Why Is String "11" Less Than String "3"

Why is string 11 less than string 3?

Strings are compared lexicographicaly. i.e. character by character until they are not equal or there aren't any characters left to compare. The first character of '11' is less than the first character of '3'.

> '11' < '3'
true
> '31' < '3'
false
> '31' < '32'
true
> '31' < '30'
false

If we use letters then, since b is not less than a, abc is not less than aaa, but since c is less than d, abc is less than abd.

> 'abc' < 'aaa'
false
> 'abc' < 'abd'
true

You can explicitly convert strings to numbers:

> +'11' < '3'
false

Why is one string greater than the other when comparing strings in JavaScript?

Because, as in many programming languages, strings are compared lexicographically.

You can think of this as a fancier version of alphabetical ordering, the difference being that alphabetic ordering only covers the 26 characters a through z.

---

This answer is in response to a java question, but the logic is exactly the same. Another good one: String Compare "Logic".

What is the cause? operators or use of strings?

I guess it is because it compare ascii values of chars and 3 had greater ascii value than 2. In string it compare char by char if 1 char is false it wont compare else

Why do JavaScript comparisons have different results for strings and for numbers?

"105.00" > "2.00"

is comparing strings instead of numbers.

Why is `12:34 123:45` equal to `false`?

Because of the ASCII code of 3 and :. We know that the comparison between strings will be done by the dictionary rule. Hence, as the first difference is in the third location of two strings, the ASCII code of 3 and : will determine the result of the comparison. As the ASCII code of : is greater than 3, you see the false as a result. See the ASCII code of them in the following.

console.log(":".charCodeAt(0));
console.log("3".charCodeAt(0));

Why in javascript 100 5 returns true?

When you use < with strings, the code points of the each index of the strings are compared. The code point for 1 is 49, and the code point for 5 is 53, so '100' < '5', because 49 < 53.

console.log(  '1'.charCodeAt(),  '5'.charCodeAt());

Why does string to number comparison work in Javascript

Because JavaScript defines >= and <= (and several other operators) in a way that allows them to coerce their operands to different types. It's just part of the definition of the operator.

In the case of <, >, <=, and >=, the gory details are laid out in §11.8.5 of the specification. The short version is: If both operands are strings (after having been coerced from objects, if necessary), it does a string comparison. Otherwise, it coerces the operands to numbers and does a numeric comparison.

Consequently, you get fun results, like that "90" > "100" (both are strings, it's a string comparison) but "90" < 100 (one of them is a number, it's a numeric comparison). :-)

Is it okay to have this comparison like this or should I use parseInt() to convert x to integer ?

That's a matter of opinion. Some people think it's totally fine to rely on the implicit coercion; others think it isn't. There are some objective arguments. For instance, suppose you relied on implicit conversion and it was fine because you had those numeric constants, but later you were comparing x to another value you got from an input field. Now you're comparing strings, but the code looks the same. But again, it's a matter of opinion and you should make your own choice.

If you do decide to explicitly convert to numbers first, parseInt may or may not be what you want, and it doesn't do the same thing as the implicit conversion. Here's a rundown of options:

• parseInt(str[, radix]) - Converts as much of the beginning of the string as it can into a whole (integer) number, ignoring extra characters at the end. So parseInt("10x") is 10; the x is ignored. Supports an optional radix (number base) argument, so parseInt("15", 16) is 21 (15 in hex). If there's no radix, assumes decimal unless the string starts with 0x (or 0X), in which case it skips those and assumes hex. Does not look for the new 0b (binary) or 0o (new style octal) prefixes; both of those parse as 0. (Some browsers used to treat strings starting with 0 as octal; that behavior was never specified, and was [specifically disallowed][2] in the ES5 specification.) Returns NaN if no parseable digits are found.

• Number.parseInt(str[, radix]) - Exactly the same function as parseInt above. (Literally, Number.parseInt === parseInt is true.)

• parseFloat(str) - Like parseInt, but does floating-point numbers and only supports decimal. Again extra characters on the string are ignored, so parseFloat("10.5x") is 10.5 (the x is ignored). As only decimal is supported, parseFloat("0x15") is 0 (because parsing ends at the x). Returns NaN if no parseable digits are found.

• Number.parseFloat(str) - Exactly the same function as parseFloat above.

