使用命名空间时,C模板函数无法使用g进行编译

#include <vector>

template <class T>
void foo(const int & from,std::vector<T> & to)
{
    for (int i = 0; i < 5; i++)
    {
        T bar;
        foo(from,bar);
        to.push_back(bar);
    }
}

struct Bar
{
    int a;
    int b;
};

struct Baz
{
    std::vector<Bar> bars;
};


void foo(const int & from,Bar & to)
{
    to.a = from;
    to.b = from - 1;
}

void foo(const int & from,Baz & to)
{
    foo(from,to.bars);
}

void fooTest()
{
    int num = 10;
    Baz baz;
    foo(num,baz);
}

int main()
{
    fooTest();
}

但是当我为Bar和Baz引入命名空间时,它无法编译. (带命名空间)

#include <vector>

template <class T>
void foo(const int & from,bar);
        to.push_back(bar);
    }
}

// When I add this namespace,it fails to compile
namespace BarBar
{
    struct Bar
    {
        int a;
        int b;
    };

    struct Baz
    {
        std::vector<Bar> bars;
    };
}


void foo(const int & from,BarBar::Bar & to)
{
    to.a = from;
    to.b = from - 1;
}

void foo(const int & from,BarBar::Baz & to)
{
    foo(from,to.bars);
}

void fooTest()
{
    int num = 10;
    BarBar::Baz baz;
    foo(num,baz);
}

int main()
{
    fooTest();
}

它显示错误：

with_namespace.cpp: In instantiation of ‘void foo(const int&,std::vector<T>&) [with T = BarBar::Bar]’:
with_namespace.cpp:37:22:   required from here
with_namespace.cpp:9:12: error: no matching function for call to ‘foo(const int&,BarBar::Bar&)’
         foo(from,bar);
            ^
with_namespace.cpp:4:6: note: candidate: template<class T> void foo(const int&,std::vector<T>&)
 void foo(const int & from,std::vector<T> & to)
      ^
with_namespace.cpp:4:6: note:   template argument deduction/substitution failed:
with_namespace.cpp:9:12: note:   ‘BarBar::Bar’ is not derived from ‘std::vector<T>’
         foo(from,bar);
        ^

另请注意,使用MSVC时,带命名空间的代码编译得很好.为什么编译器在使用命名空间时找不到定义？

我使用以下版本：g(Ubuntu 5.4.0-6ubuntu1~16.04.9)5.4.0 20160609

更新：
在@ M.M指出函数查找如何为模板和ADL工作之后,我进行了以下修复：

#include <vector>



template <class T>
void foo(const int & from,bar);
        to.push_back(bar);
    }
}

namespace BarBar
{
    struct Bar
    {
        int a;
        int b;
    };

    struct Baz
    {
        std::vector<Bar> bars;
    };
};


// Put them in the same namespace as Bar so that the templated foo find this function
namespace BarBar
{
    using ::foo; // We are going to use templated foo in the latter functions

    void foo(const int & from,BarBar::Bar & to)
    {
        to.a = from;
        to.b = from - 1;
    }

    void foo(const int & from,BarBar::Baz & to)
    {
        foo(from,to.bars);
    }

}

void fooTest()
{
    int num = 10;
    BarBar::Baz baz;
    BarBar::foo(num,baz);
}



int main()
{
    fooTest();
}