如何使用类型推断,以便删除可能的不可变/ const?

时间:2019-03-15 05:27:16

标签: d

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");
}

1 个答案:

答案 0 :(得分:3)

根据用例,有std.traits.Unqual,它删除了最外面的immutableconstshared

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。