CODEWITHSUNDEEP
typescript
Petro Ivanenko
Petro Ivanenko

Reputation: 727

Typescript type inference does not work with function type guards

In alpha I check whether the result is null and throw new Error if so, but the comiler still shows a compilation error:

 const obj = {
  objMethod: function (): string | null {
    return 'always a string';
  },
};

function alpha() {
  if (obj.objMethod() === null) {
    throw new Error('type guard, but compiler does not count it as one :( ');
  }

  const beta: string = obj.objMethod();
  console.log(beta);
}

But this code works perfectly well:

const obj = {
  objMethod: function (): string | null {
    return 'always a string';
  },
};

function alpha() {
  const result = obj.objMethod();
  if (result === null) {
    throw new Error('this DOES count as a type guard. ');
  }

  const beta: string = result; // no error
  console.log(beta);
}

Upvotes: 0

Views: 44

Answers (1)

Brian Pfretzschner
Brian Pfretzschner

Reputation: 617

Typescript cannot prove that your objMethod will always return the same value. It could return a non-null value at the first call, and null at the second. If you assign it to a const variable, the check asserts that the variable is not null. Since the variable is const, the value cannot change after the assignment, thus, it must be non-null if it passed the check.

Upvotes: 1

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