短篇小说;

fn tyof<T>(_: &T) -> String {
    std::any::type_name::<T>().into()
}

很长的故事;

trait Type {
    fn type_of(&self) -> String;
}

macro_rules! Type {
    ($($ty:ty),*) => {
        $(
            impl Type for $ty {
                fn type_of(&self) -> String {
                    stringify!($ty).into()
                }
            }
        )*
    }
}

#[rustfmt::skip]
Type!(
    u8, i8, u16, i16, u32, i32, i64, u64, i128, String, [()], (), Vec<()>, &u8, &i8, &u16, &i16, &u32, &i32, &i64, &u64, &i128, &str, &[()], &Vec<()>, &() 
    // add any struct, enum or type you want
);

macro_rules! tyof {
    ($var: expr) => {{
        $var.type_of()
    }};
}

fn main() {
    let x = "Hello world!";
    println!("{}", tyof!(x));
    // or
    println!("{}", x.type_of());

    let x = 5;
    println!("{}", tyof!(x));
    // or
    println!("{}", x.type_of());
}

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"
}

UPD以下不再工作。检查Shubham的答案以作更正。

检查std::intrinsic::get_tydesc<T>()。它现在处于“实验”状态，但如果您只是对类型系统进行了修改，那么它是OK的。

请看下面的例子:

fn print_type_of<T>(_: &T) -> () {
    let type_name =
        unsafe {
            (*std::intrinsics::get_tydesc::<T>()).name
        };
    println!("{}", type_name);
}

fn main() -> () {
    let mut my_number = 32.90;
    print_type_of(&my_number);       // prints "f64"
    print_type_of(&(vec!(1, 2, 4))); // prints "collections::vec::Vec<int>"
}

这是在内部用来实现著名的{:?}格式化程序。

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。

你可以使用std::any::type_name函数。这并不需要一个夜间编译器或外部板条箱，结果是非常正确的:

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

fn main() {
    let s = "Hello";
    let i = 42;

    print_type_of(&s); // &str
    print_type_of(&i); // i32
    print_type_of(&main); // playground::main
    print_type_of(&print_type_of::<i32>); // playground::print_type_of<i32>
    print_type_of(&{ || "Hi!" }); // playground::main::{{closure}}
}

注意:如文档中所述，此信息只能用于调试目的:

这是用于诊断用途。字符串的确切内容和格式没有指定，只是尽力描述该类型。

如果你想让你的类型表示在不同的编译器版本中保持相同，你应该使用一个trait，就像phicr的答案一样。