Develop Reference

Undefined in Split String

i have a function to split string into 2 part, front and back. Then reverse it to back and front. Here is my code
function reverseString(string) {
let splitString = ""
let firstString = ""
for(i = 0; i <= string.length/2 - 1; i++) {
firstString += string[i]
}
for(i = string.length/2; i <= string.length; i++) {
splitString += string[i]
}
return splitString + firstString
}
Sorry for bad explanation, this is test case and expected result (first one is expected result, the second one is my result)
console.log(reverseString("aaabccc")); // "cccbaaa" "undefinedundefinedundefinedundefinedaaa"
console.log(reverseString("aab")); // "baa" "undefinedundefineda"
console.log(reverseString("aaaacccc")); // "ccccaaaa" "ccccundefinedaaa"
console.log(reverseString("abcdefghabcdef")); // "habcdefabcdefg" "habcdefundefinedabcdefg"
could you help me, whats wrong with it. Thank you

You could try another approach and use the slice function
function reverseString(string)
{
if (string.length < 2) { return string; }
let stringHalfLength = string.length / 2;
let isLengthOdd = stringHalfLength % 1 !== 0;
if (isLengthOdd) {
return string.slice(Math.ceil(stringHalfLength), string.length + 1) + string[Math.floor(stringHalfLength)] + string.slice(0, Math.floor(stringHalfLength));
}
return string.slice(stringHalfLength, string.length + 1) + string.slice(0, stringHalfLength);
}
console.log(reverseString("aaabccc") === "cccbaaa");
console.log(reverseString("aab") === "baa");
console.log(reverseString("aaaacccc") === "ccccaaaa");
console.log(reverseString("abcdefghabcdef") === "habcdefabcdefg");

A more efficient way to reverse the string would be to split the string, then use the built-in reverse javascript function (which reverses the elements of the split string), and then re-join the elements using the join function.. No need to re-invent the wheel?
You can concatenate the functions in shorthand (.split.reverse.join etc...) so your function would look something like this:
function reverseString(string) {
return string.split("").reverse().join("");
}
Try it out!
function reverseString(string) {
return string.split("").reverse().join("");
}
console.log(reverseString("hello"));
console.log(reverseString("aaabbbccc"));
If there's a particular reason you're opting not to use the in-built functions (i.e. if I've missed something?) , feel free to comment.

The short version of what you need:
function reverseString(string) {
const splitPosition = Math.ceil(string.length / 2);
return string.substring(splitPosition) + string.substring(0, splitPosition);
}

The key to your question is the middle element. To accomplish that, you probably want to use Math.floor that round under.
console.log(reverseString("aaabccc")); // "cccbaaa"
console.log(reverseString("abcdefghabcdef")); // "habcdefabcdefg"
function reverseString (str) {
if (str.length<2) {
return str
}
var half = Math.floor(str.length / 2);
return (str.slice(-half) + (str.length%2?str[half]:"") + str.slice(0,half));
}
reverseString('')
> ""
reverseString('1')
> "1"
reverseString('12')
> "21"
reverseString('123')
> "321"
reverseString('1234')
> "3412"
reverseString('12345')
> "45312"
reverseString("aaabccc")
> "cccbaaa"
reverseString("abcdefghabcdef")
> "habcdefabcdefg"
So basically your problem is not to grab 2 parts of the string and rearrange, it is to grab 3 parts.
1 part: str.slice(0,half)
2 part: str.length%2 ? str[half] : ""
3 part: str.slice(-half)
The second part is empty if the string length is even and the middle character if is odd.
So the code version in long self explanatory code:
function reverseString (str) {
if (str.length<2) {
return str
}
var half = Math.floor(str.length / 2);
var firstPart = str.slice(0,half);
var midlePart = str.length % 2 ? str[half] : ""; // we could expand also this
var endPart = str.slice(-half);
return endPart + midlePart + firstPart;
}
And also, notice the precondition, so I don't have to deal with the easy cases.
Also, in your code, you got undefined because you access in the last loop to:
string[string.length] you need to change <= by <

