使用 g++ 4.8 缺少 std::vector::erase() 的 const_iterator 重载

2022-01-23 00:00:00 g++ c++ c++11 stl

以下示例将无法使用 g++ 4.8.2 进行编译:

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;
}

编译器会这样说.(它的可读性不是很好，但它抱怨 vector<int>::const_iterator 和 vector<int>::iterator 之间没有已知的转换.)

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

为什么? C++11 标准在 §23.3.6.5 中明确指出 vector::erase 函数 采用 const_iterator.(释义是 here 和 这里.)

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.)

假设我必须使用 const_iterator，有什么好的解决方法?

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

推荐答案

这是 gcc 中的一个已知错误:http://gcc.gnu.org/bugzilla/show_bug.cgi?id=57158

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

erase 需要一个迭代器，而不是当前 gcc 的 const_iterator.

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

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