函數模板，並通過引用的區別

Question

我下面這個教程 - 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; 
} 

程序編譯罰款，但在運行時，我總是得到一個分段錯誤。但是，如果我將模板更改爲按值傳遞，那麼一切正常。

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

任何人都可以請解釋嗎？

Answer 1

這

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

返回到一個臨時的參考。讓返回值

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

你應該沒問題。

順便說一下，不清楚的是，返回T2是在這裏做的最好的事情。考慮例如T1=size_t和T2=char的情況。不如歸去的操作x+y實際上產生

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

------ -----編輯

您不得引用返回一個臨時對象類型。這是一個錯誤。如果你想要一個bug，那就去做吧。你不應該使用可憐的教程。有時你想/應該返回一個引用，但這不是其中之一。這些參考是指從函數返回時不會被銷燬（或超出範圍）的對象（因爲所有自動和臨時變量都會）。

Answer 2

爲了讓它更清晰讓我們將整數包裹在結構中。

這裏是一個示範項目

#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; 
} 

它的輸出可能看起來像

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 

所有的編譯器首先警告說，該函數返回引用臨時值。即在退出功能後，臨時對象將被破壞，並且此輸出

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

證實了這一點。

結果引用將是無效的，程序有未定義的行爲。

其實你能想象的程序邏輯如下方式

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

    return 0; 
} 

它的輸出看起來在上面的程序

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 

這是參考變爲無效幾乎相同。

如果刪除函數聲明

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

參考然後程序輸出可能看起來像

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]