为什么使用 Rust 将可变结构传递给函数会导致不可变字段? [英] Why using Rust does passing a mutable struct to a function result in immutable fields?

问题描述

我正在 Win8-64 上使用 0.8 学习 Rust.我有一个测试程序，我正在处理一个处理参数输入的函数返回一个包含这些参数的结构.那工作正常.然后我修改了程序以将 &struct 传递给函数，现在我得到一个编译器错误，我试图将它分配给一个不可变字段.

I'm learning Rust using 0.8 on Win8-64. I have a test program I'm working on where a function that handles input of parameters returned a struct containing those parameters. That worked OK. I then altered the program to pass the &struct to the function, and I now get a compiler error that I'm attempting to assign to an immutable field.

我应该如何将指针/引用传递给结构以防止出现此错误?

How should I be passing a pointer/reference to the struct to prevent this error?

导致错误的代码(我尝试了一些变体):

The code that results in the error (I've tried a few variations) :

let mut ocParams : cParams = cParams::new();     //!!!!!! This is the struct passed

fInputParams(&ocParams);               // !!!!!!! this is passing the struct

if !ocParams.tContinue {
    return;
}

.......

struct cParams {
  iInsertMax : i64,
  iUpdateMax : i64,
  iDeleteMax : i64,
  iInstanceMax : i64,
  tFirstInstance : bool,
  tCreateTables : bool,
  tContinue : bool
}

impl cParams {
  fn new() -> cParams {
     cParams {iInsertMax : -1, iUpdateMax : -1, iDeleteMax : -1, iInstanceMax : -1,
              tFirstInstance : false, tCreateTables : false, tContinue : false}
  }   
}

.....

fn fInputParams(mut ocParams : &cParams) {

    ocParams.tContinue = (sInput == ~"y");    // !!!!!! this is one of the error lines

对函数中结构域的所有赋值都会导致编译时出错.编译导致的错误示例:

All of the assignments to the struct fields in the function result in an error at compile-time. An example of the errors that result from compile :

testli007.rs:240:2: 240:20 error: cannot assign to immutable field
testli007.rs:240   ocParams.tContinue = (sInput == ~"y");   

推荐答案

在你的函数声明中:

fn fInputParams(mut ocParams : &cParams) {

ocParams 是一个可变变量，包含一个借用的指向不可变结构的指针.您想要的是该结构是可变的，而不是变量.因此，函数的签名应该是:

ocParams is a mutable variable containing a borrowed pointer to an immutable struct. What you want is for that struct to be mutable, not the variable. Therefore, the signature of the function should be:

fn fInputParams(ocParams : &mut cParams) {

然后您必须将调用本身更改为fInputParams:

Then you have to change the call itself to fInputParams:

fInputParams(&mut ocParams);  // pass a pointer to mutable struct.

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