C++：模板(template)中typename的利用要领

声明template参数时, 前缀要害字class和typename可以交流;

利用要害字typename标识嵌套从属范例名称, 但不需在基类列表和成员初始化列表内利用.

从属名称(dependent names): 模板(template)内呈现的名称, 相依于某个模板(template)参数, 如T t;

嵌套从属名称(nested dependent names):从属名称在class内呈嵌套装, 如T::const_iterator ci;

非从属名称(non-dependent names): 不依赖任何template参数的名称, 如int value;

假如不特定指出typename, 嵌套从属名称, 有大概发生理会(parse)歧义.

任何时候在模板(template)中指涉一个嵌套从属范例名称, 需要在前一个位置, 添加要害字typename;

不然报错(GCC): error: need ‘typename’ before ‘T::xxx’ because ‘T’ is a dependent scope

代码:

/* 
 * BInsertSort.cpp 
 * 
 *  Created on: 2014.4.17. 
 *      Author: Spike 
 */
      
#include <iostream>  
#include <string>  
#include <vector>  
      
using namespace std;  
      
template<typename T>  
void print2nd(const T& container) {  
    typename T::const_iterator iter(container.begin()); //未加typename, 报错  
    ++iter;  
    int value = *iter;  
    std::cout << value;  
}  
      
int main () {  
    vector<int> vi = {1,2,3,4,5};  
    print2nd(vi);  
      
    return 0;  
}

输出:

2

破例:嵌套从属范例名称, 假如是基类列表(base class list)和成员初值列(member initialization list)中,不利用typename;

代码:

/* 
 * BInsertSort.cpp 
 * 
 *  Created on: 2014.4.17 
 *      Author: Spike 
 */
      
#include <iostream>  
#include <vector>  
      
using namespace std;  
      
struct Number {  
    Number(int x) {  
        std::cout << "Number = " << x << std::endl;  
    }  
};  
      
template<typename T>  
struct Base{  
    typedef Number Nested;  
};  
      
template<typename T>  
class Derived: public Base<T>::Nested { //不消typename  
public:  
    explicit Derived(int x) : Base<T>::Nested(x) { //不消typename  
        typename Base<T>::Nested temp(7); //必需利用  
    }  
};  
      
int main () {  
    Derived<int> d(5);  
      
    return 0;  
}

输出:

Number = 5  
Number = 7

当利用特性类(traits class)时, 必需利用typename, 如

代码:

/* 
 * BInsertSort.cpp 
 * 
 *  Created on: 2014.4.17 
 *      Author: Spike 
 */
      
#include <array>  
#include <iostream>  
      
using namespace std;  
      
template<typename T>  
void workWithIter(T iter) {  
    typedef typename std::iterator_traits<T>::value_type value_type; //利用typename  
    value_type temp(*iter);  
    std::cout << "temp = " << temp << std::endl;  
      
}  
      
int main () {  
    std::array<int, 5> ai = {1,2,3,4,5};  
    std::array<int, 5>::iterator aiIter = ai.begin();  
    workWithIter(aiIter);  
    return 0;  
}

输出:

temp = 1

作者：csdn博客 Spike_King