Debug
All types which want to use std::fmt
formatting traits
require an
implementation to be printable. Automatic implementations are only provided
for types such as in the std
library. All others must be manually
implemented somehow.
The fmt::Debug
trait
makes this very straightforward. All types can
derive
(automatically create) the fmt::Debug
implementation. This is
not true for fmt::Display
which must be manually implemented.
#![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); }
All std
library types are automatically printable with {:?}
too:
// 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))); }
So fmt::Debug
definitely makes this printable but sacrifices some elegance.
Rust also provides "pretty printing" with {:#?}
.
#[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); }
One can manually implement fmt::Display
to control the display.
See also:
attributes
, derive
, std::fmt
,
and struct