模板分辨率-升压测试找不到运算符<<

时间:2018-11-29 00:39:31

标签: c++ templates boost-test

以下代码可以正常工作:

#define BOOST_TEST_MODULE TestFoo
#include <boost/test/unit_test.hpp>
#include <boost/dynamic_bitset.hpp>
#include <string>

template <typename T>
std::ostream& operator<<(std::ostream& os, const std::vector<T> &v)
{
    os << "[ ";
    for ( const T& elem : v )
        os << elem << ' ';
    return os << ']';
}

typedef boost::dynamic_bitset<> BS;
static const std::vector<BS> foo = { BS(std::string("101")) };

BOOST_AUTO_TEST_CASE( test_foo )
{
    BOOST_CHECK_EQUAL( foo[0], foo[0] );
}

但是,当我用替换测试用例时

BOOST_AUTO_TEST_CASE( test_foo )
{
    BOOST_CHECK_EQUAL( foo, foo );
}

然后编译器不再找到operator<<

/usr/include/boost/test/test_tools.hpp:326:14: error: no match for ‘operator<<’ (operand types are ‘std::ostream {aka std::basic_ostream<char>}’ and ‘const std::vector<boost::dynamic_bitset<> >’)

我希望编译器实例化上面定义的operator<<模板。为什么这没有发生/如何解决?

1 个答案:

答案 0 :(得分:1)

编辑:查看评论,这是UB-似乎没有解决该问题的“好方法”。

将您的op<<包裹在namespace std {...}

#include <boost/dynamic_bitset.hpp>
#include <boost/test/unit_test.hpp>

namespace std { // THIS LINE

template <typename T, typename... Rest>
std::ostream& operator<<(std::ostream& os, const std::vector<T, Rest...> &v)
{
    os << "[ ";
    for ( const T& elem : v )
        os << elem << ' ';
    os << ']';
    return os;
}
} // THIS LINE

typedef boost::dynamic_bitset<> BS;
static const std::vector<BS> foo = { BS(std::string("101")) };


BOOST_AUTO_TEST_CASE( test_foo )
{
    BOOST_CHECK_EQUAL( foo, foo );
}

https://godbolt.org/z/xoW-IJ

否则,它没有在正确的命名空间中查找实现。可以确定这是ADL:https://en.cppreference.com/w/cpp/language/adl