在模板中键入条件 [英] Type condition in template

问题描述

是否可以仅仅构建代码中的某些部分，给定C ++中的模板类型？
这将是湖：

Is it possible to build only some part of the code given the type of the template in C++ ? It would be something lake that :

#include <iostream>

using namespace std;

template<typename T>
void printType(T param)
{
    #if T == char*
        cout << "char*" << endl;
    #elif T == int
        cout << "int" << endl;
    #else
        cout << "???" << endl;
    #endif
}

int main()
{
    printType("Hello world!");
    printType(1);
    return 0;
}

推荐答案

输入traits：

#include <iostream>
#include <type_traits> // C++0x
//#include <tr1/type_traits> // C++03, use std::tr1

template<typename T>
void printType(T param)
{
  if(std::is_same<T,char*>::value)
        std::cout << "char*" << endl;
  else if(std::is_same<T,int>::value)
        std::cout << "int" << endl;
  else
        std::cout << "???" << endl;
}

或更好的是，只需重载函数：

Or even better yet, just overload the function:

template<class T>
void printType(T partam){
  std::cout << "???" << endl;
}

void printType(char* partam){
  std::cout << "char*" << endl;
}

void printType(int partam){
  std::cout << "int" << endl;
}

部分排序会注意调用正确的函数。此外，在一般情况下，模板特殊化优先于重载，请参见此和这个artice为什么。如果您完全必须打印类型，可能不适用，因为重载函数考虑隐式转换。

Partial ordering will take care that the correct function is called. Also, overloading is preferred to template specialization in the general case, see this and this artice for why. Might not apply for you if you totally have to print the type, as implicit conversions are considered for overloaded functions.

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