I've recently begun to prefer the free functions std::next and std::prev to explicitly copying and incrementing/decrementing iterators. Now, I am seeing weird behavior in a pretty specific case, and I would appreciate any help demystifying it.
I have an interpolation/extrapolation function operating on a boost::any_range of some X_type. The full definition of the range type is:
boost::any_range <
const X_type,
boost::random_access_traversal_tag,
const X_type,
std::ptrdiff_t
>
The any_range, in this particular case, is assigned from an iterator_range holding two pointers to const X_type, which serves as an X_type view of about half of the data() area of a vector<char>.
Compiling my application in MSVC 2010, everything works just fine.
Compiling the same code in MinGW g++ 4.7.0, it seemed to hang in one particular location, which I've then narrowed down to this (slightly abbreviated):
// Previously ensured conditions:
// 1) xrange is nonempty;
// 2) yrange is the same size as xrange.
auto x_equal_or_greater =
std::lower_bound(std::begin(xrange),std::end(xrange),xval);
if (x_equal_or_greater == std::end(xrange))
{
return *yit_from_xit(std::prev(x_equal_or_greater),xrange,yrange);
}
Stepping through the code in gdb, I found out it wasn't getting stuck, just taking a very long time to return from the single std::prev call - which in libstdc++ is implemented in terms of std::advance and ultimately the += operator.
By merely replacing the return line with:
auto xprev=x_equal_or_greater;
--xprev;
return *yit_from_xit(xprev,xrange,yrange);
Performance is great again, and there's virtually no delay.
I am aware of the overhead of using type-erased iterators (those of any_range), but even so, are the two cases above really supposed to carry such different costs? Or am I doing something wrong?
Okay, after responding to SplinterOfChaos's comment, I realized something. The problem is in your use of the any_range. In particular, the 3rd argument, which indicates that the Reference argument is a const int. In the boost iterator facade, when the reference is not a real reference, it will either use std::input_iterator_tag, or not provide an STL equivalent tag.
It has to do with the fact, that strictly speaking, all forward, bidirectional, and random access STL iterators must use a real reference for their reference type. From 24.2.5 of the C++11 standard:
A class or a built-in type X satisfies the requirements of a forward iterator if
— X satisfies the requirements of an input iterator (24.2.3),
— X satisfies the DefaultConstructible requirements (17.6.3.1),
— if X is a mutable iterator, reference is a reference to T; if X is a const iterator, reference is a reference to const T,
— the expressions in Table 109 are valid and have the indicated semantics, and
— objects of type X offer the multi-pass guarantee, described below.
In this case, it's returning an std::input_iterator_tag when queried for its iterator_category, which causes the call to std::prev() veer into Undefined Behavior.
Either way, the solution is to change (if possible) your use of boost::any_range to the following:
boost::any_range <
const X_type,
boost::random_access_traversal_tag,
const X_type&,
std::ptrdiff_t
>
This will cause it to have an iterator_category of std::random_access_iterator_tag, and will perform the operation as you originally expected.
Related
Item 18 of Scott Meyers's book Effective STL: 50 Specific Ways to Improve Your Use of the Standard Template Library says to avoid vector <bool> as it's not an STL container and it doesn't really hold bools.
The following code:
vector <bool> v;
bool *pb =&v[0];
will not compile, violating a requirement of STL containers.
Error:
cannot convert 'std::vector<bool>::reference* {aka std::_Bit_reference*}' to 'bool*' in initialization
vector<T>::operator [] return type is supposed to be T&, but why is it a special case for vector<bool>?
What does vector<bool> really consist of?
The Item further says:
deque<bool> v; // is a STL container and it really contains bools
Can this be used as an alternative to vector<bool>?
Can anyone please explain this?
For space-optimization reasons, the C++ standard (as far back as C++98) explicitly calls out vector<bool> as a special standard container where each bool uses only one bit of space rather than one byte as a normal bool would (implementing a kind of "dynamic bitset"). In exchange for this optimization it doesn't offer all the capabilities and interface of a normal standard container.
