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Missing const_iterator overload of std::vector::erase() with g++ 4.8

Tags:

c++

c++11

g++

stl

The following example will not compile using g++ 4.8.2:

#include <iostream>
#include <vector>
using namespace std;

int main() {
    vector<int> v {1, 2, 3};

    v.erase(v.cbegin()); // Compiler complains

    return 0;
}

The compiler says the following. (It isn't very readable, but it's complaining that there's not a known conversion between vector<int>::const_iterator and vector<int>::iterator.)

prog.cpp: In function ‘int main()’:
prog.cpp:8:20: error: no matching function for call to ‘std::vector<int>::erase(std::vector<int>::const_iterator)’
  v.erase(v.cbegin());
                    ^
prog.cpp:8:20: note: candidates are:
In file included from /usr/include/c++/4.8/vector:69:0,
                 from prog.cpp:2:
/usr/include/c++/4.8/bits/vector.tcc:134:5: note: std::vector<_Tp, _Alloc>::iterator std::vector<_Tp, _Alloc>::erase(std::vector<_Tp, _Alloc>::iterator) [with _Tp = int; _Alloc = std::allocator<int>; std::vector<_Tp, _Alloc>::iterator = __gnu_cxx::__normal_iterator<int*, std::vector<int> >; typename std::_Vector_base<_Tp, _Alloc>::pointer = int*]
     vector<_Tp, _Alloc>::
     ^
/usr/include/c++/4.8/bits/vector.tcc:134:5: note:   no known conversion for argument 1 from ‘std::vector<int>::const_iterator {aka __gnu_cxx::__normal_iterator<const int*, std::vector<int> >}’ to ‘std::vector<int>::iterator {aka __gnu_cxx::__normal_iterator<int*, std::vector<int> >}’
/usr/include/c++/4.8/bits/vector.tcc:146:5: note: std::vector<_Tp, _Alloc>::iterator std::vector<_Tp, _Alloc>::erase(std::vector<_Tp, _Alloc>::iterator, std::vector<_Tp, _Alloc>::iterator) [with _Tp = int; _Alloc = std::allocator<int>; std::vector<_Tp, _Alloc>::iterator = __gnu_cxx::__normal_iterator<int*, std::vector<int> >; typename std::_Vector_base<_Tp, _Alloc>::pointer = int*]
     vector<_Tp, _Alloc>::
     ^
/usr/include/c++/4.8/bits/vector.tcc:146:5: note:   candidate expects 2 arguments, 1 provided

Why? The C++11 standard plainly states, in §23.3.6.5, that the vector::erase function takes a const_iterator. (Paraphrases are here and here.)

What's a good workaround, assuming that I must use a const_iterator?

like image 498
George Hilliard Avatar asked Oct 24 '13 07:10

George Hilliard


2 Answers

This is a known bug in gcc: http://gcc.gnu.org/bugzilla/show_bug.cgi?id=57158

erase requires an iterator instead of a const_iterator with current gcc's.

like image 187
mirk Avatar answered Nov 05 '22 12:11

mirk


You can get a non-const iterator via pointer arithmetic, here's a helper function to do that:

template<typename T>
  typename std::vector<T>::iterator
  const_iterator_cast(std::vector<T>& v, typename std::vector<T>::const_iterator iter)
  {
    return v.begin() + (iter - v.cbegin());
  }

used like so:

std::vector<T> v(1);
auto citer = v.cbegin();
v.erase( const_iterator_cast(v, citer) );
like image 20
Jonathan Wakely Avatar answered Nov 05 '22 13:11

Jonathan Wakely