javascript query 2

#1 December 10, 2008

With reference to the following script...

var a=("123");
document.write(a + " is the value of a when first assigned" + "</br>");

var b=(a++);
document.write(b + " is the value of b when first assigned" + "</br>");
document.write(a + " is the value of a AFTER b first assigned" + "</br>");

...and its output...

123 is the value of a when first assigned
123 is the value of b when first assigned
124 is the vaule of a AFTER b is first assigned

...can anyone tell me what the statement...

var b=(a++);

...is actually doing?

I've used var b=(a++) to "convert" the numeric string a to an integer b so that I can perform various operations on string a using its integer or b form: e.g. (b+1), (b-1) etc.

However, it also seems the assignation of a to b via var b=(a++) converts a into an "unassigned" integer, or "über a" with the value (a+1)....

The thing I can't figure out is why the "integer value" of a doesn't remain exclusively associated with the variable - in this case b - to which it was very specifically assigned? Why does a "new" value of a spill or leak out of the assignation of a to b?

Essentially, how can a become an integer if I haven't made it one? How does a become "changed" by an operation I've specifically and only assigned to b?

Thanks for any clarification!

#2 December 11, 2008

a++ stands for a+1 - this is the same in PHP too.

#3 December 11, 2008

Think of it as interesting syntax.

JavaScript is a loosely and dynamically typed programming language.

If it said

var b= (a+1);

the outcome would have been the same.

But if it said

var b=(a+"1");

then the outcome would have been "1231".

This is a wanted benefit for JavaScript as it is a scripting language, instead of a programming language, where data types do not want to be assumed.

Also,

a++

is commonly used to indicate a = a + 1 in for loops: [PHP]for ($a = 0; $a < 10; $a++) { echo '$a = '.$a; }[/PHP] and is merely used as a short form.

#4 December 12, 2008

banago:

and

altonbr:

Thanks for the clarification(s)!

I think I was clear, for the most part (save for the PHP reference), on the "meaning" of a++, generally (banago), and the "syntax" of javascript, more specifically (altonbr), re: (a+1) vs. (a+"1") and so forth....

However, what still puzzles me is...

...if...

var **a**=("**123**");

...and...

var **b**=(**a++**);

...how can the output of...

document.write(**a**);

...be 124?

Thanks for any additional thoughts....

#5 December 12, 2008

maclenin said: However, what still puzzles me is...

...if...

var **a**=("**123**");

...and...

var **b**=(**a++**);

...how can the output of...

document.write(**a**);

...be 124?

Thanks for any additional thoughts....

Read where I said:

a++;
// is the same as
var a = a + 1;

// so if
var a=("123");
// and...
var b=(a++);
// equals 124

//then,
var b=(a++);
//must also equal
var b = a + 1;

// also if
var a=("123");
// and...
var b=(a + "1"); // notice the "" around the one
// than it will equal 1231 because JavaScript is running concatenation, not arithmetic

#6 December 12, 2008

maclenin said: banago:

and

altonbr:

Thanks for the clarification(s)!

I think I was clear, for the most part (save for the PHP reference), on the "meaning" of a++, generally (banago), and the "syntax" of javascript, more specifically (altonbr), re: (a+1) vs. (a+"1") and so forth....

However, what still puzzles me is...

...if...

var **a**=("**123**");

...and...

var **b**=(**a++**);

...how can the output of...

document.write(**a**);

...be 124?

Thanks for any additional thoughts....

It can't be, can it? If a = 123 and b=a++, then writing variable a would have nothing to do with the a++ in variable b.

See what I'm saying? You add 1 to a in assigning variable b, but then you call variable a, which has had nothing added to it.

#7 December 13, 2008

altonbr and jrock612

Thanks to both for the continued reflections....

However, in order for me to make certain I am getting my latest point across, I'd like to suggest that y'all join me in a little experiment:

Pop open a simple .txt editor (e.g. mousepad)...

Copy and paste the following into the .txt file (NB: nothing malicious here)...

<html>
<title>javatest.html</title>
<body>

<script type="text/javascript">

var a=("123");
var b=(a++);

document.write(b + " is the value of var b" + "</br>");
document.write(a + " is the 'curious' value of var a and the basis for my question...." + "</br>");

</script>

</body>
</html>

Save your new .txt file as, for example, "javatest.html"...

Open this new .txt file with your web brower of choice...

Observe....

Comments?

#8 December 15, 2008

<html>
<title>javatest.html</title>
<body>

<script type="text/javascript">

var a=("123");
var b=(a++);
var c=(a + "1");
var d=(a + 1);

document.write("a = " + a + "</br>"); // a = 124
document.write("b = " + b + "</br>"); // b = 123
document.write("c = " + c + "</br>"); // c = 1241
document.write("d = " + d + "</br>"); // d = 125

</script>

</body>
</html>

Must say that's pretty weird for variable 'a'. Sorry, I can't tell you what that's called or why that happens but I don't like it.

For PHP, [PHP]<?php

$a=("123"); $b=(a++); // error here $c=(a + "1"); $d=(a + 1);

echo('$a = ' + $a + "\n"); echo('$b = ' + $b + "\n"); echo('$c = ' + $c + "\n"); echo('$d = ' + $d + "\n");

?> [/PHP] it returns

Parse error: syntax error, unexpected T_INC in /home/brett/test.php on line 4

#9 December 18, 2008

I think I've found the answer...

Increment (++)

The increment operator is used as follows: var++ or ++var

This operator increments (adds one to) its operand and returns a value. If used postfix, with operator after operand (for example x++), then it returns the value before incrementing. If used prefix with operator before operand (for example, ++x), then it returns the value after incrementing.

For example, if x is 3, then the statement

y = x++

increments x to 4 and sets y to 3.

If x is 3, then the statement

y = ++x

increments x to 4 and sets y to 4.

...courtesy: http://lib.ru/JAVA/javascr/expr.html