如何以所有可用的精度打印Rust浮点数? [英] How do I print a Rust floating-point number with all available precision?

问题描述

我正在为sin三角函数实现CORDIC算法.为此，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(已由Wolfram Alpha验证)到打印的精度，但是我希望能够打印我的f32的所有32位精度.我该怎么办?

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

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，比这两个f32中的小数点位数更多.

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； 浮点值很困难！如评论中所述，您可能希望以十六进制打印，甚至可以使用文字作为十六进制.

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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