 # How do I check if a variable is an integer in JavaScript without using Number.isInteger()?Alex K  const t="undefined"!=typeof HTMLImageElement&&"loading"in HTMLImageElement.prototype;if(t){const t=document.querySelectorAll("img[data-main-image]");for(let e of t){e.dataset.src&&(e.setAttribute("src",e.dataset.src),e.removeAttribute("data-src")),e.dataset.srcset&&(e.setAttribute("srcset",e.dataset.srcset),e.removeAttribute("data-srcset"));const t=e.parentNode.querySelectorAll("source[data-srcset]");for(let e of t)e.setAttribute("srcset",e.dataset.srcset),e.removeAttribute("data-srcset");e.complete&&(e.style.opacity=1,e.parentNode.parentNode.querySelector("[data-placeholder-image]").style.opacity=0)}}

To check if a variable is an integer in JavaScript without using`Number.isInteger()`, you can use a combination of type checking and mathematical operations. Here's an example implementation:

``````1
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``````
function isInteger(value) {
// Check if the value is a number and is finite
if (typeof value !== 'number' || !isFinite(value)) {
return false;
}

// Check if the value is a whole number
return Math.floor(value) === value;
}

// Example usage
console.log(isInteger(42)); // true
console.log(isInteger(3.14)); // false
console.log(isInteger('42')); // false
console.log(isInteger(null)); // false
``````

In this implementation, the`isInteger` function first checks if the value is a number and is finite using`typeof` and`isFinite` checks. If the value is not a number or is not finite, it immediately returns`false`. Next, if the value is a number, it checks if it is a whole number by comparing it with its floor value using`Math.floor()`. If the floor value and the original value are the same, it means the value is an integer, and the function returns`true`. Note that this implementation considers`NaN` and`Infinity` as non-integers. If you want to include`Infinity` as an integer, you can modify the`isFinite` check accordingly.