c++ - 模棱两可的重载模板

Question

我有以下模板化代码

#include <vector>
#include <array>
#include <iostream>

template<typename T1>
void foo(std::vector<T1> bar) {
    std::cout << "GENERIC" << std::endl;
}

template<typename T1>
void foo(std::vector<std::vector<T1>> bar) {
    std::cout << "SPECIFIC (vector)" << std::endl;
}

template<typename T1, int SIZE>
void foo(std::vector<std::array<T1, SIZE>> bar) {
    std::cout << "SPECIFIC (array)" << std::endl;
}

int main() {
    std::vector<std::vector<int>> a(2, std::vector<int> { 1, 2, 3});
    std::vector<std::array<int, 3>> b(2, std::array<int, 3> {4, 5, 6});

    foo(a);
    foo(b);
}

产生

SPECIFIC (vector)
GENERIC

我想知道为什么使用特定模板调用 vector-of-vector 版本，而使用泛型调用 vector-of-array 版本？

Answer 1

template<typename T1, size_t SIZE>
void foo(std::vector<std::array<T1, SIZE>> bar) {
    std::cout << "SPECIFIC (array)" << std::endl;
}

您应该使用 std::size_t而不是 int .
run here

编辑:
实际上，您的评论和我对代码的直觉使我深入研究了该主题。乍一看，标准开发人员(像我一样)希望编译器能够转换 int至 std::size_t (因为它们都是整数类型并且隐式转换非常简单)并选择 void foo(std::vector<std::array<T1, SIZE>> bar)作为最好的专业。所以在阅读时 template argument deduction我发现这个页面:

If a non-type template parameter is used in the parameter list, and the corresponding template argument is deduced, the type of the deduced template argument ( as specified in its enclosing template parameter list, meaning references are preserved) must match the type of the non-type template parameter exactly, except that cv-qualifiers are dropped, and except where the template argument is deduced from an array bound—in that case any integral type is allowed, even bool though it would always become true:

与往常一样，当然，您必须多读几遍才能理解它的含义:)

于是一个有趣的结果出来了。

已经没有选择我们想要的特化，但如果编译器被迫选择，那将是一个错误。

template<typename T1, int SIZE>
void foo(std::vector<std::array<T1, SIZE>> bar) {
    std::cout << "SPECIFIC (array)" << std::endl;
}

int main() {
    std::vector<std::array<int, 3>> b(2, std::array<int, 3> {4, 5, 6});

    foo(b); // P = std::vector<std::array<int,(int)SIZE>
            // A = std::vector<std::array<int,(unsigned_long)SIZE>>
            // error: deduced non-type template argument does not have the same
            // type as its corresponding template argument */
}

run code

另一个有趣的事情是:

如果没有推导出非类型模板参数，则不会有强制参数和模板类型相同的限制。

#include <vector>
#include <array>
#include <iostream>

template<typename T1, int SIZE>
void foo(std::vector<std::array<T1, SIZE>> bar) {
    std::cout << "SPECIFIC (array)" << std::endl;
}

int main() {
    std::vector<std::array<int, 3>> b(2, std::array<int, 3> {4, 5, 6});

    foo<int,3>(b);
}

run code