In the code below

When func1 uses Promise.all, the execution time is normal.

func2’s execution time doubles when Promise.all is used.

I want to know why the time is doubled.

I checked the results in Mac, Chrome browser.

code

const func1 = ()=>{
    return new Promise((resolve) => {
        setTimeout(() => {
            resolve("Task completed!");
        }, 1000);
    });
}

const func2 = async ()=>{
    await 1 // This code prevents js optimization in Chrome.
    for (let i = 0; i < 1e8; i++);
    return "Task completed!";
}

// Measures function execution time
const funcTimer = async (name, func)=>{
    console.time(name)
    await func()
    console.timeEnd(name)
}

await funcTimer('func1 x1 time',func1)

await funcTimer('func2 x1 time',func2)

await funcTimer('func1 x2 Promise.all time',async ()=>{
    await Promise.all([
        func1(),
        func1()
    ])
})

await funcTimer('func2 x2 Promise.all time',async ()=>{
    await Promise.all([
        func2(),
        func2()
    ])
})

result