C ++ 14标准对auto作为参数类型的说法是什么

Question

Lets take a look at the following code below:

#include <iostream>

class Test
{
private:
    int x;
public:
    Test(int _x) :
        x(_x)
    {
        std::cout << "Im being constructed" << std::endl;
    }

    int getx()
    {
        return this->x;
    }

    friend std::ostream& operator<<(std::ostream& os, Test& other)
    {
        os << other.getx();
        return os;
    }
};

auto func(auto x)
{
    std::cout << x << std::endl;
    return x.getx();
}

int main()
{
    auto y = func(20);

    return 0;
}

How does the compiler decide if (20) should be an int or Test object? The constructor for Test is not explicit so what does the standard say about it?

Answer 1

So although gcc allows auto as a parameter in functions this is not part of C++14 but part of concepts-lite which may become part of C++1z according to Herb Sutter's last trip report.

I believe gcc allows this as an extension and if we compile your program using -pedantic gcc will warn:

 warning: ISO C++ forbids use of 'auto' in parameter declaration [-Wpedantic]
 auto func(auto x)
           ^

So from concept-lite proposal we can see that:

auto func(auto x)
{
    std::cout << x << std::endl;
    return x.getx();
}

is equivalent to:

template <typename T1>
auto func(T1 x)
{
    std::cout << x << std::endl;
    return x.getx();
}

and therefore x will be deduced as int . The compiler will rightfully give you an error like the following one from gcc ( see it live ):

error: request for member 'getx' in 'x', which is of non-class type 'int'
     return x.getx();
                   ^

This is covered in the proposal section 5.1.1 [dcl.fct] :

The use of auto or a concept-name in the parameter-declaration-clause shall be interpreted as the use of a type-parameter having the same constraints and the named concept. [ Note: The exact mechanism for achieving this is unspecified. —end note ] [ Example: The generic function declared below

 auto f(auto x, const Regular& y); 

Is equivalent to the following declaration

 template<typename T1, Regular T2> auto f(T1 x, const T2&); 

—end example ]

Answer 2

The rules for auto as argument type are the same as for template arguments.

auto func(auto x) is equivalent to template<typename T> auto func(T x)

I am no lawyer to quote the standard on the specific rules, though.