Can std::iterator check, if the next element exist over that iterator?
I have a template Iterator class that contains std::iterator of container that is specified by template. I did not find any way to check if the next element exist over the iterator, without using container.
There is a control like this;
vector<int> v;
vector<int>::iterator itr;
if(itr== v.end()) { /*...*/}
but I wanted to do this control in my Iterator class and my class is like following...
template <class E, class C= vector<E> >
class Iterator {
public:
/*...*/
bool hasNext()noexcept;
/*...*/
private:
typename C::iterator itr; // is there any problem with this decleration?
};
//implementation of hasNext() function.
template<class E, class C>
bool
Iterator<E,C>::hasNext()noexcept {
return(itr!=end())?true:false; // this line is wrong. How can I fix it?
}
c++ stl iterator
edited Dec 28 '18 at 18:15
SergeyA
41.6k53783
asked Dec 28 '18 at 18:06
husnuhusnu
63
add a comment |
I have a template Iterator class that contains std::iterator of container that is specified by template. I did not find any way to check if the next element exist over the iterator, without using container.
There is a control like this;
vector<int> v;
vector<int>::iterator itr;
if(itr== v.end()) { /*...*/}
but I wanted to do this control in my Iterator class and my class is like following...
template <class E, class C= vector<E> >
class Iterator {
public:
/*...*/
bool hasNext()noexcept;
/*...*/
private:
typename C::iterator itr; // is there any problem with this decleration?
};
//implementation of hasNext() function.
template<class E, class C>
bool
Iterator<E,C>::hasNext()noexcept {
return(itr!=end())?true:false; // this line is wrong. How can I fix it?
}
c++ stl iterator
edited Dec 28 '18 at 18:15
SergeyA
41.6k53783
asked Dec 28 '18 at 18:06
husnuhusnu
63
5
You are free to do that if you want, but know that it won't be used by the standard functions that use iterators or even range-based for loops. You'll still need to provide a range for those purposes.
– François Andrieux
Dec 28 '18 at 18:08
6
This is a Java design. Importing Java notions into C++ hardly ever works well.
– Pete Becker
Dec 28 '18 at 18:12
2
No, you need a container to call it'sendfunction. Your design is unfixable.
– SergeyA
Dec 28 '18 at 18:14
1
And to add up upon @SergeyA's comment an iterator implementation should be bound to a specific container instance.
– πάντα ῥεῖ
Dec 28 '18 at 18:15
add a comment |
I have a template Iterator class that contains std::iterator of container that is specified by template. I did not find any way to check if the next element exist over the iterator, without using container.
There is a control like this;
vector<int> v;
vector<int>::iterator itr;
if(itr== v.end()) { /*...*/}
but I wanted to do this control in my Iterator class and my class is like following...
template <class E, class C= vector<E> >
class Iterator {
public:
/*...*/
bool hasNext()noexcept;
/*...*/
private:
typename C::iterator itr; // is there any problem with this decleration?
};
//implementation of hasNext() function.
template<class E, class C>
bool
Iterator<E,C>::hasNext()noexcept {
return(itr!=end())?true:false; // this line is wrong. How can I fix it?
}
c++ stl iterator
edited Dec 28 '18 at 18:15
SergeyA
41.6k53783
asked Dec 28 '18 at 18:06
husnuhusnu
63
I have a template Iterator class that contains std::iterator of container that is specified by template. I did not find any way to check if the next element exist over the iterator, without using container.
There is a control like this;
vector<int> v;
vector<int>::iterator itr;
if(itr== v.end()) { /*...*/}
but I wanted to do this control in my Iterator class and my class is like following...
template <class E, class C= vector<E> >
class Iterator {
public:
/*...*/
bool hasNext()noexcept;
/*...*/
private:
typename C::iterator itr; // is there any problem with this decleration?
};
//implementation of hasNext() function.
template<class E, class C>
bool
Iterator<E,C>::hasNext()noexcept {
return(itr!=end())?true:false; // this line is wrong. How can I fix it?
}
c++ stl iterator
c++ stl iterator
edited Dec 28 '18 at 18:15
SergeyA
41.6k53783
asked Dec 28 '18 at 18:06
husnuhusnu
63
edited Dec 28 '18 at 18:15
SergeyA
41.6k53783
asked Dec 28 '18 at 18:06
husnuhusnu
63
edited Dec 28 '18 at 18:15
SergeyA
41.6k53783
edited Dec 28 '18 at 18:15
SergeyA
41.6k53783
edited Dec 28 '18 at 18:15
SergeyA
41.6k53783
41.6k53783
asked Dec 28 '18 at 18:06
husnuhusnu
63
asked Dec 28 '18 at 18:06
husnuhusnu
63
asked Dec 28 '18 at 18:06
husnuhusnu
63
63
5
You are free to do that if you want, but know that it won't be used by the standard functions that use iterators or even range-based for loops. You'll still need to provide a range for those purposes.
