我该如何使用这种格式!在no_std环境中的宏？

Question

如何在不使用std的情况下实现以下示例？

let text = format!("example {:.1} test {:x} words {}", num1, num2, num3);

text具有类型&str和num1,num2并num3具有任何数字类型.

我已经尝试使用numtoa和itoa/dtoa显示数字,但numtoa不支持浮动,itoa不支持no_std.我觉得在字符串中显示一个数字是相当普遍的,我可能会遗漏一些明显的东西.

Answer 1

除了Shepmaster的答案,您还可以在没有分配器的情况下格式化字符串.

在core::fmt::Write你只需要实现write_str,然后你得到write_fmt免费.

使用format_args!(...)(语法相同format!),您可以准备一个core::fmt::Arguments可以传递给的值core::fmt::write.

看游乐场:

#![crate_type = "dylib"]
#![no_std]

pub mod write_to {
    use core::cmp::min;
    use core::fmt;

    pub struct WriteTo<'a> {
        buffer: &'a mut [u8],
        // on write error (i.e. not enough space in buffer) this grows beyond
        // `buffer.len()`.
        used: usize,
    }

    impl<'a> WriteTo<'a> {
        pub fn new(buffer: &'a mut [u8]) -> Self {
            WriteTo { buffer, used: 0 }
        }

        pub fn as_str(self) -> Option<&'a str> {
            if self.used <= self.buffer.len() {
                // only successful concats of str - must be a valid str.
                use core::str::from_utf8_unchecked;
                Some(unsafe { from_utf8_unchecked(&self.buffer[..self.used]) })
            } else {
                None
            }
        }
    }

    impl<'a> fmt::Write for WriteTo<'a> {
        fn write_str(&mut self, s: &str) -> fmt::Result {
            if self.used > self.buffer.len() {
                return Err(fmt::Error);
            }
            let remaining_buf = &mut self.buffer[self.used..];
            let raw_s = s.as_bytes();
            let write_num = min(raw_s.len(), remaining_buf.len());
            remaining_buf[..write_num].copy_from_slice(&raw_s[..write_num]);
            self.used += raw_s.len();
            if write_num < raw_s.len() {
                Err(fmt::Error)
            } else {
                Ok(())
            }
        }
    }

    pub fn show<'a>(buffer: &'a mut [u8], args: fmt::Arguments) -> Result<&'a str, fmt::Error> {
        let mut w = WriteTo::new(buffer);
        fmt::write(&mut w, args)?;
        w.as_str().ok_or(fmt::Error)
    }
}

pub fn test() {
    let mut buf = [0u8; 64];
    let _s: &str = write_to::show(
        &mut buf,
        format_args!("write some stuff {:?}: {}", "foo", 42),
    ).unwrap();
}

Answer 2

一般来说,你没有.format!分配a String,而no_std环境没有分配器.

如果您有分配器,则可以使用alloc crate.这个箱包含format!宏.

这个箱子不稳定,所以你需要每晚使用Rust:

#![crate_type = "dylib"]
#![no_std]

#[macro_use]
extern crate alloc;

fn thing() {
    let text = format!("example {:.1} test {:x} words {}", 1, 2, 3);
}