调试

所有想要使用 std::fmt 格式化 trait 的类型都需要实现可打印接口。自动实现仅针对 std 库中的类型提供。所有其他类型都必须以某种方式手动实现。

fmt::Debug trait 使这变得非常简单。所有类型都可以derive（自动创建）fmt::Debug 实现。但这不适用于 fmt::Display，它必须手动实现。

#![allow(unused)]
fn main() {
// This structure cannot be printed either with `fmt::Display` or
// with `fmt::Debug`.
struct UnPrintable(i32);

// The `derive` attribute automatically creates the implementation
// required to make this `struct` printable with `fmt::Debug`.
#[derive(Debug)]
struct DebugPrintable(i32);
}

所有 std 库类型也可以使用 {:?} 自动打印

// Derive the `fmt::Debug` implementation for `Structure`. `Structure`
// is a structure which contains a single `i32`.
#[derive(Debug)]
struct Structure(i32);

// Put a `Structure` inside of the structure `Deep`. Make it printable
// also.
#[derive(Debug)]
struct Deep(Structure);

fn main() {
    // Printing with `{:?}` is similar to with `{}`.
    println!("{:?} months in a year.", 12);
    println!("{1:?} {0:?} is the {actor:?} name.",
             "Slater",
             "Christian",
             actor="actor's");

    // `Structure` is printable!
    println!("Now {:?} will print!", Structure(3));

    // The problem with `derive` is there is no control over how
    // the results look. What if I want this to just show a `7`?
    println!("Now {:?} will print!", Deep(Structure(7)));
}

因此，fmt::Debug 肯定可以使其可打印，但牺牲了一些优雅性。Rust 还通过 {:#?} 提供“漂亮打印”。

#[derive(Debug)]
struct Person<'a> {
    name: &'a str,
    age: u8
}

fn main() {
    let name = "Peter";
    let age = 27;
    let peter = Person { name, age };

    // Pretty print
    println!("{:#?}", peter);
}

可以手动实现 fmt::Display 来控制显示。

另请参阅

属性、derive、std::fmt 和 结构体