Javascript: Cut string after last specific character

I'm doing some Javascript to cut strings into 140 characters, without breaking words and stuff, but now i want the text so have some sense. so i would like if you find a character (just like ., , :, ;, etc) and if the string is>110 characters and <140 then slice it, so the text has more sense. Here is what i have done:
where texto means text, longitud means length, and arrayDeTextos means ArrayText.
Thank you.
//function to cut strings
function textToCut(texto, longitud){
if(texto.length<longitud) return texto;
else {
var cortado=texto.substring(0,longitud).split(' ');
texto='';
for(key in cortado){
if(key<(cortado.length-1)){
texto+=cortado[key]+' ';
if(texto.length>110 && texto.length<140) {
alert(texto);
}
}
}
}
return texto;
}
function textToCutArray(texto, longitud){
var arrayDeTextos=[];
var i=-1;
do{
i++;
arrayDeTextos.push(textToCut(texto, longitud));
texto=texto.replace(arrayDeTextos[i],'');
}while(arrayDeTextos[i].length!=0)
arrayDeTextos.push(texto);
for(key in arrayDeTextos){
if(arrayDeTextos[key].length==0){
delete arrayDeTextos[key];
}
}
return arrayDeTextos;
}

Break the string into sentences, then check the length of the final string before appending each sentence.
var str = "Test Sentence. Test Sentence";
var arr = str.split(/[.,;:]/) //create an array of sentences delimited by .,;:
var final_str = ''
for (var s in arr) {
if (final_str.length == 0) {
final_str += arr[s];
} else if (final_str.length + s.length < 140) {
final_str += arr[s];
}
}
alert(final_str); // should have as many full sentences as possible less than 140 characters.

I think Martin Konecny's solution doesn't work well because it excludes the delimiter and so removes lots of sense from the text.
This is my solution:
var arrTextChunks = text.split(/([,:\?!.;])/g),
finalText = "",
finalTextLength = 0;
for(var i = 0; i < arrTextChunks.length; i += 2) {
if(finalTextLength + arrTextChunks[i].length + 1 < 140) {
finalText += arrTextChunks[i] + arrTextChunks[i + 1];
finalTextLength += arrTextChunks[i].length;
} else if(finalTextLength > 110) {
break;
}
}
http://jsfiddle.net/Whre/3or7j50q/3/
I'm aware of the fact that the i += 2 part does only make sense for "common" usages of punctuation (a single dot, colon etc.) and nothing like "hi!!!?!?1!1!".

Should be a bit more effective without regex splits.
var truncate = function (str, maxLen, delims) {
str = str.substring(0, maxLen);
return str.substring(0, Math.max.apply(null, delims.map(function (s) {
return str.lastIndexOf(s);
})));
};

Try this regex, you can see how it works here: http://regexper.com/#%5E(%5Cr%5Cn%7C.)%7B1%2C140%7D%5Cb
str.match(/^(\r\n|.){1,140}\b/g).join('')

Converting exponential LaTeX syntax to PHP's pow() using JavaScript

I would like to use JS to convert a nested exponential LaTeX expression such as
2^{3^{4^5}}
to PHP's pow() syntax
pow(2,pow(3,pow(4,5)))
I know JS doesn't support recursive RegExp. The expression is part of an equation, so I expect the solution to work with something like
\frac{3}{9}+\frac{2^{\sqrt{4^2}}}{6}
which should output
\frac{3}{9}+\frac{pow(2,\sqrt{pow(4,2)})}{6}
like in #AbcAeffchen's solution.
I don't need conversion for the non-exponential parts.
Notice: The solution must not require resorting to PHP 5.6, which introduced the ** operator

An ugly hack,
> foo
'2,(3^(4^5))'
> var foo = "2^(3^(4^5))".replace(/\^/g, ",");
undefined
> foo
'2,(3,(4,5))'
> var bar = foo.replace(/(\d,)(?=[^\d])/g, "$1pow");
undefined
> var foobar = bar.replace(/^(.*)$/, "pow($1)")
undefined
> foobar
'pow(2,pow(3,pow(4,5)))'

