c ++模板函数用const参数重叠参数推导(c++ template function overlading argument deduction with const argument)

 我有以下设置： 
 
template <typename T>
void foo(T& t);
void foo(const int& t);

void f()
{
    int i;
    foo(i); //Unresolved reference to "void foo<int>(int &)"
    foo(const_cast<const int&>(i)); //Unresolved reference to "void foo(int const &)"
}
 
 在第一次调用foo时，编译器尝试调用模板版本，因为非模板版本的参数与i的类型不匹配。 在第二次调用中，调用非模板版本。 我使用的是Microsoft C ++编译器版本10.这是标准行为吗？ 如果类型不完全匹配，即使它只有一个const修饰符，那么调用模板函数？ 
 
 编辑：我知道这两个函数没有定义，我只是指出链接器抱怨的内容，以便更清楚编译器想要调用的内容。 

I am having the following setup:
 
template <typename T>
void foo(T& t);
void foo(const int& t);

void f()
{
    int i;
    foo(i); //Unresolved reference to "void foo<int>(int &)"
    foo(const_cast<const int&>(i)); //Unresolved reference to "void foo(int const &)"
}
 
In the first call to foo, the compiler tries to call the template version, since the argument of the non-template one does not match the type of i. In the second call the non-template version is called. I am using the Microsoft C++ compiler version 10. Is this standard behavior? If the type is not exactly matched, even if it only has a const modifier, then the template function is called?
 
EDIT: I know those two functions don't have definition, I am just pointing out what the linker complains about, to make it more clear what the compiler wants to call.

原文：https://stackoverflow.com/questions/15601057