如何打印具有所有可用精度的 Rust 浮点数?

问题描述

我正在为 sin 三角函数实现 CORDIC 算法.为了做到这一点，我需要硬编码/计算一堆反正切值.现在我的函数 似乎 可以工作(由 Wolfram Alpha 验证)到打印的精度，但我希望能够打印我的 f32.我该怎么做?

I am implementing the CORDIC algorithm for the sin trigonometric function. In order to do this, I need to hardcode/calculate a bunch of arctangent values. Right now my function seems to work (as validated by Wolfram Alpha) to the precision that is printed, but I would like to be able to print all 32 bits of precision of my f32. How may I do that?

fn generate_table() {
    let pi: f32 = 3.1415926536897932384626;
    let k1: f32 = 0.6072529350088812561694; // 1/k
    let num_bits: uint = 32;
    let num_elms: uint = num_bits;
    let mul: uint = 1 << (num_bits - 2);

    println!("Cordic sin in rust");
    println!("num bits {}", num_bits);
    println!("pi is {}", pi);
    println!("k1 is {}", k1);

    let shift: f32 = 2.0;
    for ii in range(0, num_bits) {
        let ipow: f32 = 1.0 / shift.powi(ii as i32);
        let cur: f32 = ipow.atan();
        println!("table values {}", cur);
    }
}

推荐答案

使用 精度格式说明符;一个 . 后跟您希望看到的精度的小数点数:

Use the precision format specifier; a . followed by the number of decimal points of precision you'd like to see:

fn main() {
    let pi: f32 = 3.1415926536897932384626;
    let k1: f32 = 0.6072529350088812561694; // 1/k

    println!("pi is {:.32}", pi);
    println!("k1 is {:.32}", k1);
}

我选择了 32，这比这两个 f32s 中的小数点数多.

I chose 32, which is more than the number of decimal points in either of these f32s.

pi is 3.14159274101257324218750000000000
k1 is 0.60725295543670654296875000000000

请注意，这些值不再匹配；浮点值很难！正如评论中提到的，您可能希望打印为十六进制，甚至使用您的文字为十六进制.

Note that the values no longer match up; floating point values are difficult! As mentioned in a comment, you may wish to print as hexadecimal or even use your literals as hexadecimal.

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