Try this function I wrote. It just uses ordinary string functions, so it is a little bit longer.
Basically it works as the following
find the first occurrence of ^
find the part before ^ that belongs to the basis
find the part after ^ that belongs to the exponent
call the function recursive on the exponent and on the part after the exponent
put all parts together.
and so it looks in JavaScript:
function convertLatexPow(str)
{
// contains no pow
var posOfPow = str.indexOf('^');
if(posOfPow == -1)
return str;
var head = str.substr(0,posOfPow);
var tail = str.substr(posOfPow+1);
// find the beginning of pow
var headLen = posOfPow;
var beginning = 0;
var counter;
if(head[headLen-1] == '}') // find the opening brace
{
counter = 1;
for(i = headLen-2; i >= 0; i--)
{
if(head[i] == '}')
counter++;
else if(head[i] == '{')
counter--;
if(counter == 0)
{
beginning = i;
break;
}
}
}
else if(head[headLen-1].match('[0-9]{1}')) // find the beginning of the number
{
for(i = headLen-2; i >= 0; i--)
{
if(!head[i].match('[0-9]{1}'))
{
beginning = i+1;
break;
}
}
}
else // the string looks like ...abc^{..}.. so the basis is only one character ('c' in this case)
beginning = headLen-1;
var untouchedHead = head.substr(0,beginning);
var firstArg = head.substr(beginning);
// find the end of pow
var secondArg, untouchedTail;
if(tail[0] != '{')
{
secondArg = tail[0];
untouchedTail = tail.substr(1);
}
else
{
counter = 1;
var len = tail.length;
var end = len+1;
for(i = 1; i < len; i++)
{
if(tail[i] == '{')
counter++;
else if(tail[i] == '}')
counter--;
if(counter == 0)
{
end = i;
break;
}
}
secondArg = tail.substr(1,end-1);
if(end < len-1)
untouchedTail = tail.substr(end+1);
else
untouchedTail = '';
}
return untouchedHead
+ 'pow(' + firstArg + ',' + convertLatexPow(secondArg) + ')'
+ convertLatexPow(untouchedTail);
}
alert(convertLatexPow('2^{3^{4^5}}'));
alert(convertLatexPow('\\frac{3}{9}+\\frac{2^{\\sqrt{4^2}}}{6}'));
alert(convertLatexPow('{a + 2 \\cdot (b + c)}^2'));
Input: '2^{3^{4^5}}'
Output: pow(2,pow(3,pow(4,5)))
Input: '\\frac{3}{9}+\\frac{2^{\\sqrt{4^2}}}{6}'
Output: \frac{3}{9}+\frac{pow(2,\sqrt{pow(4,2)})}{6}
Input: '{a + 2 \\cdot (b + c)}^2'
Output: pow({a + 2 \cdot (b + c)},2)
Notice: It do not parse the \sqrt. you have to do this extra.
Feel free to improve it :)
Notice: ^ in LaTeX does not mean power. It just means superscript. So 2^3 becomes 23 (and looks like "2 to the power of 3"), but \sum_{i=1}^n just becomes better formatted. But you can extend the function above to ignore ^ directly after }.
Notice: As Lucas Trzesniewski mentioned in the comment, 2^3^4 is not converted "correct", but it is also not a valid LaTeX expression.
Edit: Improved the function to convert '{a + 2 \\cdot (b + c)}^2' right.
Notice: In LaTeX exists many ways to write a brace (e.g. (, \left(, [, \lbrace,..).
To be sure this function works fine with all this braces you should convert all that braces to { and } first. Or to normal braces (, but then the function has to be edited to look for ( instead of {.
Notice: The complexity of this function is O(n⋅k), where n is the length of the input and k is the number of ^ in the input. An worst case input would be the first test case 2^{3^{4^{...}}}. But in most cases the function will be much faster. Something about O(n).

