繼續以解一元多次方程割線法為例

use std::f64; pub fn get_next_x(com_fn:impl Fn(f64)->f64,x0:f64,x1:f64)->f64{let res1=&f_compute(&com_fn,x1);let res0=&f_compute(&com_fn,x0);x1-res1*((x1-x0)/(res1-res0)) } pub fn f_compute(com_fn:impl Fn(f64)->f64,x:f64)->f64{com_fn(x) }fn main() {let tol=1e-8_f64;let x0=-10.0_f64;let x1=10.0_f64;let mut x_i:f64;let mut x_isub1:f64;let mut x_iplus1:f64;let mut ans:f64; let mut finished;let mut fn_comp;for i in 1..20{fn_comp=move |x:f64| x.powi(i+1)+x.powi(i)-1.0_f64;x_i=x1;x_isub1=x0;ans=0.0_f64; finished=false; for _n in 1..2000 {x_iplus1=get_next_x(fn_comp,x_isub1,x_i);if (x_iplus1-x_i).abs()<tol { ans=x_iplus1;finished=true;break;} x_isub1=x_i;x_i=x_iplus1;}if finished{println!("function{},解:{}",i,ans); }else{println!("function{},找不到解",i); } } } PS F:\learn\rustlearn> rustc learn4.rs PS F:\learn\rustlearn> .\learn4.exe function1,解:0.6180339887498948 function2,解:0.7548776662466918 function3,解:0.8191725133961645 function4,解:0.0012008970334152279 function5,解:0.8812714616336657 function6,解:0.00001800013706088066 function7,解:0.9115923534820559 function8,解:-0.999999980000001 function9,解:0.9295701282320228 function10,解:-0.9999999998 function11,解:10.000000000244464 function12,解:-0.9999999999979999 function13,解:10.000000000002444 function14,解:-0.9999999999999793 function15,解:10.000000000000027 function16,解:-0.9999999999999999 function17,解:9.999999999999986 function18,解:-1 function19,解:10

下面定義了一個閉包

fn_comp=move |x:f64| x.powi(i+1)+x.powi(i)-1.0_f64;

下面閉包作為參數

pub fn f_compute(com_fn:impl Fn(f64)->f64,x:f64)->f64{com_fn(x) }

閉包作為返回值

pub fn get_fn(i:i32)->impl Fn(f64)->f64{move |x:f64| x.powi(i+1)+x.powi(i)-1.0_f64 } use std::f64; pub fn get_next_x(com_fn:impl Fn(f64)->f64,x0:f64,x1:f64)->f64{let res1=&f_compute(&com_fn,x1);let res0=&f_compute(&com_fn,x0);x1-res1*((x1-x0)/(res1-res0)) } pub fn f_compute(com_fn:impl Fn(f64)->f64,x:f64)->f64{com_fn(x) }pub fn get_fn(i:i32)->impl Fn(f64)->f64{move |x:f64| x.powi(i+1)+x.powi(i)-1.0_f64 }fn main() {let tol=1e-8_f64;let x0=-10.0_f64;let x1=10.0_f64;let mut x_i:f64;let mut x_isub1:f64;let mut x_iplus1:f64;let mut ans:f64; let mut finished;let mut fn_comp;for i in 1..20{fn_comp=get_fn(i);x_i=x1;x_isub1=x0;ans=0.0_f64; finished=false; for _n in 1..2000 {x_iplus1=get_next_x(&fn_comp,x_isub1,x_i);if (x_iplus1-x_i).abs()<tol { ans=x_iplus1;finished=true;break;} x_isub1=x_i;x_i=x_iplus1;}if finished{println!("function{},解:{}",i,ans); }else{println!("function{},找不到解",i); } } } 與50位技術專家面對面20年技術見證，附贈技術全景圖

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