由于您是从a本身创建ro，因此ro和a在内存中将具有相同的引用，因此无论何时更改其中一个，另一个都会对另一个生效。

如何在不传递引用的情况下互相创建数组？

• 可以使用.slice()

let a: number[] = [1, 2, 3, 4];
let ro: ReadonlyArray<number> = a.slice();
a[0] = 8;

console.log(ro); //expext to be [1,2,3,4] ok
console.log(a);//expext to be [8,2,3,4] ok

• 您可以通过扩展语法传递它。

let a: number[] = [1, 2, 3, 4];
let ro: ReadonlyArray<number> = [...a];
a[0] = 8;

console.log(ro); //expext to be [1,2,3,4] ok
console.log(a);//expext to be [8,2,3,4] ok

• 您可以使用Array.from()方法，该方法生成现有数组的浅层副本

let a: number[] = [1, 2, 3, 4];
let ro: ReadonlyArray<number> = Array.from(a);
a[0] = 8;

console.log(ro); //expext to be [1,2,3,4] ok
console.log(a);//expext to be [8,2,3,4] ok

使用Array.from()或Spread运算符创建副本

let a = [1, 2, 3, 4];
let ro = Array.from(a); // [...a]
a[0] = 8;

console.log(ro);
console.log(a);

对于只读，请尝试使用Object.freeze()

let a = [1, 2, 3, 4];
let ro = [...a];

Object.freeze(ro);

a[0] = 8;
ro[0] = 8;

console.log(a);
console.log(ro);