可变参数模板示例 [英] Variadic Templates example

问题描述

考虑以下代码，我不明白为什么必须定义空白打印功能。

Consider following code, I don't understand why empty function of print must be defined.

#include <iostream>
using namespace std;

void print()
{   
}   

    template <typename T, typename... Types>
void print (const T& firstArg, const Types&... args)
{   
    cout << firstArg << endl; // print first argument
    print(args...); // call print() for remaining arguments
}

int main()
{   
    int i=1;
    int  j=2;
    char c = 'T';
    print(i,"hello",j,i,c,"word");

}   

推荐答案

正确的方法：

CORRECT WAY:

可变参数模板与归纳严格相关

Variadic templates is strictly related to induction, a mathematical concept.

编译器解析以下函数调用

The compiler resolves the following function call

print('a', 3, 4.0f);

进入

std::cout<< 'a' <<std::endl;
print(3, 4.0f);

已解析为

std::cout<< 'a' <<std::endl;
std::cout<< 3 <<std::endl;
print( 4.0f);

已解析为

std::cout<< 'a' <<std::endl;
std::cout<< 3 <<std::endl;
std::cout<< 4.0f <<std::endl;
print();

此时，它会搜索匹配为空函数的函数重载。

At this point it searches for a function overload whose match is the empty function.

• 所有具有1个或多个参数的函数都与可变参数模板匹配


• 所有没有参数的函数都将匹配到空函数

罪魁祸首是，对于每种可能的参数组合，您只能拥有一个函数。

The culprit is that you must have, for every possible combination of parameters, only 1 function.

错误1：

ERROR 1:

执行以下操作将出错

template< typename T>
void print( const T& arg) // first version
{   
    cout<< arg<<endl;
}   

template <typename T, typename... Types>
void print (const T& firstArg, const Types&... args) // second version
{   
    cout << firstArg << endl; // print first argument
    print(args...); // call print() for remaining arguments
}

因为调用 print 编译器不知道要调用哪个函数。

Because when you call print the compiler doesn't know which function to call.

执行 print（3）是指第一还是第二版本？两者都是有效的，因为第一个参数有1个参数，第二个参数也可以接受一个参数。

Does print(3) refers to "first" or "second" version? Both would be valid because the first has 1 parameter, and the second can accept one parameter too.

print(3); // error, ambiguous, which one you want to call, the 1st or the 2nd?

---

错误2：

---

ERROR 2:

以下内容仍然是错误

// No empty function

template <typename T, typename... Types>
void print (const T& firstArg, const Types&... args) 
{   
    cout << firstArg << endl; // print first argument
    print(args...); // call print() for remaining arguments
}

实际上，如果您使用它

 print('k', 0, 6.5);

已解析为

 std::cout<<'k'<<std::endl;
 print(0, 6.5);

已解析为

 std::cout<<'k'<<std::endl;
 std::cout<< 0 <<std::endl;
 print( 6.5);

已解析为

 std::cout<<'k'<<std::endl;
 std::cout<< 0 <<std::endl;
 std::cout<< 6.5 <<std::endl;
 print(); //Oops error, no function 'print' to call with no arguments

尝试，编译器尝试不带任何参数调用 print（）。但是，如果不存在这样的函数，则不会调用它，这就是为什么您应该提供该空函数（不用担心，编译器会优化代码，因此空函数不会降低性能）。

As you see in the last attempt, the compiler tries to call print() with no arguments. However if such a function does not exists it is not called, and that's why you should provide that empty function (don't worry, the compiler will optimize code so empty functions don't decrease performance).

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