You can do it iteratively:
var foo = "2^(3^(4^5))";
while (/\([^^]+\^[^^]+\)/.test(foo)) {
foo = foo.replace(/\(([^^]+)\^([^^]+)\)/, "pow($1,$2)");
}
if (/(.+)\^(.+)/.test(foo)) {
foo = "pow(" + RegExp.$1 + "," + RegExp.$2 + ")";
}
# foo == "pow(2,pow(3,pow(4,5)))"

JavaScript - How can I replace string with certain char in to a function

How can I do this using regex, replacing every string with ! wrapped in function:
Examples:
3! => fact(3)
2.321! => fact(2.321)
(3.2+1)! => fact(3.2+1)
(sqrt(2)+2/2^2)! => fact(sqrt(2)+2/2^2)

Given your examples, you don't need a regex at all:
var s = "3!"; //for example
if (s[s.length-1] === "!")
s = "fact(" + s.substr(0, s.length-1) + ")";
Not doubling the parentheses for the last case just requires another test:
var s = "(sqrt(2)+2/2^2)!"; //for example
if (s[s.length-1] === "!") {
if(s.length > 1 && s[0] === "(" && s[s.length-2] === ")")
s = "fact" + s.substr(0, s.length-1);
else
s = "fact(" + s.substr(0, s.length-1) + ")";
}

My own answer i just found out was:
Number.prototype.fact = function(n) {return fact(this,2)}
str = str.replace(/[\d|\d.\d]+/g, function(n) {return "(" + n + ")"}).replace(/\!/g, ".fact()")
But I'll see if the other answers might be much better, think they are

Here is as the op requested; using regexp:
"3*(2+1)!".replace(/([1-9\.\(\)\*\+\^\-]+)/igm,"fact($1)");
You might end up with double parentheses:
"(2+1)!".replace(/([1-9\.\(\)\*\+\^\-]+)/igm,"fact($1)");

"(sqrt(2)+2/2^2)!".replace(/(.*)!/g, "fact($1)");
Fiddle with it!
(.*)!
Match the regular expression below and capture its match into backreference number 1 (.*)
Match any single character that is not a line break character .
Between zero and unlimited times, as many times as possible, giving back as needed (greedy) *
Match the character “!” literally !

var testArr = [];
testArr.push("3!");
testArr.push("2.321!");
testArr.push("(3.2+1)!");
testArr.push("(sqrt(2)+2/2^2)!");
//Have some fun with the name. Why not?
function ohIsThatAFact(str) {
if (str.slice(-1)==="!") {
str = str.replace("!","");
if(str[0]==="(" && str.slice(-1)===")")
str = "fact"+str;
else
str = "fact("+str+")";
}
return str;
}
for (var i = 0; i < testArr.length; i++) {
var testCase = ohIsThatAFact(testArr[i]);
document.write(testCase + "<br />");
}
Fiddle example

Converting VBScript to Javascript, What is the right way of parsing source code?

