我有以下几点:

let mut my_number = 32.90;

如何打印my_number的类型?

使用type和type_of不起作用。有其他方法可以打印数字的类型吗?


当前回答

**更新**最近没有被验证工作。

我把一个小板条箱一起做这个基于vbo的答案。它提供了一个宏来返回或打印类型。

把这个放在你的货物里。toml文件:

[dependencies]
t_bang = "0.1.2"

然后你可以这样使用它:

#[macro_use] extern crate t_bang;
use t_bang::*;

fn main() {
  let x = 5;
  let x_type = t!(x);
  println!("{:?}", x_type);  // prints out: "i32"
  pt!(x);                    // prints out: "i32"
  pt!(5);                    // prints out: "i32"
}

其他回答

您还可以使用println中的变量!("{:?}”,var)。如果没有为该类型实现Debug,则可以在编译器的错误消息中看到该类型:

mod some {
    pub struct SomeType;
}

fn main() {
    let unknown_var = some::SomeType;
    println!("{:?}", unknown_var);
}

(游戏围栏)

虽然很脏,但很管用。

最好使用这个:

fn print_type_of<T>(_: &T) -> String {
    format!("{}", std::any::type_name::<T>())
}

fn main() {
    let s = &"hello world".to_string();
    let cloned_s = s.clone();
    println!("{:?}", print_type_of(&s));
    println!("{:?}", print_type_of(&cloned_s));
}

来自https://stackoverflow.com/a/29168659/6774636的推论

在稳定rust中有一个@ChrisMorgan答案可以获得近似类型(“float”),在夜间rust中有一个@ShubhamJain答案可以通过不稳定函数获得精确类型(“f64”)。

现在有一种方法可以得到精确的类型(即在f32和f64之间决定)在稳定的rust:

fn main() {
    let a = 5.;
    let _: () = unsafe { std::mem::transmute(a) };
}

结果

error[E0512]: cannot transmute between types of different sizes, or dependently-sized types
 --> main.rs:3:27
  |
3 |     let _: () = unsafe { std::mem::transmute(a) };
  |                           ^^^^^^^^^^^^^^^^^^^
  |
  = note: source type: `f64` (64 bits)
  = note: target type: `()` (0 bits)

更新

涡轮鱼的变异

fn main() {
    let a = 5.;
    unsafe { std::mem::transmute::<_, ()>(a) }
}

略短,但可读性稍差。

其他一些答案不工作,但我发现typename crate工作。

Create a new project: cargo new test_typename Modify the Cargo.toml [dependencies] typename = "0.1.1" Modify your source code use typename::TypeName; fn main() { assert_eq!(String::type_name(), "std::string::String"); assert_eq!(Vec::<i32>::type_name(), "std::vec::Vec<i32>"); assert_eq!([0, 1, 2].type_name_of(), "[i32; 3]"); let a = 65u8; let b = b'A'; let c = 65; let d = 65i8; let e = 65i32; let f = 65u32; let arr = [1,2,3,4,5]; let first = arr[0]; println!("type of a 65u8 {} is {}", a, a.type_name_of()); println!("type of b b'A' {} is {}", b, b.type_name_of()); println!("type of c 65 {} is {}", c, c.type_name_of()); println!("type of d 65i8 {} is {}", d, d.type_name_of()); println!("type of e 65i32 {} is {}", e, e.type_name_of()); println!("type of f 65u32 {} is {}", f, f.type_name_of()); println!("type of arr {:?} is {}", arr, arr.type_name_of()); println!("type of first {} is {}", first, first.type_name_of()); }

输出结果为:

type of a 65u8  65 is u8
type of b b'A'  65 is u8
type of c 65    65 is i32
type of d 65i8  65 is i8
type of e 65i32 65 is i32
type of f 65u32 65 is u32
type of arr [1, 2, 3, 4, 5] is [i32; 5]
type of first 1 is i32

有一个不稳定的函数std::intrinsic::type_name可以获取类型的名称,尽管您必须使用Rust的夜间构建(这在稳定的Rust中不太可能工作)。这里有一个例子:

#![feature(core_intrinsics)]

fn print_type_of<T>(_: &T) {
    println!("{}", unsafe { std::intrinsics::type_name::<T>() });
}

fn main() {
    print_type_of(&32.90);          // prints "f64"
    print_type_of(&vec![1, 2, 4]);  // prints "std::vec::Vec<i32>"
    print_type_of(&"foo");          // prints "&str"
}