Using regular expressions is probably the best way. You can see a bunch of tests here (taken from chromium)
function validateEmail(email) {
const re = /^(([^<>()[\]\\.,;:\s@"]+(\.[^<>()[\]\\.,;:\s@"]+)*)|(".+"))@((\[[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\])|(([a-zA-Z\-0-9]+\.)+[a-zA-Z]{2,}))$/;
return re.test(String(email).toLowerCase());
}
Here's the example of regular expresion that accepts unicode:
const re = /^(([^<>()[\]\.,;:\s@\"]+(\.[^<>()[\]\.,;:\s@\"]+)*)|(\".+\"))@(([^<>()[\]\.,;:\s@\"]+\.)+[^<>()[\]\.,;:\s@\"]{2,})$/i;
But keep in mind that one should not rely only upon JavaScript validation. JavaScript can easily be disabled. This should be validated on the server side as well.
Here's an example of the above in action:
function validateEmail(email) {
const re = /^(([^<>()[\]\\.,;:\s@\"]+(\.[^<>()[\]\\.,;:\s@\"]+)*)|(\".+\"))@((\[[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\])|(([a-zA-Z\-0-9]+\.)+[a-zA-Z]{2,}))$/;
return re.test(email);
}
function validate() {
const $result = $("#result");
const email = $("#email").val();
$result.text("");
if (validateEmail(email)) {
$result.text(email + " is valid :)");
$result.css("color", "green");
} else {
$result.text(email + " is not valid :(");
$result.css("color", "red");
}
return false;
}
$("#email").on("input", validate);
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<label for=email>Enter an email address:</label>
<input id="email">
<h2 id="result"></h2>
Setting a bit
Use the bitwise OR operator (|
) to set a bit.
number |= 1UL << n;
That will set the n
th bit of number
. n
should be zero, if you want to set the 1
st bit and so on upto n-1
, if you want to set the n
th bit.
Use 1ULL
if number
is wider than unsigned long
; promotion of 1UL << n
doesn't happen until after evaluating 1UL << n
where it's undefined behaviour to shift by more than the width of a long
. The same applies to all the rest of the examples.
Clearing a bit
Use the bitwise AND operator (&
) to clear a bit.
number &= ~(1UL << n);
That will clear the n
th bit of number
. You must invert the bit string with the bitwise NOT operator (~
), then AND it.
Toggling a bit
The XOR operator (^
) can be used to toggle a bit.
number ^= 1UL << n;
That will toggle the n
th bit of number
.
Checking a bit
You didn't ask for this, but I might as well add it.
To check a bit, shift the number n to the right, then bitwise AND it:
bit = (number >> n) & 1U;
That will put the value of the n
th bit of number
into the variable bit
.
Changing the nth bit to x
Setting the n
th bit to either 1
or 0
can be achieved with the following on a 2's complement C++ implementation:
number ^= (-x ^ number) & (1UL << n);
Bit n
will be set if x
is 1
, and cleared if x
is 0
. If x
has some other value, you get garbage. x = !!x
will booleanize it to 0 or 1.
To make this independent of 2's complement negation behaviour (where -1
has all bits set, unlike on a 1's complement or sign/magnitude C++ implementation), use unsigned negation.
number ^= (-(unsigned long)x ^ number) & (1UL << n);
or
unsigned long newbit = !!x; // Also booleanize to force 0 or 1
number ^= (-newbit ^ number) & (1UL << n);
It's generally a good idea to use unsigned types for portable bit manipulation.
or
number = (number & ~(1UL << n)) | (x << n);
(number & ~(1UL << n))
will clear the n
th bit and (x << n)
will set the n
th bit to x
.
It's also generally a good idea to not to copy/paste code in general and so many people use preprocessor macros (like the community wiki answer further down) or some sort of encapsulation.
Best Answer
Keep your 64-bit number as separate high and low partitions. To rotate left N when N < 32:
hi_rot = ((hi << N) | (lo >>> (32-N))) & (0xFFFFFFFF)
lo_rot = ((lo << N) | (hi >>> (32-N))) & (0xFFFFFFFF)
If N >= 32, then subtract 32 from N, swap hi and lo, and then do the above.