In this case, since you can't take the address of a bit within a byte, things such as operator[] can't return a bool& but instead return a proxy object that allows to manipulate the particular bit in question. Since this proxy object is not a bool&, you can't assign its address to a bool* like you could with the result of such an operator call on a "normal" container. In turn this means that bool *pb =&v[0]; isn't valid code.
On the other hand deque doesn't have any such specialization called out so each bool takes a byte and you can take the address of the value return from operator[].
Finally note that the MS standard library implementation is (arguably) suboptimal in that it uses a small chunk size for deques, which means that using deque as a substitute isn't always the right answer.
The problems is that vector<bool> returns a proxy reference object instead of a true reference, so that C++98 style code bool * p = &v[0]; won't compile. However, modern C++11 with auto p = &v[0]; can be made to compile if operator& also returns a proxy pointer object. Howard Hinnant has written a blog post detailing the algorithmic improvements when using such proxy references and pointers.
Scott Meyers has a long Item 30 in More Effective C++ about proxy classes. You can come a long way to almost mimic the builtin types: for any given type T, a pair of proxies (e.g. reference_proxy<T> and iterator_proxy<T>) can be made mutually consistent in the sense that reference_proxy<T>::operator&() and iterator_proxy<T>::operator*() are each other's inverse.
However, at some point one needs to map the proxy objects back to behave like T* or T&. For iterator proxies, one can overload operator->() and access the template T's interface without reimplementing all the functionality. However, for reference proxies, you would need to overload operator.(), and that is not allowed in current C++ (although Sebastian Redl presented such a proposal on BoostCon 2013). You can make a verbose work-around like a .get() member inside the reference proxy, or implement all of T's interface inside the reference (this is what is done for vector<bool>::bit_reference), but this will either lose the builtin syntax or introduce user-defined conversions that do not have builtin semantics for type conversions (you can have at most one user-defined conversion per argument).
TL;DR: no vector<bool> is not a container because the Standard requires a real reference, but it can be made to behave almost like a container, at least much closer with C++11 (auto) than in C++98.
vector<bool> contains boolean values in compressed form using only one bit for value (and not 8 how bool[] arrays do). It is not possible to return a reference to a bit in c++, so there is a special helper type, "bit reference", which provides you a interface to some bit in memory and allows you to use standard operators and casts.
Many consider the vector<bool> specialization to be a mistake.
In a paper "Deprecating Vestigial Library Parts in C++17"
There is a proposal to
Reconsider vector Partial Specialization.
There has been a long history of the bool partial specialization of
std::vector not satisfying the container requirements, and in
particular, its iterators not satisfying the requirements of a random
access iterator. A previous attempt to deprecate this container was
rejected for C++11, N2204.
One of the reasons for rejection is that it is not clear what it would
mean to deprecate a particular specialization of a template. That
could be addressed with careful wording. The larger issue is that the
(packed) specialization of vector offers an important
optimization that clients of the standard library genuinely seek, but
would not longer be available. It is unlikely that we would be able to
deprecate this part of the standard until a replacement facility is
proposed and accepted, such as N2050. Unfortunately, there are no such
revised proposals currently being offered to the Library Evolution
Working Group.
Look at how it is implemented. the STL builds vastly on templates and therefore the headers do contain the code they do.
for instance look at the stdc++ implementation here.
also interesting even though not an stl conforming bit vector is the llvm::BitVector from here.
the essence of the llvm::BitVector is a nested class called reference and suitable operator overloading to make the BitVector behaves similar to vector with some limitations. The code below is a simplified interface to show how BitVector hides a class called reference to make the real implementation almost behave like a real array of bool without using 1 byte for each value.
class BitVector {
public:
class reference {
reference &operator=(reference t);
reference& operator=(bool t);
operator bool() const;
};
reference operator[](unsigned Idx);
bool operator[](unsigned Idx) const;
};
this code here has the nice properties:
BitVector b(10, false); // size 10, default false
BitVector::reference &x = b[5]; // that's what really happens
bool y = b[5]; // implicitly converted to bool
assert(b[5] == false); // converted to bool
assert(b[6] == b[7]); // bool operator==(const reference &, const reference &);
b[5] = true; // assignment on reference
assert(b[5] == true); // and actually it does work.
