c++ - Typeid()检查模板化函数的传递参数

Question

我不想使用函数重载来对函数进行小的更改。相反，我想使用typeid()检查下面的模板化函数的传递参数。但是，如果我没有在下面的代码中注释掉该行，则会产生以下编译错误:

Severity    Code    Description Project File    Line    Suppression State
Error       invalid operands to binary expression ('basic_ostream<char, std::char_traits<char> >' and 'std::vector<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::allocator<std::basic_string<char, std::char_traits<char>, std::allocator<char> > > >')

据我了解，编译器不知道如何运行。有解决方案吗？

代码是:

#include <iostream>
#include <vector>

using namespace std;


template <class T>
void Test(T A)
{
    if (typeid(T) == typeid(vector<string>)) {
        cout << "The type of A is vector<string>" << endl << endl;
        //cout << "First element of A is:" << A[0] << endl;    // If I don't comment out this line, it gives the compiler error.
    }

    if (typeid(T) == typeid(vector<vector<string>>)) {
        cout << "The type of A is vector<vector<string>>" << endl;
        cout << "First row first element of A is:" << A[0][0] << endl;
    }
}

int main()
{
    Test(vector<string> {"1", "2", "3"});

    Test(vector<vector<string>> { {"11", "12", "13"}, { "21", "22", "23" }});

    return 0;
}

Answer 1

问题出在普通的if中，对于条件为给定的Test类型的T的每个实例化，无论条件是什么结果，语句true(如果存在则为false)都必须是有效的语句。

自C++ 17起，可以使用constexpr if(带有std::is_same)。

In a constexpr if statement, the value of condition must be a contextually converted constant expression of type bool. If the value is true, then statement-false is discarded (if present), otherwise, statement-true is discarded.

例如

if constexpr (std::is_same_v<T, vector<string>>) {
    cout << "The type of A is vector<string>" << endl << endl;
    cout << "First element of A is:" << A[0] << endl;
} 

if constexpr (std::is_same_v<T, vector<vector<string>>>) {
    cout << "The type of A is vector<vector<string>>" << endl;
    cout << "First row first element of A is:" << A[0][0] << endl;
}

LIVE

在C++ 17之前，您可以使用 SFINAE或 specialization(带有模板)，或者仅使用 overloading(甚至不带模板)。