Function type guards in Typescript do not support type inference

When checking for null in alpha, I validate the result and use throw new Error if needed. However, even after doing so, the compiler still indicates 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);
}

Surprisingly, the following code works without any issues:

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);
}

Answer №1

Typescript lacks the ability to guarantee that your function objMethod will consistently return the same value. It may return a non-null value on one call and null on another. However, by assigning it to a const variable, the compiler ensures that the variable is not null. Because the variable is declared as a constant, its value remains unchanged after the initial assignment, thus ensuring it must be non-null if it passes the check.

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