这是@Boiethios回答的简化版。我已经从原始解决方案中删除了一些“&”符号。

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}}
}

Rust游乐场的景观

宏形式允许使用“无处不在”，而函数需要一个对象来解析。

宏表单(一行):

macro_rules! ty {($type:ty) => {std::any::type_name::<$type>()}}

形成的宏观形式:

macro_rules! ty {
    ($type:ty) => {
        std::any::type_name::<$type>()
    };
}

函数形式(借用是为了不破坏已解析的变量):

fn type_of<T>(_: &T) -> &'static str {std::any::type_name::<T>()}

fn type_of<T>(_: &T) -> &'static str {
    std::any::type_name::<T>()
}

例子:

macro_rules! ty {($type:ty) => {std::any::type_name::<$type>()}}
fn type_of<T>(_: &T) -> &'static str {std::any::type_name::<T>()}

struct DontMater<T>(T);

impl<T: std::fmt::Debug> std::fmt::Debug for DontMater<T> {
    fn fmt(&self, fmt: &mut std::fmt::Formatter<'_>) -> std::fmt::Result {
        fmt.write_fmt(format_args!("DontMater<{}>({:?})", ty!(T), self.0))
    }
}

fn main() {
    type µ = [Vec<String>; 7];
    println!("{:?}", DontMater(5_usize));
    println!("{:?}", DontMater("¤"));
    println!("{}", ty!(char));
    println!("{:?}", ty!(µ));
    println!("{}", type_of(&DontMater(72_i8)));
    println!("{:?}", type_of(&15_f64));
}

返回:

DontMater<usize>(5)
DontMater<&str>("¤")
char
"[alloc::vec::Vec<alloc::string::String>; 7]"
env_vars::DontMater<i8>
"f64"

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());
}

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。

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。

1.38版新增std::any::type_name

use std::any::type_name;

fn type_of<T>(_: T) -> &'static str {
    type_name::<T>()
}
fn main() {
    let x = 21;
    let y = 2.5;
    println!("{}", type_of(&y));
    println!("{}", type_of(x));
}