我正在尝试编写constexpr查找函数,该函数将返回包含特定值的std :: array的索引。下面的函数似乎可以正常工作,除非包含的类型为 const char * :
I am trying to write a constexpr find function that will return the index of a std::array containing a certain value. The function below seems to work OK except when the contained type is const char*:
#include <array> constexpr auto name1() { return "name1"; } constexpr auto name2() { return "name2"; } template <class X, class V> constexpr auto find(X& x, V key) { std::size_t i = 0; while(i < x.size()) { if(x[i] == key) return i; ++i; } return i; } int main() { constexpr std::array<const char*, 2> x{{name1(), name2()}}; constexpr auto f1 = find(x, name1()); // this compiles constexpr auto f2 = find(x, name2()); // this doesn't... }奇怪的是, find(x,name1())可以干净地编译,但 find(x,name2())失败并显示错误:
The weird thing is that find(x, name1()) compiles cleanly but find(x, name2()) fails with the error:
subexpression not valid in a constant expression if(x[i] == key) return i; `与 name1(),但与 name2()一起使用会失败?
How can this expression work when used with name1() but fail when used with name2()?
我也发现了这个答案,但是用户从头开始构建数组类,我不想这样做。
I have also found this answer, but the user builds the array class from scratch and I do not want to do that.
推荐答案似乎是编译器错误。 f1 和 f2 应该都无法以相同的方式编译。
Seems like a compiler bug. Both f1 and f2 should fail to compile in the same way.
主要问题是它是一个假设,其中 name1 == name1 和 name1 != name2 。该标准实际上不提供此类保证,请参见 [lex.string] / 16 :
The main issue is that it's an assumption that "name1" == "name1" and "name1" != "name2". The standard in fact provides no such guarantees, see [lex.string]/16:
是否所有字符串文字都是不同的(即存储在非重叠对象中)以及是否连续对 string-literal 产生的对象相同或不同
Whether all string literals are distinct (that is, are stored in nonoverlapping objects) and whether successive evaluations of a string-literal yield the same or a different object is unspecified.
即使假设最有可能成立,也要比较 constexpr 明确不允许,请参见 [expr.const] /2.23 :
Even though the assumption most likely holds, comparing unspecified values inside constexpr is expressly not allowed, see [expr.const]/2.23:
—关系( [expr.rel] )或相等( [expr.eq] )运算符,未指定结果;
— a relational ([expr.rel]) or equality ([expr.eq]) operator where the result is unspecified;
一种解决方法(正确的做法)是不依赖字符串文字的地址,而是比较实际的字符串。例如:
A workaround (and the right thing to do) would be to not rely on addresses of string literals and instead compare the actual strings. For example:
constexpr bool equals(const char* a, const char* b) { for (std::size_t i = 0; ; ++i) { if (a[i] != b[i]) return false; if (a[i] == 0) break; } return true; } template <class X, class V> constexpr auto find(X& x, V key) { std::size_t i = 0; while(i < x.size()) { if(equals(x[i], key)) return i; ++i; } return i; }更多推荐
Constexpr使用C ++ 17查找数组
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