如何将值映射到 Rust 中的类型? [英] How can I map a value to a type in Rust?
问题描述
我正在编写一个程序,它接受命令行上的数字类型,然后使用该类型调用通用函数.例如,我这样运行程序:
I am writing a program which accepts a numerical type on the command line, and then calls a generic function using that type. For example, I run program thus:
my_programme u32
...在我的代码中我有:
...and in my code I have:
match cli_type
{
"u32" =>
{
my_generic_function::<u32>(args);
},
"u16" =>
...
如果我想对 8、16、32 和 64 位整数(有符号和无符号)执行此操作,则需要大量调用 my_generic_function()
.它看起来很乱,似乎没有必要.
If I want to do this for 8, 16, 32, and 64 bit integers, both signed and unsigned, that's a lot of calls to my_generic_function()
. It looks messy and seems needless.
我可以在 &str
或 String
值和类型 T
之间定义映射,或者编写一个函数来返回类型 T
,所以我可以不写匹配语句:
Can I define a map between &str
or String
values and type T
, or write a function to return type T
, so instead of the match statement, I could just write:
my_generic_function::<get_type(cli_type)>(args);
推荐答案
Rust 在运行时不保留任何类型信息.标准库中有一个函数,std::any::type_name
,它可以给你一个类型的名称,但它只是一个字符串,没有办法回到类型的世界.这一切都发生在编译时,以后无法更改.
Rust doesn't keep any type information at runtime. There is a function in the standard library, std::any::type_name
, which can give you the name of a type, but it's just a string and there is no way to get back into the world of types. That all happens at compile time and can't be changed later.
但是,您可以使用宏保存一些代码:
However, you can save some code with a macro:
macro_rules! make_call_typed {
($($t: ty),*) => {
fn call_typed(input: &str, args: &str) {
match input {
$(
stringify!($t) => {
my_generic_function::<$t>(args);
}
),*
other => panic!("unexpected type: {}", other)
}
}
}
}
当你这样称呼它时:
make_call_typed!(u32, u16, i32, i16);
它会生成这样的代码:
fn call_typed(input: &str, args: &str) {
match input {
"u32" => {
my_generic_function::<u32>(args);
}
"u16" => {
my_generic_function::<u16>(args);
}
"i32" => {
my_generic_function::<i32>(args);
}
"i16" => {
my_generic_function::<i16>(args);
}
other => panic!("unexpected type: {}", other),
}
}
这篇关于如何将值映射到 Rust 中的类型?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!