This code actually has a flaw, try to run:
std::for_each(&b[5], &b[6], some_func); // address of reference not an iterator
will not work because assert( (&b[5] - &b[3]) == (5 - 3) ); will fail (within llvm::BitVector)
this is the very simple llvm version. std::vector<bool> has also working iterators in it.
thus the call for(auto i = b.begin(), e = b.end(); i != e; ++i) will work. and also std::vector<bool>::const_iterator.
However there are still limitations in std::vector<bool> that makes it behave differently in some cases.
This comes from http://www.cplusplus.com/reference/vector/vector-bool/
Vector of bool This is a specialized version of vector, which is used
for elements of type bool and optimizes for space.
It behaves like the unspecialized version of vector, with the
following changes:
The storage is not necessarily an array of bool values, but the library implementation may optimize storage so that each value is
stored in a single bit.
Elements are not constructed using the allocator object, but their value is directly set on the proper bit in the internal storage.
Member function flip and a new signature for member swap.
A special member type, reference, a class that accesses individual bits in the container's internal storage with an interface that
emulates a bool reference. Conversely, member type const_reference is
a plain bool.
The pointer and iterator types used by the container are not necessarily neither pointers nor conforming iterators, although they
shall simulate most of their expected behavior.
These changes provide a quirky interface to this specialization and
favor memory optimization over processing (which may or may not suit
your needs). In any case, it is not possible to instantiate the
unspecialized template of vector for bool directly. Workarounds to
avoid this range from using a different type (char, unsigned char) or
container (like deque) to use wrapper types or further specialize for
specific allocator types.
bitset is a class that provides a similar functionality for fixed-size
arrays of bits.
Item 18 of Scott Meyers's book Effective STL: 50 Specific Ways to Improve Your Use of the Standard Template Library says to avoid vector <bool> as it's not an STL container and it doesn't really hold bools.
The following code:
vector <bool> v;
bool *pb =&v[0];
will not compile, violating a requirement of STL containers.
Error:
cannot convert 'std::vector<bool>::reference* {aka std::_Bit_reference*}' to 'bool*' in initialization
vector<T>::operator [] return type is supposed to be T&, but why is it a special case for vector<bool>?
What does vector<bool> really consist of?
The Item further says:
deque<bool> v; // is a STL container and it really contains bools
Can this be used as an alternative to vector<bool>?
Can anyone please explain this?
For space-optimization reasons, the C++ standard (as far back as C++98) explicitly calls out vector<bool> as a special standard container where each bool uses only one bit of space rather than one byte as a normal bool would (implementing a kind of "dynamic bitset"). In exchange for this optimization it doesn't offer all the capabilities and interface of a normal standard container.
In this case, since you can't take the address of a bit within a byte, things such as operator[] can't return a bool& but instead return a proxy object that allows to manipulate the particular bit in question. Since this proxy object is not a bool&, you can't assign its address to a bool* like you could with the result of such an operator call on a "normal" container. In turn this means that bool *pb =&v[0]; isn't valid code.
On the other hand deque doesn't have any such specialization called out so each bool takes a byte and you can take the address of the value return from operator[].
Finally note that the MS standard library implementation is (arguably) suboptimal in that it uses a small chunk size for deques, which means that using deque as a substitute isn't always the right answer.
The problems is that vector<bool> returns a proxy reference object instead of a true reference, so that C++98 style code bool * p = &v[0]; won't compile. However, modern C++11 with auto p = &v[0]; can be made to compile if operator& also returns a proxy pointer object. Howard Hinnant has written a blog post detailing the algorithmic improvements when using such proxy references and pointers.
Scott Meyers has a long Item 30 in More Effective C++ about proxy classes. You can come a long way to almost mimic the builtin types: for any given type T, a pair of proxies (e.g. reference_proxy<T> and iterator_proxy<T>) can be made mutually consistent in the sense that reference_proxy<T>::operator&() and iterator_proxy<T>::operator*() are each other's inverse.
