问题描述
例如,如果我们有一个std::array
,并且使用constexpr
实例化了一个超出范围的元素,则编译器将不会报告错误:
For example if we have an std::array
and we instantiate an element that is out of bound using constexpr
the compiler wouldn't report error:
constexpr int EvaluateSpecialArrayIndex(int a)
{ return a * sizeof(int); }
array<int, 5> arr;
cout << arr[98] << endl; //compiles fine
cout << arr[EvaluateSpecialArrayIndex(4)] << endl; //the same as above
我们不能以某种方式限制这一点吗?
Can't we restrict this somehow?
推荐答案
为确保在编译时对constexpr
函数进行求值,必须通过使它们的结果为constexpr
来强制使用它们.例如:
To ensure that constexpr
functions are evaluated at compile time, you must force them to be by making their result constexpr
. For example:
#include <array>
int
main()
{
constexpr std::array<int, 5> arr{1, 2, 3, 4, 5};
int i = arr[6]; // run time error
}
但是:
#include <array>
int
main()
{
constexpr std::array<int, 5> arr{1, 2, 3, 4, 5};
constexpr int i = arr[6]; // compile time error
}
不幸的是,要使其真正起作用,std::array
必须符合C ++ 14规范,而不是C ++ 11规范.由于C ++ 11规范未将std::array::operator[]
的const
重载标记为constexpr
.
Unfortunately, for this to actually work, std::array
must conform to the C++14 specification, not the C++11 specification. As the C++11 specification does not mark the const
overload of std::array::operator[]
with constexpr
.
所以在C ++ 11中,您很不走运.在C ++ 14中,您可以使其工作,但前提是同时声明array
和调用索引运算符的结果都声明为constexpr
.
So in C++11 you're out of luck. In C++14, you can make it work, but only if both the array
and the result of calling the index operator are declared constexpr
.
说明
用于数组索引的C ++ 11规范为:
The C++11 specification for array indexing reads:
reference operator[](size_type n);
const_reference operator[](size_type n) const;
对于数组索引的C ++ 14规范为:
And the C++14 specification for array indexing reads:
reference operator[](size_type n);
constexpr const_reference operator[](size_type n) const;
即constexpr
已添加到C ++ 14的const
重载中.
I.e. constexpr
was added to the const
overload for C++14.
更新
对于数组索引的C ++ 17规范为:
And the C++17 specification for array indexing reads:
constexpr reference operator[](size_type n);
constexpr const_reference operator[](size_type n) const;
该循环现已完成.可以在编译时计算Universe. ;-)
The cycle is now complete. The universe can be computed at compile-time. ;-)
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