c++ - 使用 std::array::size 实例化 std::array 时出错

Question

示例代码test.cpp

#include <array>
#include <string>

int main ()
{
  // OK
  const std::array<int, 2> array_int = {42, 1337};

  std::array<float, array_int.size()> array_float_ok;

  // Error
  const std::array<std::string, 2> array_string = {"foo", "bar"};

  std::array<float, array_string.size()> array_float_error;

  return 0;
}

使用 g++ 4.8.4 (Ubuntu 14.04) 编译

g++ -Wall -std=c++0x test.cpp -o test

给出如下错误信息

test.cpp: In function ‘int main()’:
test.cpp:14:39: error: call to non-constexpr function ‘constexpr std::array<_Tp, _Nm>::size_type std::array<_Tp, _Nm>::size() const [with _Tp = std::basic_string<char>; long unsigned int _Nm = 2ul; std::array<_Tp, _Nm>::size_type = long unsigned int]’
   std::array<float, array_string.size()> array_float_error;
                                       ^
In file included from test.cpp:1:0:
/usr/include/c++/4.8/array:162:7: note: ‘constexpr std::array<_Tp, _Nm>::size_type std::array<_Tp, _Nm>::size() const [with _Tp = std::basic_string<char>; long unsigned int _Nm = 2ul; std::array<_Tp, _Nm>::size_type = long unsigned int]’ is not usable as a constexpr function because:
       size() const noexcept { return _Nm; }
       ^
/usr/include/c++/4.8/array:162:7: error: enclosing class of constexpr non-static member function ‘constexpr std::array<_Tp, _Nm>::size_type std::array<_Tp, _Nm>::size() const [with _Tp = std::basic_string<char>; long unsigned int _Nm = 2ul; std::array<_Tp, _Nm>::size_type = long unsigned int]’ is not a literal type
/usr/include/c++/4.8/array:81:12: note: ‘std::array<std::basic_string<char>, 2ul>’ is not literal because:
     struct array
            ^
/usr/include/c++/4.8/array:81:12: note:   ‘std::array<std::basic_string<char>, 2ul>’ has a non-trivial destructor
test.cpp:14:39: error: call to non-constexpr function ‘constexpr std::array<_Tp, _Nm>::size_type std::array<_Tp, _Nm>::size() const [with _Tp = std::basic_string<char>; long unsigned int _Nm = 2ul; std::array<_Tp, _Nm>::size_type = long unsigned int]’
   std::array<float, array_string.size()> array_float_error;
                                       ^
test.cpp:14:40: note: in template argument for type ‘long unsigned int’
   std::array<float, array_string.size()> array_float_error;
                                        ^
test.cpp:14:59: error: invalid type in declaration before ‘;’ token
   std::array<float, array_string.size()> array_float_error;
                                                           ^
test.cpp:9:39: warning: unused variable ‘array_float_ok’ [-Wunused-variable]
   std::array<float, array_int.size()> array_float_ok;
                                       ^
test.cpp:14:42: warning: unused variable ‘array_float_error’ [-Wunused-variable]
   std::array<float, array_string.size()> array_float_error;
                                          ^

有人可以解释这个错误吗？为什么第一个示例可以运行而第二个示例无法编译？

Answer 1

std::string 类型不是文字类型，这意味着它不能在编译时作为 constexpr 函数的一部分进行操作。在编译时，编译器会尝试评估 array_string 的 size() 函数。您在第一个错误中看到的函数第一个类型参数设置为 std::basic_string < char >(又名 std::string)；因此，由于 std::string 不是字面量类型，因此无法在编译时将该函数作为 constexpr 函数求值，并且会出现错误。

我建议您引用以下内容以了解有关 constexpr 的更多信息。

http://en.cppreference.com/w/cpp/language/constexpr

我建议您引用以下内容以了解文字类型。

http://en.cppreference.com/w/cpp/concept/LiteralType

最后，试试下面的简单代码，你会发现 int 和 float 是字面量类型，而 std::string 不是。您可以尝试使用其他类型来查看哪些是或不是文字类型。

#include <iostream>
int main(int argc, char** argv) 
{ 
    std::cout << std::is_literal_type<int>::value << std::endl;
    std::cout << std::is_literal_type<float>::value << std::endl;
    std::cout << std::is_literal_type<std::string>::value << std::endl;
    return 0;
}                                  

希望对您有所帮助。

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