/others can do something like:

int main()
{
MyClass<type>::iterator it;
return 0;
}

Instead of:

int main()
{
MyClass<type>::my_vector::iterator it; // Or even vector<type>::iterator
}

Would you agree that it is practical? regards Fratelli

Nibu babu thomas

AndreFratelli wrote:

Would you agree that it is practical?

Yeah, I thought you were letting the iterator typedef float around, if it's in a class then should be fine. :)

Nibu babu thomas Microsoft MVP for VC++ Code must be written to be read, not by the compiler, but by another human being. Programming Blog: http://nibuthomas.wordpress.com

AndreFratelli

I actually get another error =/

e:\projects\netsim\netsim\grafo.cpp(64) : error C2440: 'initializing' : cannot convert from 'const std::vector<_Ty>' to 'std::vector<_Ty> &'
with
[
_Ty=Ligacao *
]
Conversion loses qualifiers

The code is:

std::vector &ref = this->_m_graf_map[_onde];

Regards Fratelli

BadKarma

Hi, I think i got it. The function where you use this list is probably defined as const. Therefore all calls are made to the constant versions of the vector object. Either change the function not to be const (not good) or use

const std::vector<ligacao*>& ref = lstTT[1];
for (std::vector<ligacao*>::const_iterator it = ref.begin(); it != ref.end(); ++it)

Learn from the mistakes of others, you may not live long enough to make them all yourself.

AndreFratelli

It works! :-D You were right, the function is const. Just two things I don't get, though: 1) Why does the iterator must be constant too? 2) Why do you say that making the function not const is not good? I'm asking this because I've seen different opinions in this matter. On one hand, they would argue that making a function const allows the compiler to optimize the function in several ways. On the other hand, I remember reading an article that said that "marking a function const is only a guarantee you're giving to the user that the method will not change the class's attributes" (nor its input, I think...). What do you think? Regards

Fratelli

BadKarma

AndreFratelli wrote:

Why do you say that making the function not const is not good?

- When marking a member function as const. You're telling the user of the class that this call to will not change the internal data of the class. This is 'guaranteed'. - You're alsoo telling the compiler to help you the writer too fulfill this design requirement. The compiler will modify for example a member int m_x; to const int m_x;

AndreFratelli wrote:

Why does the iterator must be constant too?

- The compiler has modified the constantness of this member, meaning that this member cannot be changed. Hence you need to use the const version of the iterator;

Learn from the mistakes of others, you may not live long enough to make them all yourself.

AndreFratelli

All clear now =D Thanks a lot ! ;) Best regards

Fratelli

BadKarma

Glad I could help

Learn from the mistakes of others, you may not live long enough to make them all yourself.