However, at some point one needs to map the proxy objects back to behave like T* or T&. For iterator proxies, one can overload operator->() and access the template T's interface without reimplementing all the functionality. However, for reference proxies, you would need to overload operator.(), and that is not allowed in current C++ (although Sebastian Redl presented such a proposal on BoostCon 2013). You can make a verbose work-around like a .get() member inside the reference proxy, or implement all of T's interface inside the reference (this is what is done for vector<bool>::bit_reference), but this will either lose the builtin syntax or introduce user-defined conversions that do not have builtin semantics for type conversions (you can have at most one user-defined conversion per argument).
TL;DR: no vector<bool> is not a container because the Standard requires a real reference, but it can be made to behave almost like a container, at least much closer with C++11 (auto) than in C++98.
vector<bool> contains boolean values in compressed form using only one bit for value (and not 8 how bool[] arrays do). It is not possible to return a reference to a bit in c++, so there is a special helper type, "bit reference", which provides you a interface to some bit in memory and allows you to use standard operators and casts.
Many consider the vector<bool> specialization to be a mistake.
In a paper "Deprecating Vestigial Library Parts in C++17"
There is a proposal to
Reconsider vector Partial Specialization.
There has been a long history of the bool partial specialization of
std::vector not satisfying the container requirements, and in
particular, its iterators not satisfying the requirements of a random
access iterator. A previous attempt to deprecate this container was
rejected for C++11, N2204.
One of the reasons for rejection is that it is not clear what it would
mean to deprecate a particular specialization of a template. That
could be addressed with careful wording. The larger issue is that the
(packed) specialization of vector offers an important
optimization that clients of the standard library genuinely seek, but
would not longer be available. It is unlikely that we would be able to
deprecate this part of the standard until a replacement facility is
proposed and accepted, such as N2050. Unfortunately, there are no such
revised proposals currently being offered to the Library Evolution
Working Group.
Look at how it is implemented. the STL builds vastly on templates and therefore the headers do contain the code they do.
for instance look at the stdc++ implementation here.
also interesting even though not an stl conforming bit vector is the llvm::BitVector from here.
the essence of the llvm::BitVector is a nested class called reference and suitable operator overloading to make the BitVector behaves similar to vector with some limitations. The code below is a simplified interface to show how BitVector hides a class called reference to make the real implementation almost behave like a real array of bool without using 1 byte for each value.
class BitVector {
public:
class reference {
reference &operator=(reference t);
reference& operator=(bool t);
operator bool() const;
};
reference operator[](unsigned Idx);
bool operator[](unsigned Idx) const;
};
this code here has the nice properties:
BitVector b(10, false); // size 10, default false
BitVector::reference &x = b[5]; // that's what really happens
bool y = b[5]; // implicitly converted to bool
assert(b[5] == false); // converted to bool
assert(b[6] == b[7]); // bool operator==(const reference &, const reference &);
b[5] = true; // assignment on reference
assert(b[5] == true); // and actually it does work.
This code actually has a flaw, try to run:
std::for_each(&b[5], &b[6], some_func); // address of reference not an iterator
will not work because assert( (&b[5] - &b[3]) == (5 - 3) ); will fail (within llvm::BitVector)
this is the very simple llvm version. std::vector<bool> has also working iterators in it.
thus the call for(auto i = b.begin(), e = b.end(); i != e; ++i) will work. and also std::vector<bool>::const_iterator.
However there are still limitations in std::vector<bool> that makes it behave differently in some cases.
This comes from http://www.cplusplus.com/reference/vector/vector-bool/
Vector of bool This is a specialized version of vector, which is used
for elements of type bool and optimizes for space.
It behaves like the unspecialized version of vector, with the
following changes:
The storage is not necessarily an array of bool values, but the library implementation may optimize storage so that each value is
stored in a single bit.
Elements are not constructed using the allocator object, but their value is directly set on the proper bit in the internal storage.
Member function flip and a new signature for member swap.
A special member type, reference, a class that accesses individual bits in the container's internal storage with an interface that
emulates a bool reference. Conversely, member type const_reference is
a plain bool.