I was asked to convert some VB6/VBScript code to javascript so after googling it and not finding anything I can use,I wrote a small javascript function to help me do the conversion; it's so crude and only converts (some of) the synatx, but it worked for me for the job I had...now I'm thinking of improving it but the method I used is so primitive (Regular Expression matching and replacing).
So...my question is:
What is the proper way to parse source code? is there any (not so complicated) way of doing it? and I don't want to use Exe's, it must be done entirely in Javascript. I'm not searching for ready-to-use source code (I don't think it exists!) but I want to learn how to be able to start with source code and turn it into objects (the opposite of serialization, I think?).
//here is the code:
var strs=[];
function vbsTojs(vbs){
var s = vbs;
s = HideStrings(s);
//only function block
s = s.match(/Function[\w\W]+End\s+Function/gim)[0];
//line-continuation char
s = s.replace(/_\n/gm,"");
//replace ":" with CRLF
s = s.replace(/:/gm,"\n");
//move inline comment to its own line
s = s.replace(/^(.+)'(.*)$/gim,"'$2\n$1");
//single line if -> multiple line
s = s.replace(/\bthen\b[ \t](.+)/gi,"then\n$1\nEnd If");
//alert(s);
var Vars='';
var Fx='';
var FxHead='';
var Args = '';
a=s.split('\n');
//trim
for(i=0;i<a.length;i++){
a[i]=a[i].replace(/^\s+|\s+$/,"");
}
//remove empty items
a=a.filter(function(val) { return val !== ""; });
//alert(a.join('\n'));
//function
a[0]=a[0].replace(/function\s+/i,"");
Fx = a[0].match(/^\w+/)[0];
a[0]=a[0].replace(Fx,"").replace(/[\(\)]/g,"");
a[0]=a[0].replace(/\bbyval\b/gi,"").replace(/\bbyref\b/gi,"").replace(/\boptional\b/gi,"");
a[0]=a[0].replace(/\bas\s+\w+\b/gi,"");
a[0]=a[0].replace(/\s+/g,"");
a[0]=a[0].replace(/,/gi,", ");
FxHead = "function " + Fx+ " ("+ a[0] + "){";
a[0]="";
//end function
a.length = a.length-1;
for(i=1;i<a.length;i++){
//Vars
if(a[i].search(/^dim\s+/i)>-1){
a[i]=a[i].replace(/dim\s*/i,"");
Vars += a[i] + ",";
a[i]='';
//FOR
}else if(a[i].search(/^\bFOR\b\s+/i)>-1){
a[i]=a[i].replace(/^\bFOR\b\s+/i,"");
counter = a[i].match(/^\w+/)[0];
from = a[i].match(/=\s*[\w\(\)]+/)[0];
from=from.replace(/=/,"").replace(/\s+/g,"");
a[i]=a[i].replace(counter,"").replace(from,"").replace(/\bTO\b/i,"");
to = a[i].match(/\s*[\w\(\)]+\s*/)[0];
to=to.replace(/=/,"").replace(/\s+/g,"");
a[i] = "for(" + counter + "=" + from + "; " + counter + "<=" + to + "; " + counter + "++){"
//NEXT
}else if(a[i].search(/^NEXT\b/i)>-1){
a[i] = "}";
//EXIT FOR
}else if(a[i].search(/\bEXIT\b\s*\bFOR\b/i)>-1){
a[i] = "break";
//IF
}else if(a[i].search(/^\bIF\b\s+/i)>-1){
a[i]=a[i].replace(/^\bIF\b\s+/i,"");
a[i]=a[i].replace(/\bTHEN$\b/i,"");
a[i]=a[i].replace(/=/g,"==").replace(/<>/g,"!="); //TODO: it should not replace if inside a string! <---------------
a[i]=a[i].replace(/\bOR\b/gi,"||").replace(/\bAND\b/gi,"&&"); //TODO: it should not replace if inside a string! <---------------
a[i] = "if(" + a[i] + "){";
//ELSE
}else if(a[i].search(/^ELSE/i)>-1){
a[i] = "}else{";
//END IF
}else if(a[i].search(/^END\s*IF/i)>-1){
a[i] = "}";
//WHILE
}else if(a[i].search(/^WHILE\s/i)>-1){
a[i] = a[i].replace(/^WHILE(.+)/i,"while($1){");
//WEND
}else if(a[i].search(/^WEND/i)>-1){
a[i] = "}";
//DO WHILE
}else if(a[i].search(/^DO\s+WHILE\s/i)>-1){
a[i] = a[i].replace(/^DO\s+WHILE(.+)/i,"while($1){");
//LOOP
}else if(a[i].search(/^LOOP$/i)>-1){
a[i] = "}";
//EXIT FUNCTION
}else if(a[i].search(/\bEXIT\b\s*\bFUNCTION\b/i)>-1){
a[i] = "return";
//SELECT CASE
}else if(a[i].search(/^SELECT\s+CASE(.+$)/i)>-1){
a[i]=a[i].replace(/^SELECT\s+CASE(.+$)/i,"switch($1){");
}else if(a[i].search(/^END\s+SELECT/i)>-1){
a[i] = "}";
}else if(a[i].search(/^CASE\s+ELSE/i)>-1){
