Can I compare function parameters within my if statement?

Question

Here is the code:

function howdy_doody(person) {
  let person = 'dog'
  if (person == { name: 'dog' }) {
    return 'hello'
  } else {
    return 'goodbye'
  }
}
howdy_doody({ name: 'dog' }); 

I'm expecting the output hello. I've tried declaring the person parameter (see code below) but get the identifier person has already been declared syntax error. I'm confused on if I can compare function parameters within an if statement or not.

Thanks

Answer 1

You mean this?

function howdy_doody(person) {
  let compareValue = 'dog'
  if (compareValue == person.name) {
    return 'hello'
  } else {
    return 'goodbye'
  }
}

howdy_doody({ name: 'dog' });

Answer 2

For an exact match of { name: 'dog' } one could e.g. utilize Object.entries which for any given object returns an array of this object's own key-value pairs / entries. Each entry gets provided as a key-value tuple.

Thus the entries array length value should be exyctly 1 and the key of this sole tuple has to equal 'name' whereas the value has to equal 'dog'.

function isValidHowdyType(value) {
  const entries = Object.entries(value ?? {});
  return (
    entries.length === 1 &&
    entries[0][0] === 'name' &&
    entries[0][1] === 'dog'
  );
}
function howdy_doody(value) {
  let phrase = 'goodbye';

  if (isValidHowdyType(value)) {
    phrase = 'hello';
  }
  return phrase;
}

console.log(
  howdy_doody({ name: 'dog' })
);
console.log(
  howdy_doody({ name: 'dog', owner: 'Jane' })
);
console.log(
  howdy_doody({ name: 'cat' })
);
console.log(
  howdy_doody({})
);
console.log(
  howdy_doody(null)
);
console.log(
  howdy_doody()
);

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