Why Does My Async/Await Function Still Return a Promise?

Async/Await: Understanding Promise Return Values

In your code, the getJSON function is an asynchronous function that uses the async/await syntax. As the question correctly states, every asynchronous function returns a Promise object. The await operator is then used to pause the execution of the function until the Promise is resolved or rejected.

However, you cannot log the result of the asynchronous function directly with console.log(getJSON()). This is because await only un-suspends the function execution and allows the Promise to resolve with the result. However, the function itself still returns the underlying Promise, not the unwrapped result.

To access the result of the Promise, you must either use the then() method, as you demonstrated, or use await again within another asynchronous function. In your case, getJSON().then(json => console.log(json)) logs the result { foo: 'bar' } because it chains a then handler onto the Promise returned by getJSON.

As mentioned in the answer, this behavior is fundamental to the Promise model. Promises represent asynchronous operations that may resolve at an unspecified time in the future. The then() method allows you to specify what actions should be taken when the Promise resolves.

By understanding that asynchronous functions always return Promises, even when using await, you can correctly handle and consume asynchronous code in your applications.

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