如何使用类型推断来删除变量声明中的不可变/ const?可能吗?
immutable uint immutableX = 42;
// keep the type (uint) but remove the immutability
/* compiler magic */ mutableX = immutableX;
非类型推断解决方案是:
uint mutableX = immutableX;
完整示例:
void main()
{
immutable uint immutableX = 42;
pragma(msg, "immutableX: ", typeof(immutableX));
assert(typeof(immutableX).stringof == "immutable(uint)");
// how to use type inference so that possible immutable/const is removed ?
// the expected type of mutableX is uint
auto mutableX = immutableX;
pragma(msg, "mutableX: ", typeof(immutableX));
// this should be true
assert(typeof(immutableX).stringof == "uint");
}
答案 0 :(得分:3)
根据用例,有std.traits.Unqual
,它删除了最外面的immutable
,const
,shared
等
import std.traits : Unqual;
immutable int a = 3;
Unqual!(typeof(a)) b = a;
static assert(is(typeof(b) == int));
一个更简单的解决方案可能是cast()
:
immutable int a = 3;
auto b = cast()a;
static assert(is(typeof(b) == int));
哪种正确取决于您在哪里以及如何使用它。