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
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"
其他回答
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"
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"
}
我非常喜欢@Coautose之前的回答,但如果有人只想要没有名称空间的类型名称,例如C而不是a::b::C,这里是一个修改后的宏版本,看起来像预期的那样工作:
macro_rules! ty {
($type:ty) => {{
let result = std::any::type_name::<$type>();
match result.rsplit_once(':') {
Some((_, s)) => s,
None => result,
}
}};
}
用法:
debug!("Testing type name: {}", ty!(A));
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