let a: {
m?: string
};
let b = a = {};
b.m = ''; // Property 'm' does not exist on type '{}'.
let a: {
m?: string
} = {};
let b = a;
b.m = ''; // It's OK
What occurs when using assignment chaining? How can this type error be resolved?
Let's dive into the initial scenario:
let a: {
m?: string
};
let b = a = {};
The type of b is actually inferred from {}, not from a. Hence, accessing m from b is not possible.
Now, let's consider the second situation:
let a: {
m?: string
} = {};
let b = a;
In this case, the type of b is inferred from a, which contains the m property.
Why does this happen?
Consider this example:
let x = y = z;
The result is z because the assignment functions as an expression.
TypeScript determines the type of z (in our case {}) and assigns it to x (or b in our context)
To resolve the issue in the first case, both a and b need to be declared as { m?: string }.
type Foo = {
m?: string;
}
let a: Foo;
let b: Foo = a = {}
Check out the playground here
When using assignment chaining, the variable b is assigned the value of {}, but it does not inherit the type of a.
To properly assign the type to variable b, you need to create a separate type and apply it.
type A = {
m?: string;
}
let a: A;
let b: A = a = {};
b.m = '';
One way to assign the same value to multiple variables is by chaining the assignment operator (link).
For example, using let b = a = {}; is equivalent to let b = {}; a = {};. It's important to note that in this scenario, a gets assigned first before b.
To achieve the desired outcome, you can utilize the following code snippet:
let a: {
m?: string
};
let b = a;
b = a = {};
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