按值和按引用差异的函数模板

Question

我正在关注本教程 - http://www.learncpp.com/cpp-tutorial/132-function-template-instances/

// passing all parameters by references
template <typename T1, typename T2>
const T2& add_two_objects(const T1& x,const T2& y) {
        return x+y;
};

int main() {
    using std::cout;
    int x(0),y(0);
    std::cout << "Please enter number 1" << std::endl;
    std::cin >> x;
    std::cout << "Please enter number 2" << std::endl;
    std::cin >> y;
    cout<< "sum of two integers is " << add_two_objects(x,y+1.2)    << '\n';
    cout<< "sum of two double is " << add_two_objects(x,y+2.52424324) << '\n';
    return 0;
}

程序编译得很好，但在运行时，我总是遇到分段错误。但是，如果我将模板更改为按值传递，那么一切正常。

// passing all parameters by value
template <typename T1, typename T2>
const T2 add_two_objects(const T1 x,const T2 y) {
        return x+y;
};

谁能解释一下？

Answer 1

这个

template <typename T1, typename T2>
const T2& add_two_objects(const T1& x, const T2& y) {
        return x+y;
};

返回对临时对象的引用。将返回值设为值

template <typename T1, typename T2>
T2 add_two_objects(const T1& x, const T2& y) {
        return x+y;
};

你应该没事的。

顺便说一句，目前尚不清楚返回T2 是不是最好的做法。例如，考虑T1=size_t 和T2=char 的情况。最好返回操作x+y实际生成的类型

template <typename T1, typename T2>
auto add_two_objects(const T1& x, const T2& y)
-> decltype(x+y)
{
    return x+y;
};

-----编辑-----

您不得返回对临时对象的引用。这是一个BUG。如果你想要一个错误，那就去吧。你不应该使用糟糕的教程。在某些情况下您想要/应该返回参考，但这不是其中之一。这些是指对从函数返回时不会被销毁（或超出范围）的对象的引用（就像所有自动和临时变量一样）。

Answer 2

为了更清楚，让我们将整数包装在结构中。

这是一个演示程序

#include <iostream>

struct A
{
    A( int x ) : x( x ) {}
    ~A() { std::cout << "[A::~A() is called for x = " << x << ']' << std::endl; }
    int x;
};

A operator +( const A &lhs, const A &rhs )
{
    return A( lhs.x + rhs.x );
}

std::ostream & operator <<( std::ostream &os, const A &a )
{
    return os << a.x;
}

template <typename T1, typename T2>
const T2& add_two_objects(const T1& x,const T2& y) {
        return x+y;
};

int main() 
{
    std::cout<< "sum of two integers is " << add_two_objects( A( 1 ), A( 2 ) ) << '\n';

    return 0;
}

它的输出可能看起来像

prog.cc:22:18: warning: returning reference to temporary [-Wreturn-local-addr]
         return x+y;
                  ^
sum of two integers is [A::~A() is called for x = 3]
Segmentation fault

首先编译器警告函数返回一个临时值的引用。即退出函数后临时对象将被销毁并输出

[A::~A() is called for x = 3]

证实了这一点。

因此引用将无效并且程序具有未定义的行为。

其实你可以这样想象程序逻辑

int main() 
{
    const A &r = add_two_objects( A( 1 ), A( 2 ) );
    std::cout<< "sum of two integers is " << r << '\n';

    return 0;
}

它的输出看起来和上面的程序差不多

prog.cc:22:18: warning: returning reference to temporary [-Wreturn-local-addr]
         return x+y;
                  ^
[A::~A() is called for x = 3]
[A::~A() is called for x = 1]
[A::~A() is called for x = 2]
Segmentation fault

即引用失效。

如果要删除函数声明中的引用

template <typename T1, typename T2>
const T2/*&*/ add_two_objects(const T1& x,const T2& y) {
        return x+y;
};

那么程序输出可能看起来像

sum of two integers is 3
[A::~A() is called for x = 3]
[A::~A() is called for x = 1]
[A::~A() is called for x = 2]