[英]What am I doing wrong here? Or is this a clang++ bug?
The following code fails to compile on my Mac 以下代码无法在我的Mac上编译
#include <iostream>
#include <array>
template <typename T, unsigned int N>
using Vector = std::array<T, N>;
template <typename T, unsigned int N>
T dot(const Vector<T, N> &l, const Vector<T, N> &r) {
T result{0};
for (auto i = 0; i < N; ++i) {
result += l[i] * r[i];
}
return result;
}
using Vector3f = Vector<float, 3>;
int main(int argc, const char * argv[]) {
Vector3f u{1.0f, 2.0f, 3.0f};
Vector3f v{6.0f, 5.0f, 4.0f};
std::cout << dot(u, v) << std::endl;
return 0;
}
Here is how I'm compiling from Terminal: 以下是我从终端编译的方式:
clang++ -std=c++11 -stdlib=libc++ repro.cpp -o repro
Here is the error I get: 这是我得到的错误:
repro.cpp:24:18: error: no matching function for call to 'dot'
std::cout << dot(u, v) << std::endl;
^~~
repro.cpp:10:3: note: candidate template ignored: substitution failure [with T = float]: deduced non-type template
argument does not have the same type as the its corresponding template parameter
('unsigned long' vs 'unsigned int')
T dot(const Vector<T, N> &l, const Vector<T, N> &r) {
^
1 error generated.
The code compiles fine in Visual Studio 2015 Preview. 该代码在Visual Studio 2015 Preview中编译良好。
And it compiles fine from Terminal when I replace the dot call by: 当我通过以下方式替换点呼叫时,它从终端编译得很好:
std::cout << dot<float, 3>(u, v) << std::endl;
PS: clang++ version I'm using: clang++ --version Apple LLVM version 6.0 (clang-600.0.57) (based on LLVM 3.5svn) PS:clang ++我正在使用的版本:clang ++ --version Apple LLVM 6.0版(clang-600.0.57)(基于LLVM 3.5svn)
Replace all instances of unsigned int
with std::size_t
. 用
std::size_t
替换unsigned int
所有实例。 The std::array
class template is declared as. std::array
类模板声明为。
template< class T, std::size_t N > struct array;
During template argument deduction, if the non-type template parameter does not match the corresponding argument, a deduction failure occurs. 在模板参数推导期间,如果非类型模板参数与相应的参数不匹配,则会发生推消失败。 On your system,
std::size_t
happens to alias long unsigned int
Changing the code to the following should work: 在您的系统上,
std::size_t
碰巧别名为long unsigned int
将代码更改为以下内容应该有效:
template <typename T, std::size_t N> // <--
using Vector = std::array<T, N>;
template <typename T, std::size_t N> // <--
T dot(const Vector<T, N> &l, const Vector<T, N> &r);
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