Program 16 Self Invoking function | Event looping | JavaScript Interview Questions | setTimeout
This code uses a self-invoking function that contains several setTimeout calls and console.log statements. Here's what it does step by step:
The function is immediately invoked as soon as the script is loaded.
The first setTimeout call is made with a delay of 2000 milliseconds (2 seconds). This means that the function passed as the first argument (which logs "1") will be executed after a 2-second delay.
The second console.log statement logs "2" immediately, before any of the setTimeout functions have been executed.
The second setTimeout call is made with a delay of 0 milliseconds. This may seem odd, but it actually means that the function passed as the first argument will be added to the event queue immediately, but it won't be executed until all the currently executing code has finished. This is often used to defer execution of a function until the next event loop iteration.
The third console.log statement logs "4" immediately after the second setTimeout call.
After all the synchronous code has finished executing, the event loop will start processing the queued events. The second setTimeout function (which logs "3") will be executed at this point.
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