• Unary +, e.g. +str - (E.g., implicit conversion) Converts the entire string to a number using floating point and JavaScript's standard number notation (just digits and a decimal point = decimal; 0x prefix = hex; 0b = binary [ES2015+]; 0o prefix = octal [ES2015+]; some implementations extend it to treat a leading 0 as octal, but not in strict mode). +"10x" is NaN because the x is not ignored. +"10" is 10, +"10.5" is 10.5, +"0x15" is 21, +"0o10" is 8 [ES2015+], +"0b101" is 5 [ES2015+]. Has a gotcha: +"" is 0, not NaN as you might expect.

• Number(str) - Exactly like implicit conversion (e.g., like the unary + above), but slower on some implementations. (Not that it's likely to matter.)

• Bitwise OR with zero, e.g. str|0 - Implicit conversion, like +str, but then it also converts the number to a 32-bit integer (and converts NaN to 0 if the string cannot be converted to a valid number).

So if it's okay that extra bits on the string are ignored, parseInt or parseFloat are fine. parseInt is quite handy for specifying radix. Unary + is useful for ensuring the entire string is considered. Takes your choice. :-)

For what it's worth, I tend to use this function:

const parseNumber = (str) => str ? +str : NaN;

(Or a variant that trims whitespace.) Note how it handles the issue with +"" being 0.

And finally: If you're converting to number and want to know whether the result is NaN, you might be tempted to do if (convertedValue === NaN). But that won't work, because as Rick points out below, comparisons involving NaN are always false. Instead, it's if (isNaN(convertedValue)).

Compare two strings with '' and '' operators in JavaScript

As said above, the formal specification is in the standard: http://www.ecma-international.org/ecma-262/7.0/#sec-abstract-relational-comparison , in layman's terms the logic is like this:

1) String vs String

Split both strings into 16-bit code units and compare them numerically. Note that code units != characters, e.g. "cafè" < "cafè" is true (really).

2) String vs other primitive

Convert both to numbers. If one of them is NaN, return false, otherwise compare numerically. +0 and -0 are considered equal, +/-Infinity is bigger/smaller than anything else.

3) String vs Object

Try to convert the object to a primitive, attempting, in order, [Symbol.toPrimitive]("number"), valueOf and toString. If we've got string, proceed to 1), otherwise proceed to 2). For arrays specifically, this will invoke toString which is the same as join.

Differences between C++ string == and compare()?

This is what the standard has to say about operator==

21.4.8.2 operator==

template<class charT, class traits, class Allocator>
bool operator==(const basic_string<charT,traits,Allocator>& lhs,
                const basic_string<charT,traits,Allocator>& rhs) noexcept;

Returns: lhs.compare(rhs) == 0.

Seems like there isn't much of a difference!

How are strings compared?

From the docs:

The comparison uses lexicographical
ordering: first the first two items
are compared, and if they differ this
determines the outcome of the
comparison; if they are equal, the
next two items are compared, and so
on, until either sequence is
exhausted.

Also:

Lexicographical ordering for strings uses the Unicode code point number to order individual characters.

or on Python 2:

Lexicographical ordering for strings uses the ASCII ordering for individual characters.

As an example:

>>> 'abc' > 'bac'
False
>>> ord('a'), ord('b')
(97, 98)

The result False is returned as soon as a is found to be less than b. The further items are not compared (as you can see for the second items: b > a is True).

Be aware of lower and uppercase:

>>> [(x, ord(x)) for x in abc]
[('a', 97), ('b', 98), ('c', 99), ('d', 100), ('e', 101), ('f', 102), ('g', 103), ('h', 104), ('i', 105), ('j', 106), ('k', 107), ('l', 108), ('m', 109), ('n', 110), ('o', 111), ('p', 112), ('q', 113), ('r', 114), ('s', 115), ('t', 116), ('u', 117), ('v', 118), ('w', 119), ('x', 120), ('y', 121), ('z', 122)]
>>> [(x, ord(x)) for x in abc.upper()]
[('A', 65), ('B', 66), ('C', 67), ('D', 68), ('E', 69), ('F', 70), ('G', 71), ('H', 72), ('I', 73), ('J', 74), ('K', 75), ('L', 76), ('M', 77), ('N', 78), ('O', 79), ('P', 80), ('Q', 81), ('R', 82), ('S', 83), ('T', 84), ('U', 85), ('V', 86), ('W', 87), ('X', 88), ('Y', 89), ('Z', 90)]

---

Related Topics