How do I convert a string to an integer in JavaScript?
The simplest way would be to use the native Number function:
var x = Number("1000")
If that doesn't work for you, then there are the parseInt, unary plus, parseFloat with floor, and Math.round methods.
parseInt()
var x = parseInt("1000", 10); // You want to use radix 10
// So you get a decimal number even with a leading 0 and an old browser ([IE8, Firefox 20, Chrome 22 and older][1])
Unary plus
If your string is already in the form of an integer:
var x = +"1000";
floor()
If your string is or might be a float and you want an integer:
var x = Math.floor("1000.01"); // floor() automatically converts string to number
Or, if you're going to be using Math.floor several times:
var floor = Math.floor;
var x = floor("1000.01");
parseFloat()
If you're the type who forgets to put the radix in when you call parseInt, you can use parseFloat and round it however you like. Here I use floor.
var floor = Math.floor;
var x = floor(parseFloat("1000.01"));
round()
Interestingly, Math.round (like Math.floor) will do a string to number conversion, so if you want the number rounded (or if you have an integer in the string), this is a great way, maybe my favorite:
var round = Math.round;
var x = round("1000"); // Equivalent to round("1000", 0)
Try parseInt function:
var number = parseInt("10");
But there is a problem. If you try to convert "010" using parseInt function, it detects as octal number, and will return number 8. So, you need to specify a radix (from 2 to 36). In this case base 10.
parseInt(string, radix)
Example:
var result = parseInt("010", 10) == 10; // Returns true
var result = parseInt("010") == 10; // Returns false
Note that parseInt ignores bad data after parsing anything valid.
This guid will parse as 51:
var result = parseInt('51e3daf6-b521-446a-9f5b-a1bb4d8bac36', 10) == 51; // Returns true
There are two main ways to convert a string to a number in JavaScript. One way is to parse it and the other way is to change its type to a Number. All of the tricks in the other answers (e.g., unary plus) involve implicitly coercing the type of the string to a number. You can also do the same thing explicitly with the Number function.
Parsing
var parsed = parseInt("97", 10);
parseInt and parseFloat are the two functions used for parsing strings to numbers. Parsing will stop silently if it hits a character it doesn't recognise, which can be useful for parsing strings like "92px", but it's also somewhat dangerous, since it won't give you any kind of error on bad input, instead you'll get back NaN unless the string starts with a number. Whitespace at the beginning of the string is ignored. Here's an example of it doing something different to what you want, and giving no indication that anything went wrong:
var widgetsSold = parseInt("97,800", 10); // widgetsSold is now 97
It's good practice to always specify the radix as the second argument. In older browsers, if the string started with a 0, it would be interpreted as octal if the radix wasn't specified which took a lot of people by surprise. The behaviour for hexadecimal is triggered by having the string start with 0x if no radix is specified, e.g., 0xff. The standard actually changed with ECMAScript 5, so modern browsers no longer trigger octal when there's a leading 0 if no radix has been specified. parseInt understands radixes up to base 36, in which case both upper and lower case letters are treated as equivalent.
Changing the Type of a String to a Number
All of the other tricks mentioned above that don't use parseInt, involve implicitly coercing the string into a number. I prefer to do this explicitly,
var cast = Number("97");
This has different behavior to the parse methods (although it still ignores whitespace). It's more strict: if it doesn't understand the whole of the string than it returns NaN, so you can't use it for strings like 97px. Since you want a primitive number rather than a Number wrapper object, make sure you don't put new in front of the Number function.
Obviously, converting to a Number gives you a value that might be a float rather than an integer, so if you want an integer, you need to modify it. There are a few ways of doing this:
var rounded = Math.floor(Number("97.654")); // other options are Math.ceil, Math.round
var fixed = Number("97.654").toFixed(0); // rounded rather than truncated
var bitwised = Number("97.654")|0; // do not use for large numbers
Any bitwise operator (here I've done a bitwise or, but you could also do double negation as in an earlier answer or a bit shift) will convert the value to a 32 bit integer, and most of them will convert to a signed integer. Note that this will not do want you want for large integers. If the integer cannot be represented in 32 bits, it will wrap.
~~"3000000000.654" === -1294967296
// This is the same as
Number("3000000000.654")|0
"3000000000.654" >>> 0 === 3000000000 // unsigned right shift gives you an extra bit
"300000000000.654" >>> 0 === 3647256576 // but still fails with larger numbers
To work correctly with larger numbers, you should use the rounding methods
Math.floor("3000000000.654") === 3000000000
// This is the same as
Math.floor(Number("3000000000.654"))
Bear in mind that coercion understands exponential notation and Infinity, so 2e2 is 200 rather than NaN, while the parse methods don't.
