在比较中使用具有隐式转换的用户定义转换

Question

我很难理解为什么以下代码不允许隐式转换发生.

#include <string>
using namespace std;

struct HasConversionToString {
  HasConversionToString(const string& s_) : s{s_} {}
  string s;
  operator const string&() const { return s; }
};

int main() {
  string s{"a"};
  HasConversionToString obj{"b"};
  return s < obj;
}

clang和gcc都找不到一种有效的方法来比较两个对象的错误:

clang++ -std=c++14 -Wall -Wextra -pedantic conversion.cpp -o test
conversion.cpp:13:12: error: invalid operands to binary expression ('string' (aka 'basic_string<char>') and 'HasConversionToString')
  return s < obj;
         ~ ^ ~~~
/usr/bin/../lib/gcc/x86_64-linux-gnu/5.3.0/../../../../include/c++/5.3.0/bits/stl_pair.h:220:5: note: candidate template ignored: could not match
      'pair' against 'basic_string'
    operator<(const pair<_T1, _T2>& __x, const pair<_T1, _T2>& __y)
    ^
/usr/bin/../lib/gcc/x86_64-linux-gnu/5.3.0/../../../../include/c++/5.3.0/bits/stl_iterator.h:298:5: note: candidate template ignored: could not match
      'reverse_iterator' against 'basic_string'
    operator<(const reverse_iterator<_Iterator>& __x,
    ^
/usr/bin/../lib/gcc/x86_64-linux-gnu/5.3.0/../../../../include/c++/5.3.0/bits/stl_iterator.h:348:5: note: candidate template ignored: could not match
      'reverse_iterator' against 'basic_string'
    operator<(const reverse_iterator<_IteratorL>& __x,
    ^
/usr/bin/../lib/gcc/x86_64-linux-gnu/5.3.0/../../../../include/c++/5.3.0/bits/stl_iterator.h:849:5: note: candidate template ignored: could not match
      '__normal_iterator' against 'basic_string'
    operator<(const __normal_iterator<_IteratorL, _Container>& __lhs,
    ^
/usr/bin/../lib/gcc/x86_64-linux-gnu/5.3.0/../../../../include/c++/5.3.0/bits/stl_iterator.h:856:5: note: candidate template ignored: could not match
      '__normal_iterator' against 'basic_string'
    operator<(const __normal_iterator<_Iterator, _Container>& __lhs,
    ^
/usr/bin/../lib/gcc/x86_64-linux-gnu/5.3.0/../../../../include/c++/5.3.0/bits/stl_iterator.h:1089:5: note: candidate template ignored: could not match
      'move_iterator' against 'basic_string'
    operator<(const move_iterator<_IteratorL>& __x,
    ^
/usr/bin/../lib/gcc/x86_64-linux-gnu/5.3.0/../../../../include/c++/5.3.0/bits/stl_iterator.h:1095:5: note: candidate template ignored: could not match
      'move_iterator' against 'basic_string'
    operator<(const move_iterator<_Iterator>& __x,
    ^
/usr/bin/../lib/gcc/x86_64-linux-gnu/5.3.0/../../../../include/c++/5.3.0/bits/basic_string.h:4989:5: note: candidate template ignored: could not match
      'basic_string<type-parameter-0-0, type-parameter-0-1, type-parameter-0-2>' against 'HasConversionToString'
    operator<(const basic_string<_CharT, _Traits, _Alloc>& __lhs,
    ^
/usr/bin/../lib/gcc/x86_64-linux-gnu/5.3.0/../../../../include/c++/5.3.0/bits/basic_string.h:5001:5: note: candidate template ignored: could not match
      'const _CharT *' against 'HasConversionToString'
    operator<(const basic_string<_CharT, _Traits, _Alloc>& __lhs,
    ^
/usr/bin/../lib/gcc/x86_64-linux-gnu/5.3.0/../../../../include/c++/5.3.0/bits/basic_string.h:5013:5: note: candidate template ignored: could not match
      'const _CharT *' against 'string' (aka 'basic_string<char>')
    operator<(const _CharT* __lhs,
    ^
1 error generated.

当我明确地将对象转换为字符串时,以下代码工作正常.

#include <string>
using namespace std;

struct HasConversionToString {
  HasConversionToString(const string& s_) : s{s_} {}
  string s;
  operator const string&() const { return s; }
};

int main() {
  string s{"a"};
  HasConversionToString obj{"b"};
  return s < static_cast<string>(obj);
}

基于cppreference上列出的隐式强制转换的规则和示例,我认为没有理由这不起作用.我认为clang和gcc都没有搞砸同样的事情,所以我想我有一个概念上的误解.

Answer 1

您要调用的是函数模板:

template<class charT, class Traits, class Alloc>
bool operator<(std::basic_string<charT, Traits, Alloc> const& lhs,
               std::basic_string<charT, Traits, Alloc> const& rhs);

第二个参数的扣除失败,因为a HasConversionToString不是std::basic_string- 模板参数推导不会通过隐式转换查看.因此,该功能模板将从重载决策中删除.

std::experimental::basic_string_view有一个类似的问题,这是通过"足够的额外重载"规则解决的(库必须添加足够的重载,以便比较a basic_string_view和可转换为一个的工作).

basic_string尽管如此,你真的不想要这样的事情- <可能会默默地导致去堆的旅行并不是一个好主意.