C++ - 模板参数推导/替换失败 [英] C++ - Template argument deduction/substitution failed

问题描述

我的目标是能够在 std::vector 上使用算术运算符.考虑以下示例:

My goal is to be able to use arithmetic operators on std::vector. Consider the following example:

#include <vector>

using namespace std;

template <class S, class T, class U> vector<U> operator+(const vector<S> &a, const vector<T> &b){
    vector<U> result;
    result.reserve(a.size());
    for(int i = 0; i < a.size();++i){
        result[i] = a[i] + b[i];
    }
    return result;
}

int main(int argc, char** argv) {
    vector<double> bla;
    bla = bla + bla;
    return 0;
}

此代码无法编译，因为编译器无法推导出模板参数 U(它不是 MWE，但我试图提供一个有意义的示例).为什么会这样?我知道在这里使用三个模板参数可能没有意义.我的想法是，在类型 S 和 T 都提供具有不同返回类型的匹配+"实现的情况下，我可以同时处理这两种情况.还是会出现歧义的问题?我只是想知道编译器是否应该不能推断出 U.当然，下面的代码工作正常:

This code does not compile because the compiler is not able to deduce template argument U (it's not an MWE but I tried to provide an example that makes sense). Why is this the case? I know that it might not make sense to use three template arguments here. My idea was that in the case where types S and T both provide a matching '+'-implementation with different return types, I could handle both cases at once. Or would then be a problem with ambiguity? I'm just wondering if the compiler shouldn't be able to deduce U. Of course the following code just works fine:

#include <vector>

using namespace std;

template <class S, class T> vector<S> operator+(const vector<S> &a, const vector<T> &b){
    vector<S> result;
    result.reserve(a.size());
    for(int i = 0; i < a.size();++i){
        result[i] = a[i] + b[i];
    }
    return result;
}

int main(int argc, char** argv) {
    vector<double> bla;
    bla = bla + bla;
    return 0;
}

推荐答案

您可以使用 常见类型

#include <vector>
#include <type_traits>

using namespace std;

template <class S, class T> 
vector<typename std::common_type<S, T>::type> operator+(const vector<S> &a, const vector<T> &b)
{
    vector<typename std::common_type<S, T>::type> result;
    result.reserve(a.size());
    for(unsigned int i = 0; i < a.size();++i){
        result[i] = a[i] + b[i];
    }
    return result;
}

int main() {
    vector<double> bla;
    bla = bla + bla;
    return 0;
}

现场示例

正如 Jarod42 建议的那样，您还可以使用 vector 作为另一种可能的返回类型(可能与 common_type 不同).请记住，后者需要尾随返回类型或 std::declval (C++11)

as Jarod42 suggested, you might also use vector<decltype(a[0] + b[0])> as another possible return type (which may be different than the common_type). Keep in mind that this latter one requires trailing return type or std::declval (C++11)

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