The pointer and iterator types used by the container are not necessarily neither pointers nor conforming iterators, although they
shall simulate most of their expected behavior.
These changes provide a quirky interface to this specialization and
favor memory optimization over processing (which may or may not suit
your needs). In any case, it is not possible to instantiate the
unspecialized template of vector for bool directly. Workarounds to
avoid this range from using a different type (char, unsigned char) or
container (like deque) to use wrapper types or further specialize for
specific allocator types.
bitset is a class that provides a similar functionality for fixed-size
arrays of bits.
Item 18 of Scott Meyers's book Effective STL: 50 Specific Ways to Improve Your Use of the Standard Template Library says to avoid vector <bool> as it's not an STL container and it doesn't really hold bools.
The following code:
vector <bool> v;
bool *pb =&v[0];
will not compile, violating a requirement of STL containers.
Error:
cannot convert 'std::vector<bool>::reference* {aka std::_Bit_reference*}' to 'bool*' in initialization
vector<T>::operator [] return type is supposed to be T&, but why is it a special case for vector<bool>?
What does vector<bool> really consist of?
The Item further says:
deque<bool> v; // is a STL container and it really contains bools
Can this be used as an alternative to vector<bool>?
Can anyone please explain this?
For space-optimization reasons, the C++ standard (as far back as C++98) explicitly calls out vector<bool> as a special standard container where each bool uses only one bit of space rather than one byte as a normal bool would (implementing a kind of "dynamic bitset"). In exchange for this optimization it doesn't offer all the capabilities and interface of a normal standard container.
In this case, since you can't take the address of a bit within a byte, things such as operator[] can't return a bool& but instead return a proxy object that allows to manipulate the particular bit in question. Since this proxy object is not a bool&, you can't assign its address to a bool* like you could with the result of such an operator call on a "normal" container. In turn this means that bool *pb =&v[0]; isn't valid code.
On the other hand deque doesn't have any such specialization called out so each bool takes a byte and you can take the address of the value return from operator[].
Finally note that the MS standard library implementation is (arguably) suboptimal in that it uses a small chunk size for deques, which means that using deque as a substitute isn't always the right answer.
The problems is that vector<bool> returns a proxy reference object instead of a true reference, so that C++98 style code bool * p = &v[0]; won't compile. However, modern C++11 with auto p = &v[0]; can be made to compile if operator& also returns a proxy pointer object. Howard Hinnant has written a blog post detailing the algorithmic improvements when using such proxy references and pointers.
Scott Meyers has a long Item 30 in More Effective C++ about proxy classes. You can come a long way to almost mimic the builtin types: for any given type T, a pair of proxies (e.g. reference_proxy<T> and iterator_proxy<T>) can be made mutually consistent in the sense that reference_proxy<T>::operator&() and iterator_proxy<T>::operator*() are each other's inverse.
However, at some point one needs to map the proxy objects back to behave like T* or T&. For iterator proxies, one can overload operator->() and access the template T's interface without reimplementing all the functionality. However, for reference proxies, you would need to overload operator.(), and that is not allowed in current C++ (although Sebastian Redl presented such a proposal on BoostCon 2013). You can make a verbose work-around like a .get() member inside the reference proxy, or implement all of T's interface inside the reference (this is what is done for vector<bool>::bit_reference), but this will either lose the builtin syntax or introduce user-defined conversions that do not have builtin semantics for type conversions (you can have at most one user-defined conversion per argument).
TL;DR: no vector<bool> is not a container because the Standard requires a real reference, but it can be made to behave almost like a container, at least much closer with C++11 (auto) than in C++98.
vector<bool> contains boolean values in compressed form using only one bit for value (and not 8 how bool[] arrays do). It is not possible to return a reference to a bit in c++, so there is a special helper type, "bit reference", which provides you a interface to some bit in memory and allows you to use standard operators and casts.
Many consider the vector<bool> specialization to be a mistake.
In a paper "Deprecating Vestigial Library Parts in C++17"
There is a proposal to
Reconsider vector Partial Specialization.