Custom
It's unlikely that either of these methods do exactly what you want. For example, usually I would want an error thrown if parsing fails, and I don't need support for Infinity, exponentials or leading whitespace. Depending on your use case, sometimes it makes sense to write a custom conversion function.
Always check that the output of Number or one of the parse methods is the sort of number you expect. You will almost certainly want to use isNaN to make sure the number is not NaN (usually the only way you find out that the parse failed).
ParseInt() and + are different
parseInt("10.3456") // returns 10
+"10.3456" // returns 10.3456
Fastest
var x = "1000"*1;
Test
Here is little comparison of speed (macOS only)... :)
For Chrome, 'plus' and 'mul' are fastest (>700,000,00 op/sec), 'Math.floor' is slowest. For Firefox, 'plus' is slowest (!) 'mul' is fastest (>900,000,000 op/sec). In Safari 'parseInt' is fastest, 'number' is slowest (but results are quite similar, >13,000,000 <31,000,000). So Safari for cast string to int is more than 10x slower than other browsers. So the winner is 'mul' :)
You can run it on your browser by this link
https://jsperf.com/js-cast-str-to-number/1
I also tested var x = ~~"1000";. On Chrome and Safari, it is a little bit slower than var x = "1000"*1 (<1%), and on Firefox it is a little bit faster (<1%).
I use this way of converting string to number:
var str = "25"; // String
var number = str*1; // Number
So, when multiplying by 1, the value does not change, but JavaScript automatically returns a number.
But as it is shown below, this should be used if you are sure that the str is a number (or can be represented as a number), otherwise it will return NaN - not a number.
You can create simple function to use, e.g.,
function toNumber(str) {
return str*1;
}
Try parseInt.
var number = parseInt("10", 10); //number will have value of 10.
I love this trick:
~~"2.123"; //2
~~"5"; //5
The double bitwise negative drops off anything after the decimal point AND converts it to a number format. I've been told it's slightly faster than calling functions and whatnot, but I'm not entirely convinced.
Another method I just saw here (a question about the JavaScript >>> operator, which is a zero-fill right shift) which shows that shifting a number by 0 with this operator converts the number to a uint32 which is nice if you also want it unsigned. Again, this converts to an unsigned integer, which can lead to strange behaviors if you use a signed number.
"-2.123" >>> 0; // 4294967294
"2.123" >>> 0; // 2
"-5" >>> 0; // 4294967291
"5" >>> 0; // 5
In JavaScript, you can do the following:
ParseInt
parseInt("10.5") // Returns 10
Multiplying with 1
var s = "10";
s = s*1; // Returns 10
Using the unary operator (+)
var s = "10";
s = +s; // Returns 10
Using a bitwise operator
(Note: It starts to break after 2140000000. Example: ~~"2150000000" = -2144967296)
var s = "10.5";
s = ~~s; // Returns 10
Using Math.floor() or Math.ceil()
var s = "10";
s = Math.floor(s) || Math.ceil(s); // Returns 10
Please see the below example. It will help answer your question.
Example Result
parseInt("4") 4
parseInt("5aaa") 5
parseInt("4.33333") 4
parseInt("aaa"); NaN (means "Not a Number")
By using parseint function, it will only give op of integer present and not the string.
Beware if you use parseInt to convert a float in scientific notation!
For example:
parseInt("5.6e-14")
will result in
5
instead of
0
Also as a side note: MooTools has the function toInt() which is used on any native string (or float (or integer)).
"2".toInt() // 2
"2px".toInt() // 2
2.toInt() // 2
We can use +(stringOfNumber) instead of using parseInt(stringOfNumber).
Example: +("21") returns int of 21, like the parseInt("21").
We can use this unary "+" operator for parsing float too...
To convert a String into Integer, I recommend using parseFloat and not parseInt. Here's why:
Using parseFloat:
parseFloat('2.34cms') //Output: 2.34
parseFloat('12.5') //Output: 12.5
parseFloat('012.3') //Output: 12.3
Using parseInt:
parseInt('2.34cms') //Output: 2
parseInt('12.5') //Output: 12
parseInt('012.3') //Output: 12
So if you have noticed parseInt discards the values after the decimals, whereas parseFloat lets you work with floating point numbers and hence more suitable if you want to retain the values after decimals. Use parseInt if and only if you are sure that you want the integer value.