a[i] = "default:";
}else if(a[i].search(/^CASE[\w\W]+$/i)>-1){
a[i] = a[i] + ":" ;
}
//CONST
else if(a[i].search(/^CONST/i)>-1){
a[i] = a[i].replace(/^CONST/i,"const");
}
else{
//alert(a[i]);
}
//COMMENT
if(a[i].search(/^\'/)>-1){
a[i]=a[i].replace(/^\'/,"//");
}else if(a[i].search(/\'.*$/)>-1){
a[i]=a[i].replace(/\'(.*)$/,"//$1");
}
}
//alert(a.join("*"));
Vars = Vars.replace(/\s*AS\s+\w+\s*/gi,"");
if(Vars!="") Vars = "var " + Vars.replace(/,$/,";").replace(/,/g,", ");
FxHead + '\n' + Vars;
a=a.filter(function(val) { return val !== ""; }) //remove empty items
for(i=0;i<a.length;i++){
if (a[i].search(/[^}{:]$/)>-1) a[i]+=";";
}
ss = FxHead + '\n' + Vars + '\n' + a.join('\n') + '\n}';
ss = ss.replace(new RegExp(Fx+"\\s*=\\s*","gi"),"return ");
ss = UnHideStrings(ss);
return jsIndenter(ss);
}
//-----------------------------------------------------
function jsIndenter(js){
var a=js.split('\n');
var margin=0;
var s = '';
//trim
for(i=0;i<a.length;i++){ a[i]=a[i].replace(/^\s+|\s+$/,""); }
//remove empty items
a=a.filter(function(val) { return val !== ""; });
for(var i=1;i<a.length;i++){
if(a[i-1].indexOf("{")>-1) margin += 4 ;
if(a[i].indexOf("}")>-1) { margin -= 4; }
if(margin<0) margin = 0;
a[i] = StrFill(margin," ") + a[i] ;
}
return a.join('\n');
}
function StrFill(Count,StrToFill){
var objStr,idx;
if(StrToFill=="" || Count==0){
return "";
}
objStr="";
for(idx=1;idx<=Count;idx++){
objStr += StrToFill;
}
return objStr;
}
function HideStrings(text){
const x = String.fromCharCode(7);
const xxx = String.fromCharCode(8);
text = text.replace(/"""/gim, '"'+xxx); //hide 3 quotes " " "
var idx=0, f=0;
while(f>-1){
f = text.search(/".+?"/gim);
if(f>-1){
strs.push(text.match(/".+?"/)[0]);
//alert(strs[idx]);
text = text.replace(/".+?"/, x+idx+x);
idx++;
}
}
//alert(text);
return text;
}
function UnHideStrings(text){
for(var i=0; i<strs.length; i++){
text = text.replace(new RegExp("\\x07"+i+"\\x07"), strs[i]);
}
//Unhide 3 quotes " " " ***BUG: causes unterminated string if triple-quotes are at the end of the string
text = text.replace(/\x08/gim,'\\"');
text = text.replace(/""/gi,'\\"');
return text;
}

The proper way to parse source code for any programming language is to use a parser. Regular expressions are a useful part of (some) parsers, but a parser is a different sort of thing. There is quite a body of research and techniques in the Computer Science literature on the subject of parsing, and it's a fascinating pursuit to study.
"Converting" a bunch of Visual Basic code to Javascript is a project that seems inherently fraught with peril and mystery. A Visual Basic parser will be just the first significant hurdle to conquer. After that, you'll need to figure out how to semantically represent the Visual Basic operations in Javascript. Depending on the original context of the code, that could be somewhat weird. (You don't mention anything about where this code all runs.)
As enriching a learning experience as this might be, it's not unlikely that translating the code by hand will (in the end) take less time and produce better results. That's particularly true if you're just now finding out that there is such a thing as a "parser".

Nice job. Sounds like you did something that might not be perfect, but it did the job.
I'd recommend looking into parsers and grammars if you want to make it more sophisticated. There are lots of parser generators that would be able to help you. You'd have to come up with a grammar for the source language, generate the lexer/parser, and then use that to generate an abstract syntax tree (AST). Once you have that, you can walk the AST and ask it to emit any code you want.
It's doable but, as Oded says, it's not trivial.