CODEWITHSUNDEEP
c++c++11stdconst-correctnessstdset
atomicpirate
atomicpirate

Reputation: 695

Why does std::set seem to force the use of a const_iterator?

Consider the simple program below, which attempts to iterate through the values of a set using NON-const references to the elements in it:

#include <set>
#include <iostream>

class Int
{
public:
   Int(int value) : value_(value) {}
   int value() const { return value_; }
   bool operator<(const Int& other) const { return value_ < other.value(); }
private:
   int value_;
};

int
main(int argc, char** argv) {
   std::set<Int> ints;
   ints.insert(10);
   for (Int& i : ints) {
      std::cout << i.value() << std::endl;
   }
   return 0;
}

When compiling this, I get an error from gcc:

test.c: In function ‘int main(int, char**)’:
test.c:18:18: error: invalid initialization of reference of type ‘Int&’ from expression of type ‘const Int’  
for (Int& i : ints) {  
              ^

Yes, I know I'm not actually trying to modify the elements in the for loop. But the point is that I should be able to get a non-const reference to use inside the loop, since the set itself is not const qualified. I get the same error if I create a setter function and use that in the loop.

Upvotes: 47

Views: 9259

Answers (5)

manlio
manlio

Reputation: 18902

Adding on nate's answer:

A set is like a map with no values, only keys. Since those keys are used for a tree that accelerates operations on the set, they cannot change. Thus all elements must be const to keep the constraints of the underlying tree from being broken.

With C++17 there is the new extract member function, so an alternative to const_cast could be:

#include <iostream>
#include <string_view>
#include <set>

struct S
{
  int used_for_sorting;
  bool not_used_for_sorting;

  bool operator<(const S &rhs) const
  { return used_for_sorting < rhs.used_for_sorting; }
};

void print(std::string_view comment, const std::set<S> &data)
{
  std::cout << comment;
  for (auto datum : data)
    std::cout << " {" << datum.used_for_sorting
              << ',' << datum.not_used_for_sorting
              << '}';

  std::cout << '\n';
}

int main()
{
  std::set<S> cont = {{1, false}, {2, true}, {3, false}};

  print("Start:", cont);

  // Extract node handle and change key
  auto nh = cont.extract({1, false});
  nh.value().not_used_for_sorting = true;

  print("After extract and before insert:", cont);

  // Insert node handle back
  cont.insert(std::move(nh));

  print("End:", cont);
}

Probably useful as hot-fix. In general it's hard to see any advantage over a std::map.

Upvotes: 0

Ohad Eytan
Ohad Eytan

Reputation: 8464

From the cpp reference:

In a set, the value of an element also identifies it (the value is itself the key, of type T), and each value must be unique. The value of the elements in a set cannot be modified once in the container (the elements are always const), but they can be inserted or removed from the container.

Upvotes: 12

lisyarus
lisyarus

Reputation: 15522

std::set uses the contained values to form a fast data structure (usually, a red-black tree). Changing a value means the whole structure needs to be altered. So, forcing constness, std::set prevents you from pushing it into a non-usable state.

Upvotes: 13

Fitzwilliam Bennet-Darcy
Fitzwilliam Bennet-Darcy

Reputation: 1342

The behaviour is by design.

Giving you a non-const iterator could inspire you to change the element in the set; the subsequent iterating behaviour would then be undefined.

Note that the C++ standard says that set<T>::iterator is const so the old-fashioned pre C++11 way still wouldn't work.

Upvotes: 6

nate
nate

Reputation: 1871

A set is like a map with no values, only keys. Since those keys are used for a tree that accelerates operations on the set, they cannot change. Thus all elements must be const to keep the constraints of the underlying tree from being broken.

Upvotes: 59

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