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stagefright5
stagefright5

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Wierd Async/Await Behaviour in JS πŸ€”

Apparently the non async code inside an async function will execute synchronously even if you don't await.
Consider this code example:

async function f() {
    g('adsasdasd'); // note that it is not `await`ed
    console.log('000');
}

async function g(m) {
    console.log('g', m);
}

f() // calling `f`
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I always thought it would produce this output:

000
g adsasdasd
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BUT, the actual output is like this

g adsasdasd // function `g` is executed first!!
000
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Can anyone explain this?

Top comments (1)

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vishalraj82 profile image
Vishal Raj

The return value of an async function is always an Promise. Here function g is technically returning undefined, but before that its executing console.log which is a simple log statement. Hence it prints the string. The subsequent call to console.log(000) prints another value.