英文:
template<int N> std::ostream& operator << (...)
问题
为什么这个应用程序无法编译？
#include <iostream>
#include <array>
template<int N>
std::ostream& operator << (std::ostream& out, std::array<int, N> const& arr) {
for(auto& a: arr) std::cout << a << ' ';
return out;
}
int main(int argc, char const* argv[]) {
std::array<int, 10> arr {1,2,3,4,5,6,7,8,9,10};
std::cout << arr << '\n';
return 0;
}
为什么它无法解析 N？错误消息是：
main.cpp:13:15: error: invalid operands to binary expression ('std::ostream' (aka 'basic_ostream<char>') and 'std::array<int, 10>')
std::cout << arr << '\n';
英文:
Why this application doesn't compile?
#include <iostream>
#include <array>
template<int N>
std::ostream& operator << (std::ostream& out, std::array<int, N> const& arr) {
for(auto& a:arr) std::cout << a << ' ';
return out;
}
int main(int argc, char const* argv[]) {
std::array<int, 10> arr {1,2,3,4,5,6,7,8,9,10};
std::cout << arr << '\n';
return 0;
}
Why it cannot resolve N? The error message is
main.cpp:13:15: error: invalid operands to binary expression ('std::ostream' (aka 'basic_ostream<char>') and 'std::array<int, 10>')
std::cout << arr << '\n';
答案1
得分: 10
因为std::array的第二个模板参数的类型是std::size_t，而不是int。类型不匹配导致模板参数推导失败。
如果函数模板的非类型模板参数用于函数参数列表的模板参数（该参数也是一个模板），并且相应的模板参数被推导出来，那么推导出的模板参数的类型（如其封闭模板参数列表中指定的那样，即保留了引用）必须与非类型模板参数的类型完全匹配，除了cv限定符被去除，并且除非模板参数从数组界限中推导出来，在这种情况下，任何整数类型都是允许的，甚至
bool，尽管它总是变为true：
您可以将操作符模板更改为：
template<std::size_t N>
std::ostream& operator << (std::ostream& out, std::array<int, N> const& arr) {
for(auto& a:arr) std::cout << a << ' ';
return out;
}
英文:
Because the type of the 2nd template parameter of std::array is std::size_t, not int. Type mismatch causes template argument duduction failing.
> If a non-type template parameter of function template is used in the template parameter list of function parameter (which is also a template), 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:
You can change the operator template to:
template<std::size_t N>
std::ostream& operator << (std::ostream& out, std::array<int, N> const& arr) {
for(auto& a:arr) std::cout << a << ' ';
return out;
}
答案2
得分: 2
我在VS 2022中使用C++的最新std和Microsoft编译器尝试了这段代码，int N版本可以编译成功，哈哈。似乎它进行了某种转换。无论如何，它可以处理不匹配的情况。
但是，正如@songyuanyao已经提到的那样，你应该使用std::size_t。
英文:
I tried the code in VS 2022 using C++'s latest std and the Microsoft compiler compiles the int N version without problems haha. It seems like it does a sort of conversion. In any case, it could deal with the mismatch.
But yes, std::size_t is what you should be using instead, as @songyuanyao already stated.
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