There has been a long history of the bool partial specialization of
std::vector not satisfying the container requirements, and in
particular, its iterators not satisfying the requirements of a random
access iterator. A previous attempt to deprecate this container was
rejected for C++11, N2204.
One of the reasons for rejection is that it is not clear what it would
mean to deprecate a particular specialization of a template. That
could be addressed with careful wording. The larger issue is that the
(packed) specialization of vector offers an important
optimization that clients of the standard library genuinely seek, but
would not longer be available. It is unlikely that we would be able to
deprecate this part of the standard until a replacement facility is
proposed and accepted, such as N2050. Unfortunately, there are no such
revised proposals currently being offered to the Library Evolution
Working Group.
Look at how it is implemented. the STL builds vastly on templates and therefore the headers do contain the code they do.
for instance look at the stdc++ implementation here.
also interesting even though not an stl conforming bit vector is the llvm::BitVector from here.
the essence of the llvm::BitVector is a nested class called reference and suitable operator overloading to make the BitVector behaves similar to vector with some limitations. The code below is a simplified interface to show how BitVector hides a class called reference to make the real implementation almost behave like a real array of bool without using 1 byte for each value.
class BitVector {
public:
class reference {
reference &operator=(reference t);
reference& operator=(bool t);
operator bool() const;
};
reference operator[](unsigned Idx);
bool operator[](unsigned Idx) const;
};
this code here has the nice properties:
BitVector b(10, false); // size 10, default false
BitVector::reference &x = b[5]; // that's what really happens
bool y = b[5]; // implicitly converted to bool
assert(b[5] == false); // converted to bool
assert(b[6] == b[7]); // bool operator==(const reference &, const reference &);
b[5] = true; // assignment on reference
assert(b[5] == true); // and actually it does work.
This code actually has a flaw, try to run:
std::for_each(&b[5], &b[6], some_func); // address of reference not an iterator
will not work because assert( (&b[5] - &b[3]) == (5 - 3) ); will fail (within llvm::BitVector)
this is the very simple llvm version. std::vector<bool> has also working iterators in it.
thus the call for(auto i = b.begin(), e = b.end(); i != e; ++i) will work. and also std::vector<bool>::const_iterator.
However there are still limitations in std::vector<bool> that makes it behave differently in some cases.
This comes from http://www.cplusplus.com/reference/vector/vector-bool/
Vector of bool This is a specialized version of vector, which is used
for elements of type bool and optimizes for space.
It behaves like the unspecialized version of vector, with the
following changes:
The storage is not necessarily an array of bool values, but the library implementation may optimize storage so that each value is
stored in a single bit.
Elements are not constructed using the allocator object, but their value is directly set on the proper bit in the internal storage.
Member function flip and a new signature for member swap.
A special member type, reference, a class that accesses individual bits in the container's internal storage with an interface that
emulates a bool reference. Conversely, member type const_reference is
a plain bool.
The pointer and iterator types used by the container are not necessarily neither pointers nor conforming iterators, although they
shall simulate most of their expected behavior.
These changes provide a quirky interface to this specialization and
favor memory optimization over processing (which may or may not suit
your needs). In any case, it is not possible to instantiate the
unspecialized template of vector for bool directly. Workarounds to
avoid this range from using a different type (char, unsigned char) or
container (like deque) to use wrapper types or further specialize for
specific allocator types.
bitset is a class that provides a similar functionality for fixed-size
arrays of bits.
C++20 introduces std::common_reference. What is its purpose? Can someone give an example of using it?
common_reference came out of my efforts to come up with a conceptualization of STL's iterators that accommodates proxy iterators.
In the STL, iterators have two associated types of particular interest: reference and value_type. The former is the return type of the iterator's operator*, and the value_type is the (non-const, non-reference) type of the elements of the sequence.
Generic algorithms often have a need to do things like this:
value_type tmp = *it;
... so we know that there must be some relationship between these two types. For non-proxy iterators the relationship is simple: reference is always value_type, optionally const and reference qualified. Early attempts at defining the InputIterator concept required that the expression *it was convertible to const value_type &, and for most interesting iterators that is sufficient.
