The JavaScript loop appears to be malfunctioning

The expectation was for the code snippet to output 100 times with the value of i.

https://i.sstatic.net/4XRXE.png

However, it only prints out once as shown below.

What could I be overlooking here?

let i = 0;
for (; i++; i < 100) {
  console.log('Loop ==>' + i);
}
console.log('Loop Done');
console.log('Value of i ==>' + i);

Answer №1

To create a loop, you need to follow the structure shown below:

<script>
    let count = 0;
                        for(; count < 100; count++)
                        {
                              console.log('Loop iteration ==>'+count);
                        }
                        console.log('Loop Completed');
                        console.log('Final value of count ==>'+count);        
    </script> 

Answer №2

There seems to be an issue with the structure of your for loop. It is important to ensure that you are using the correct syntax, such as this example.

for (; i < 100; i++)

for loops consist of three statements: initialization, condition, and updation, in that specific order. The loop will continue to run as long as the second statement (the condition) holds true. The third statement (updation) will then be executed after all the code within the block has been executed.

It appears that you have inadvertently switched the positions of the third (updation) and the second (condition) statements within your for loop.

Answer №3

You have mistakenly switched the positions of the "condition" and "change" in your for loop. The condition should come second, while the change should come third.

let i = 0;

for(; i < 100; i++) {
    console.log('Loop ==>'+i);
}

console.log('Loop Done');
console.log('Value of i ==>'+i);

Answer №4

Replace ; i++; i < 100 with ; i < 100; i++. When the former case is used, the value of i is incremented and then tested if it is less than 100, which is not the desired behavior.

let i = 0;
for (; i < 100; i++) {
  console.log('Loop ==>' + i);
}
console.log('Loop Done');
console.log('Value of i ==>' + i);

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