– François Andrieux
Dec 28 '18 at 18:08
6
This is a Java design. Importing Java notions into C++ hardly ever works well.
– Pete Becker
Dec 28 '18 at 18:12
2
No, you need a container to call it'sendfunction. Your design is unfixable.
– SergeyA
Dec 28 '18 at 18:14
1
And to add up upon @SergeyA's comment an iterator implementation should be bound to a specific container instance.
– πάντα ῥεῖ
Dec 28 '18 at 18:15
add a comment |
5
You are free to do that if you want, but know that it won't be used by the standard functions that use iterators or even range-based for loops. You'll still need to provide a range for those purposes.
– François Andrieux
Dec 28 '18 at 18:08
6
This is a Java design. Importing Java notions into C++ hardly ever works well.
– Pete Becker
Dec 28 '18 at 18:12
2
No, you need a container to call it'sendfunction. Your design is unfixable.
– SergeyA
Dec 28 '18 at 18:14
1
And to add up upon @SergeyA's comment an iterator implementation should be bound to a specific container instance.
– πάντα ῥεῖ
Dec 28 '18 at 18:15
5
5
You are free to do that if you want, but know that it won't be used by the standard functions that use iterators or even range-based for loops. You'll still need to provide a range for those purposes.
– François Andrieux
Dec 28 '18 at 18:08
You are free to do that if you want, but know that it won't be used by the standard functions that use iterators or even range-based for loops. You'll still need to provide a range for those purposes.
– François Andrieux
Dec 28 '18 at 18:08
6
6
This is a Java design. Importing Java notions into C++ hardly ever works well.
– Pete Becker
Dec 28 '18 at 18:12
This is a Java design. Importing Java notions into C++ hardly ever works well.
– Pete Becker
Dec 28 '18 at 18:12
2
2
No, you need a container to call it's
end function. Your design is unfixable.– SergeyA
Dec 28 '18 at 18:14
No, you need a container to call it's
end function. Your design is unfixable.– SergeyA
Dec 28 '18 at 18:14
1
1
And to add up upon @SergeyA's comment an iterator implementation should be bound to a specific container instance.
– πάντα ῥεῖ
Dec 28 '18 at 18:15
And to add up upon @SergeyA's comment an iterator implementation should be bound to a specific container instance.
– πάντα ῥεῖ
Dec 28 '18 at 18:15
add a comment |
1 Answer
1
active
oldest
votes
An iterator represents a position within a sequence of items. An iterator knows how to get to the next element in that sequence, but the nature of C++'s iterator model is based on the idea that there exists a "past-the-end" iterator after the end of such a sequence. Such an iterator does not represent a valid item in the sequence; it merely represents the end of the sequence, and iterators can be tested against it.
This construct is useful, as it allows you to talk about sub-ranges of elements within a sequence. For example, if a container has 10 elements, you can pass the begin/end iterators to the std::sort algorithm to sort them. However, you could also sort the first 10 elements by passing the begin iterator and the begin() + 10 iterator to the same function. The sort algorithm will treat the given ending iterator as the "past-the-end" iterator, not an iterator to a valid element.
What you are trying to do is combine two distinct ideas: position and range. Some iteration models do things that way, but that is not the STL-derived model that C++'s standard library is based on.
Now to be fair, there are some cases where position inherently carries range information. stream-based iterators know whether they are at the end of the stream or not, because to function as an iterator, they have to store a reference to the stream they iterate over. And the stream knows whether it is out of data. But overall, a single iterator is not meant to know whether it is at the end of its range or not.
answered Dec 28 '18 at 18:19
Nicol BolasNicol Bolas
283k33468643
add a comment |
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An iterator represents a position within a sequence of items. An iterator knows how to get to the next element in that sequence, but the nature of C++'s iterator model is based on the idea that there exists a "past-the-end" iterator after the end of such a sequence. Such an iterator does not represent a valid item in the sequence; it merely represents the end of the sequence, and iterators can be tested against it.
This construct is useful, as it allows you to talk about sub-ranges of elements within a sequence. For example, if a container has 10 elements, you can pass the begin/end iterators to the std::sort algorithm to sort them. However, you could also sort the first 10 elements by passing the begin iterator and the begin() + 10 iterator to the same function. The sort algorithm will treat the given ending iterator as the "past-the-end" iterator, not an iterator to a valid element.
What you are trying to do is combine two distinct ideas: position and range. Some iteration models do things that way, but that is not the STL-derived model that C++'s standard library is based on.