I wanted iterators in C++20 to be more powerful than this. For example, consider the needs of a zip_iterator that iterates two sequences in lock-step. When you dereference a zip_iterator, you get a temporary pair of the two iterators' reference types. So, zip'ing a vector<int> and a vector<double> would have these associated types:
zip iterator's reference : pair<int &, double &>
zip iterator's value_type: pair<int, double>
As you can see, these two types are not related to each other simply by adding top-level cv- and ref qualification. And yet letting the two types be arbitrarily different feels wrong. Clearly there is some relationship here. But what is the relationship, and what can generic algorithms that operate on iterators safely assume about the two types?
The answer in C++20 is that for any valid iterator type, proxy or not, the types reference && and value_type & share a common reference. In other words, for some iterator it there is some type CR which makes the following well-formed:
void foo(CR) // CR is the common reference for iterator I
{}
void algo( I it, iter_value_t<I> val )
{
foo(val); // OK, lvalue to value_type convertible to CR
foo(*it); // OK, reference convertible to CR
}
CR is the common reference. All algorithms can rely on the fact that this type exists, and can use std::common_reference to compute it.
So, that is the role that common_reference plays in the STL in C++20. Generally, unless you are writing generic algorithms or proxy iterators, you can safely ignore it. It's there under the covers ensuring that your iterators are meeting their contractual obligations.
EDIT: The OP also asked for an example. This is a little contrived, but imagine it's C++20 and you are given a random-access range r of type R about which you know nothing, and you want to sort the range.
Further imagine that for some reason, you want to use a monomorphic comparison function, like std::less<T>. (Maybe you've type-erased the range, and you need to also type-erase the comparison function and pass it through a virtual? Again, a stretch.) What should T be in std::less<T>? For that you would use common_reference, or the helper iter_common_reference_t which is implemented in terms of it.
using CR = std::iter_common_reference_t<std::ranges::iterator_t<R>>;
std::ranges::sort(r, std::less<CR>{});
That is guaranteed to work, even if range r has proxy iterators.
cppreference says that the iterators for the vector<bool> specialization are implementation defined and many not support traits like ForwardIterator (and therefore RandomAccessIterator).
cplusplus adds a mysterious "most":
The pointer and iterator types used by the container are not
necessarily neither pointers nor conforming iterators, although they
shall simulate most of their expected behavior.
I don't have access to the official specification. Are there any iterator behaviors guaranteed for the vector<bool> iterators?
More concretely, how would one write standards-compliant code to insert an item in the middle of a vector<bool>? The following works on several compilers that I tried:
std::vector<bool> v(4);
int k = 2;
v.insert(v.begin() + k, true);
Will it always?
The fundamental problem with vector<bool>'s iterators is that they are not ForwardIterators. C++14 [forward.iterators]/1 requires that ForwardIterators' reference type be T& or const T&, as appropriate.
Any function which takes a forward iterator over a range of Ts is allowed to do this:
T &t = *it;
t = //Some value.
However, vector<bool>'s reference types are not bool&; they're a proxy object that is convertible to and assignable from a bool. They act like a bool, but they are not a bool. As such, this code is illegal:
bool &b = *it;
It would be attempting to get an lvalue reference to a temporary created from the proxy object. That's not allowed.
Therefore, you cannot use vector<bool>'s iterators in any function that takes ForwardIterators or higher.
However, your code doesn't necessarily have to care about that. As long as you control what code you pass those vector<bool> iterators to, and you don't do anything that violates how they behave, then you're fine.
As far as their interface is concerned, they act like RandomAccessIterators, except for when they don't (see above). So you can offset them with integers with constant time complexity and so forth.
vector<bool> is fine, so long as you don't treat it like a vector that contains bools. Your code will work because it uses vector<bool>'s own interface, which it obviously accepts.
It would not work if you passed a pair of vector<bool> iterators to std::sort.
C++14 [vector.bool]/2:
Unless described below, all operations have the same requirements and
semantics as the primary vector template, except that operations
dealing with the bool value type map to bit values in the container
storage and allocator_traits::construct (20.7.8.2) is not used to
construct these values.