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Home/ Questions/Q 1086891
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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T22:52:08+00:00 2026-05-16T22:52:08+00:00

Can someone help me out here? Compiling this code: void test() { std::set<int> test;

  • 0

Can someone help me out here?

Compiling this code:

void test()
{
  std::set<int> test;
  test.insert(42);
  test.erase(std::remove(test.begin(), test.end(), 30), test.end());  // <- Line 33
}

Is generating the following error when compiling:

$ make
g++ -c -Wall -pedantic-errors -Wextra -Wunused -Werror a_star.cpp
/usr/lib/gcc/i686-pc-cygwin/4.3.4/include/c++/bits/stl_algo.h: In function `_FIter std::remove(_FIter, _FIter, const _Tp&) [with _FIter = std::_Rb_tree_const_iterator<int>, _Tp = int]':
a_star.cpp:33:   instantiated from here
/usr/lib/gcc/i686-pc-cygwin/4.3.4/include/c++/bits/stl_algo.h:779: error: assignment of read-only location `__result.std::_Rb_tree_const_iterator<_Tp>::operator* [with _Tp = int]()'
make: *** [a_star.o] Error 1
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-16T22:52:09+00:00Added an answer on May 16, 2026 at 10:52 pm

    In std::set, the elements are not modifiable. So, the std::set::iterator is also unmodifiable. From this tutorial, section 27.3.2.1:

    In simple associative containers,
    where the elements are the keys, the
    elements are completely immutable; the
    nested types iterator and
    const_iterator are therefore the same.

    Hence, the erase-remove idiom cannot be applied as is. You have to write a for loop, and use the member function std::set::erase inside it. See this question and this accepted answer and another answer for exact details, but in short, the loop is like the following

    typename std::set::iterator set_iter; 
    
    for( set_iter it = s.begin(); it != s.end(); /* blank */ ) {
        if( some_condition() ) {
            s.erase( it++ );       // Note the subtlety here
        }
        else {
            ++it;
        }
    }
    
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