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?
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
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) => {{
$var.type_of()
}};
}
fn main() {
let x = "Hello world!";
println!("{}", tyof!(x));
// or
println!("{}", x.type_of());
let x = 5;
println!("{}", tyof!(x));
// or
println!("{}", x.type_of());
}
其他回答
1.38版新增std::any::type_name
use std::any::type_name;
fn type_of<T>(_: T) -> &'static str {
type_name::<T>()
}
fn main() {
let x = 21;
let y = 2.5;
println!("{}", type_of(&y));
println!("{}", type_of(x));
}
**更新**最近没有被验证工作。
我把一个小板条箱一起做这个基于vbo的答案。它提供了一个宏来返回或打印类型。
把这个放在你的货物里。toml文件:
[dependencies]
t_bang = "0.1.2"
然后你可以这样使用它:
#[macro_use] extern crate t_bang;
use t_bang::*;
fn main() {
let x = 5;
let x_type = t!(x);
println!("{:?}", x_type); // prints out: "i32"
pt!(x); // prints out: "i32"
pt!(5); // prints out: "i32"
}
短篇小说;
fn tyof<T>(_: &T) -> String {
std::any::type_name::<T>().into()
}
很长的故事;
trait Type {
fn type_of(&self) -> String;
}
macro_rules! Type {
($($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) => {{
$var.type_of()
}};
}
fn main() {
let x = "Hello world!";
println!("{}", tyof!(x));
// or
println!("{}", x.type_of());
let x = 5;
println!("{}", tyof!(x));
// or
println!("{}", x.type_of());
}
您还可以使用println中的变量!("{:?}”,var)。如果没有为该类型实现Debug,则可以在编译器的错误消息中看到该类型:
mod some {
pub struct SomeType;
}
fn main() {
let unknown_var = some::SomeType;
println!("{:?}", unknown_var);
}
(游戏围栏)
虽然很脏,但很管用。
宏形式允许使用“无处不在”,而函数需要一个对象来解析。
宏表单(一行):
macro_rules! ty {($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"