Now to be fair, there are some cases where position inherently carries range information. stream-based iterators know whether they are at the end of the stream or not, because to function as an iterator, they have to store a reference to the stream they iterate over. And the stream knows whether it is out of data. But overall, a single iterator is not meant to know whether it is at the end of its range or not.
answered Dec 28 '18 at 18:19
Nicol BolasNicol Bolas
283k33468643
add a comment |
An iterator represents a position within a sequence of items. An iterator knows how to get to the next element in that sequence, but the nature of C++'s iterator model is based on the idea that there exists a "past-the-end" iterator after the end of such a sequence. Such an iterator does not represent a valid item in the sequence; it merely represents the end of the sequence, and iterators can be tested against it.
This construct is useful, as it allows you to talk about sub-ranges of elements within a sequence. For example, if a container has 10 elements, you can pass the begin/end iterators to the std::sort algorithm to sort them. However, you could also sort the first 10 elements by passing the begin iterator and the begin() + 10 iterator to the same function. The sort algorithm will treat the given ending iterator as the "past-the-end" iterator, not an iterator to a valid element.
What you are trying to do is combine two distinct ideas: position and range. Some iteration models do things that way, but that is not the STL-derived model that C++'s standard library is based on.
Now to be fair, there are some cases where position inherently carries range information. stream-based iterators know whether they are at the end of the stream or not, because to function as an iterator, they have to store a reference to the stream they iterate over. And the stream knows whether it is out of data. But overall, a single iterator is not meant to know whether it is at the end of its range or not.
answered Dec 28 '18 at 18:19
Nicol BolasNicol Bolas
283k33468643
add a comment |
An iterator represents a position within a sequence of items. An iterator knows how to get to the next element in that sequence, but the nature of C++'s iterator model is based on the idea that there exists a "past-the-end" iterator after the end of such a sequence. Such an iterator does not represent a valid item in the sequence; it merely represents the end of the sequence, and iterators can be tested against it.
This construct is useful, as it allows you to talk about sub-ranges of elements within a sequence. For example, if a container has 10 elements, you can pass the begin/end iterators to the std::sort algorithm to sort them. However, you could also sort the first 10 elements by passing the begin iterator and the begin() + 10 iterator to the same function. The sort algorithm will treat the given ending iterator as the "past-the-end" iterator, not an iterator to a valid element.
What you are trying to do is combine two distinct ideas: position and range. Some iteration models do things that way, but that is not the STL-derived model that C++'s standard library is based on.
Now to be fair, there are some cases where position inherently carries range information. stream-based iterators know whether they are at the end of the stream or not, because to function as an iterator, they have to store a reference to the stream they iterate over. And the stream knows whether it is out of data. But overall, a single iterator is not meant to know whether it is at the end of its range or not.
answered Dec 28 '18 at 18:19
Nicol BolasNicol Bolas
283k33468643
An iterator represents a position within a sequence of items. An iterator knows how to get to the next element in that sequence, but the nature of C++'s iterator model is based on the idea that there exists a "past-the-end" iterator after the end of such a sequence. Such an iterator does not represent a valid item in the sequence; it merely represents the end of the sequence, and iterators can be tested against it.
This construct is useful, as it allows you to talk about sub-ranges of elements within a sequence. For example, if a container has 10 elements, you can pass the begin/end iterators to the std::sort algorithm to sort them. However, you could also sort the first 10 elements by passing the begin iterator and the begin() + 10 iterator to the same function. The sort algorithm will treat the given ending iterator as the "past-the-end" iterator, not an iterator to a valid element.
What you are trying to do is combine two distinct ideas: position and range. Some iteration models do things that way, but that is not the STL-derived model that C++'s standard library is based on.
Now to be fair, there are some cases where position inherently carries range information. stream-based iterators know whether they are at the end of the stream or not, because to function as an iterator, they have to store a reference to the stream they iterate over. And the stream knows whether it is out of data. But overall, a single iterator is not meant to know whether it is at the end of its range or not.
answered Dec 28 '18 at 18:19
Nicol BolasNicol Bolas
283k33468643
answered Dec 28 '18 at 18:19
Nicol BolasNicol Bolas
283k33468643
answered Dec 28 '18 at 18:19
Nicol BolasNicol Bolas
283k33468643
answered Dec 28 '18 at 18:19
Nicol BolasNicol Bolas
283k33468643
283k33468643
add a comment |
add a comment |
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5
You are free to do that if you want, but know that it won't be used by the standard functions that use iterators or even range-based for loops. You'll still need to provide a range for those purposes.
– François Andrieux
Dec 28 '18 at 18:08
6
This is a Java design. Importing Java notions into C++ hardly ever works well.
– Pete Becker
Dec 28 '18 at 18:12
2
No, you need a container to call it's
endfunction. Your design is unfixable.– SergeyA
Dec 28 '18 at 18:14
1
And to add up upon @SergeyA's comment an iterator implementation should be bound to a specific container instance.
– πάντα ῥεῖ
Dec 28 '18 at 18:15