why javascript's charAt() with a string returns the first letter

Question

I am doing some exercises in my object-oriented javascript book, I notice that this:

var a = "hello";
a.charAt('e'); // 'h'
a.charAt('adfadf'); //'h'

Why is the string in the argument seemingly evaluated to the integer 0 for the charAt() method for strings?

Edit: I was aware that the charAt()'s usage usually takes an integer, and the exercise feeds charAt() with a string, and I also was aware that the string is likely then to be coerced into an integer first, which I did verify to be NaN. Thanks Kendall, for suggesting putting this missing bit of information in the question proper

Thanks!

Answer 1

Because Number('e') is NaN, and <any nonempty string>.charAt(NaN) just returns the first character. This behavior is exactly what is laid out in the spec:

15.5.4.4 String.prototype.charAt (pos)

When the charAt method is called with one argument pos, the following steps are taken:

1. Call CheckObjectCoercible passing the this value as its argument.
2. Let S be the result of calling ToString, giving it the this value as its argument.
3. Let position be ToInteger(pos).
4. Let size be the number of characters in S.
5. If position < 0 or position ≥ size, return the empty String.
6. Return a String of length 1, containing one character from S, namely the character at position position, where the first (leftmost) character in S is considered to be at position 0, the next one at position 1, and so on.

Step 3 is the crux of the matter. ToInteger of both 'e' and 'adfadf' is 0. Why? Again, time to hit the spec:

9.4 ToInteger

The abstract operation ToInteger converts its argument to an integral numeric value. This abstract operation functions as follows:

1. Let number be the result of calling ToNumber on the input argument.
2. If number is NaN, return +0.
3. If number is +0, -0, +∞, or −∞, return number.
4. Return the result of computing sign(number) × floor(abs(number)).

We need to go deeper! What is ToNumber('e'), and what is ToNumber('adfadf')? If you're surprised that I'm once again about to quote the spec, I'm doing something wrong:

9.3.1 ToNumber Applied to the String Type

ToNumber applied to Strings applies the following grammar to the input String. If the grammar cannot interpret the String as an expansion of StringNumericLiteral, then the result of ToNumber is NaN.

...I'm not going to quote the entire grammar for StringNumericLiteral. Because 'e' and 'adfadf' are neither StrDecimalLiteral s nor HexIntegerLiteral s, ToNumber of both of those values is NaN. Finally we have the conversion: from string to NaN to 0, which brings us back up the chain to charAt: position is 0, so charAt('e') and charAt('adfadf') both return the leftmost character in S.

Now, if those strings were instead valid StrNumericLiteral s, such as '0xe' and '0xadfadf':

> 'hello'.charAt('0xe')
  ""
> 'hello'.charAt('0xadfadf')
  ""

well, that's a different story for a different answer.

Answer 2

The JavaScript core string.charAt(idx) method takes an integer argument as the index for which character to return.

'abc'.charAt(0); // => 'a'

If you give it a non-integer argument then it will presumably attempt to convert the argument to a number using either the Number(arg) converter or possibly parseInt(arg, 10). Both of these functions return NaN if given a string that doesn't parse as an integer, so the charAt() function must convert NaN to zero automatically:

Number('e'); // => NaN
parseInt('e', 10); // => NaN
'abc'.charAt(NaN); // => 'a'

Perhaps directly related, calling charAt() with no argument returns the first character:

'abc'.charAt(); // => 'a'

Answer 3

When you call String.charAt(pos), it first evaluates the toInteger value of pos, and since you gave it 'e', that evaluates to 0, resulting in your answer. For more information, see below:

---

According to http://ecma262-5.com/ELS5_HTML.htm#Section_15.5.4.4

String.prototype.charAt (pos)

Returns a String containing the character at position pos in the String resulting from converting this object to a String.

If pos is a value of Number type that is an integer, then the result of x.charAt(pos) is equal to the result of x.substring(pos, pos+1).

When the charAt method is called with one argument pos, the following steps are taken:

1. Call CheckObjectCoercible passing the this value as its argument.

2. Let S be the result of calling ToString, giving it the this value as its argument.

   3. Let position be ToInteger(pos).

3. Let size be the number of characters in S.
4. If position < 0 or position ≥ size, return the empty String.

According to http://ecma262-5.com/ELS5_HTML.htm#Section_9.4

9.4 ToInteger

The abstract operation ToInteger converts its argument to an integral numeric value. This abstract operation functions as follows:

1. Let number be the result of calling ToNumber on the input argument.

   2. If number is NaN, return +0.

2. If number is +0, −0, +∞ or –∞, return number.
3. Return the result of computing sign(number) * floor(abs(number)).

Answer 4

The charAt() method takes an integer as its parameter. Passing a string will evaluate to NaN, false, 0. Therefore, the first character in the string will always be returned.

Answer 5

Because charAt tries to convert what you give it to a number. Converting "adfadf" to a number gives you NaN, or Not A Number. The defined behavior for charAt with NaN is to return the first character.

Answer 6

Both 'e' and 'adfadf' are strings. Strings when are inputed as integer values are always returning 0 as a value.

Since "hello" is an array of chars itself, it will return you the first index which is 'h' (starting from 0).

Answer 7

The charAt function takes an integer, not a string, as an argument:

"hello".charAt(0) // 'h'
"hello".charAt(1) // 'e'
"hello".charAt(2) // 'l'

And when you give it a string, it basically ends up interpreting it as a 0 and returns the first character.

Answer 8

Friend charAt method is like

string.charAt(index)

i.e. you need to pass a index to method

if you want to do same thing than use code like

string.charAt(string.indexOf('a'));

hope this will help you.