Category: JavaScript Answers

Sometimes, we may run into the ‘ReferenceError: deprecated caller or arguments usage’ when we’re developing JavaScript apps.

In this article, we’ll look at how to fix the ‘ReferenceError: deprecated caller or arguments usage’ when we’re developing JavaScript apps.

Fix the ‘ReferenceError: deprecated caller or arguments usage’ When Developing JavaScript Apps

To fix the ‘ReferenceError: deprecated caller or arguments usage’ when we’re developing JavaScript apps, we should make sure we aren’t trying to access the Function.caller or arguments.callee.callers properties in our JavaScript code.

They’re both deprecated properties that will be removed in the future.

For instance, we should write code like:

'use strict';

function myFunc() {
  if (myFunc.caller == null) {
    return "The function was called from the top";
  } else {
    return myFunc.caller;
  }
}

myFunc();

Also, we shouldn’t access the Function.arguments property in our code:

"use strict";

function f(n) {
  g(n - 1);
}

function g(n) {
  console.log("before: " + g.arguments[0]);
  if (n > 0) {
    f(n);
  }
  console.log("after: " + g.arguments[0]);
}

f(2);

We should get function arguments from the signature of the function directly or with the rest operator:

"use strict";

function f(n) {
  g(n - 1);
}

function g(...args) {
  console.log("before: " + args[0]);
  if (n > 0) {
    f(n);
  }
  console.log("after: " + args[0]);
}

f(2);

args is an array with all the arguments g is called with.

Conclusion

To fix the ‘ReferenceError: deprecated caller or arguments usage’ when we’re developing JavaScript apps, we should make sure we aren’t trying to access the Function.caller or arguments.callee.callers properties in our JavaScript code.

They’re both deprecated properties that will be removed in the future.

Sometimes, we may run into the ‘ReferenceError: can’t access lexical declaration "X" before initialization’ when we’re developing JavaScript apps.

In this article, we’ll look at how to fix the ‘ReferenceError: can’t access lexical declaration "X" before initialization’ when we’re developing JavaScript apps.

Fix the ‘ReferenceError: can’t access lexical declaration "X" before initialization’ When Developing JavaScript Apps

To fix the ‘ReferenceError: can’t access lexical declaration "X" before initialization’ when we’re developing JavaScript apps, we should make sure that we don’t try to access variables declared with let or const before they’re defined.

The error message for the same error in Edge is ReferenceError: Use before declaration.

And the error message for the same error in Chrome is ReferenceError: Use before declaration.

For instance, if we have code like:

const bar = () => {
  let foo = 33;
  if (true) {
    let foo = (foo + 55);
  }
}
bar ();

We declared foo inside and outside the if block.

In the if block, we would be trying to reference the foo variable inside the block since we have let foo inside the block.

However, on the right side, we’re trying to access it before the declaration is done.

Therefore, this error will be thrown.

To fix this, we should write:

const bar = () => {
  let foo = 33;
  if (true) {
    foo = (foo + 55);
  }
}
bar ();

This way, we reassign the value to the foo variable that’s declared before the if block is run.

And therefore, foo is available for reassignment in the if block.

Conclusion

To fix the ‘ReferenceError: can’t access lexical declaration "X" before initialization’ when we’re developing JavaScript apps, we should make sure that we don’t try to access variables declared with let or const before they’re defined.

The error message for the same error in Edge is ReferenceError: Use before declaration.

And the error message for the same error in Chrome is ReferenceError: Use before declaration.

Sometimes, we may run into the ‘ReferenceError: assignment to undeclared variable "x"’ when we’re developing JavaScript apps.

In this article, we’ll look at how to fix the ‘ReferenceError: assignment to undeclared variable "x"’ when we’re developing JavaScript apps.

Fix the ‘ReferenceError: assignment to undeclared variable "x"’ When Developing JavaScript Apps

To fix the ‘ReferenceError: assignment to undeclared variable "x"’ when we’re developing JavaScript apps, we should make sure that we’re assigning values to variables that have already been declared.

The error message for this error is ReferenceError: "x" is not defined in Chrome.

And in Edge, the error message for this error is ReferenceError: Variable undefined in strict mode.

For instance, we shouldn’t write code like:

const foo = () => {
  "use strict";
  bar = true;
};
foo();

since bar hasn’t been declared in the foo function.

Instead, we should declare the bar variable by writing:

const foo = () => {
  "use strict";
  const bar = true;
};
foo();

Now the error should be fixed.

Conclusion

To fix the ‘ReferenceError: assignment to undeclared variable "x"’ when we’re developing JavaScript apps, we should make sure that we’re assigning values to variables that have already been declared.

The error message for this error is ReferenceError: "x" is not defined in Chrome.

And in Edge, the error message for this error is ReferenceError: Variable undefined in strict mode.

Sometimes, we may run into the ‘ReferenceError: "x" is not defined’ when we’re developing JavaScript apps.

In this article, we’ll look at how to fix the ‘ReferenceError: "x" is not defined’ when we’re developing JavaScript apps.

Fix the ‘ReferenceError: "x" is not defined’ When Developing JavaScript Apps

To fix the ‘ReferenceError: "x" is not defined’ when we’re developing JavaScript apps, we should make sure that we do operations on variables that have already been declared.

We should also make sure the variable is available in the scope that we want to do the operation to.

For instance, instead of writing:

foo.substring(1); 

which will throw the error since foo isn’t defined before we call substring, we write:

const foo = 'bar';
foo.substring(1)

to declare foo before calling substring.

We should also make sure the variable is actually available in the given scope before we do any operation with it.

For instance, we write:

const add = () =>{
  const num1 = 2,
      num2 = 3;
  return num1 + num2;
}

console.log(num1);

which will throw the error since num1 is only available in the add function.

Instead, we write:

const add = () => {
  const num1 = 2,
    num2 = 3;

  console.log(num1);
  return num1 + num2;
};

to log the num1 value since it’s available in the add function.

Conclusion

To fix the ‘ReferenceError: "x" is not defined’ when we’re developing JavaScript apps, we should make sure that we do operations on variables that have already been declared.

We should also make sure the variable is available in the scope that we want to do the operation to.

Sometimes, we may run into the ‘RangeError: repeat count must be non-negative’ when we’re developing JavaScript apps.

In this article, we’ll look at how to fix the ‘RangeError: repeat count must be non-negative’ when we’re developing JavaScript apps.

Fix the ‘RangeError: repeat count must be non-negative’ When Developing JavaScript Apps

To fix the ‘RangeError: repeat count must be non-negative’ when we’re developing JavaScript apps, we should make sure we call the String.prototype.repeat method with an argument that’s a valid JavaScript integer.

For instance, instead of writing:

'abc'.repeat(-100);
'a'.repeat(-28);    

which has negative numbers as arguments, we write:

'abc'.repeat(0);
'abc'.repeat(1);
'abc'.repeat(2);
'abc'.repeat(3.5);

which all are called with valid numbers as arguments.

3.5 will be converted to an integer when repeat is run.

Conclusion

To fix the ‘RangeError: repeat count must be less than non-negative’ when we’re developing JavaScript apps, we should make sure we call the String.prototype.repeat method with an argument that’s a valid JavaScript integer.