There are many ways in JavaScript to convert a string to a number value... All are simple and handy. Choose the way which one works for you:
var num = Number("999.5"); //999.5
var num = parseInt("999.5", 10); //999
var num = parseFloat("999.5"); //999.5
var num = +"999.5"; //999.5
Also, any Math operation converts them to number, for example...
var num = "999.5" / 1; //999.5
var num = "999.5" * 1; //999.5
var num = "999.5" - 1 + 1; //999.5
var num = "999.5" - 0; //999.5
var num = Math.floor("999.5"); //999
var num = ~~"999.5"; //999
My prefer way is using + sign, which is the elegant way to convert a string to number in JavaScript.
Try str - 0 to convert string to number.
> str = '0'
> str - 0
0
> str = '123'
> str - 0
123
> str = '-12'
> str - 0
-12
> str = 'asdf'
> str - 0
NaN
> str = '12.34'
> str - 0
12.34
Here are two links to compare the performance of several ways to convert string to int
https://jsperf.com/number-vs-parseint-vs-plus
http://phrogz.net/js/string_to_number.html
Here is the easiest solution
let myNumber = "123" | 0;
More easy solution
let myNumber = +"123";
In my opinion, no answer covers all edge cases as parsing a float should result in an error.
function parseInteger(value) {
if(value === '') return NaN;
const number = Number(value);
return Number.isInteger(number) ? number : NaN;
}
parseInteger("4") // 4
parseInteger("5aaa") // NaN
parseInteger("4.33333") // NaN
parseInteger("aaa"); // NaN
The easiest way would be to use + like this
const strTen = "10"
const numTen = +strTen // string to number conversion
console.log(typeof strTen) // string
console.log(typeof numTen) // number
I actually needed to "save" a string as an integer, for a binding between C and JavaScript, so I convert the string into an integer value:
/*
Examples:
int2str( str2int("test") ) == "test" // true
int2str( str2int("t€st") ) // "t¬st", because "€".charCodeAt(0) is 8364, will be AND'ed with 0xff
Limitations:
maximum 4 characters, so it fits into an integer
*/
function str2int(the_str) {
var ret = 0;
var len = the_str.length;
if (len >= 1) ret += (the_str.charCodeAt(0) & 0xff) << 0;
if (len >= 2) ret += (the_str.charCodeAt(1) & 0xff) << 8;
if (len >= 3) ret += (the_str.charCodeAt(2) & 0xff) << 16;
if (len >= 4) ret += (the_str.charCodeAt(3) & 0xff) << 24;
return ret;
}
function int2str(the_int) {
var tmp = [
(the_int & 0x000000ff) >> 0,
(the_int & 0x0000ff00) >> 8,
(the_int & 0x00ff0000) >> 16,
(the_int & 0xff000000) >> 24
];
var ret = "";
for (var i=0; i<4; i++) {
if (tmp[i] == 0)
break;
ret += String.fromCharCode(tmp[i]);
}
return ret;
}
String to Number in JavaScript:
Unary + (most recommended)
+numStr is easy to use and has better performance compared with others
Supports both integers and decimals
console.log(+'123.45') // => 123.45
Some other options:
Parsing Strings:
parseInt(numStr) for integers
parseFloat(numStr) for both integers and decimals
console.log(parseInt('123.456')) // => 123
console.log(parseFloat('123')) // => 123
JavaScript Functions
Math functions like round(numStr), floor(numStr), ceil(numStr) for integers
Number(numStr) for both integers and decimals
console.log(Math.floor('123')) // => 123
console.log(Math.round('123.456')) // => 123
console.log(Math.ceil('123.454')) // => 124
console.log(Number('123.123')) // => 123.123
Unary Operators
All basic unary operators, +numStr, numStr-0, 1*numStr, numStr*1, and numStr/1
All support both integers and decimals
Be cautious about numStr+0. It returns a string.
console.log(+'123') // => 123
console.log('002'-0) // => 2
console.log(1*'5') // => 5
console.log('7.7'*1) // => 7.7
console.log(3.3/1) // =>3.3
console.log('123.123'+0, typeof ('123.123' + 0)) // => 123.1230 string
Bitwise Operators
Two tilde ~~numStr or left shift 0, numStr<<0
Supports only integers, but not decimals
console.log(~~'123') // => 123
console.log('0123'<<0) // => 123
console.log(~~'123.123') // => 123
console.log('123.123'<<0) // => 123
// Parsing
console.log(parseInt('123.456')) // => 123
console.log(parseFloat('123')) // => 123
// Function
console.log(Math.floor('123')) // => 123
console.log(Math.round('123.456')) // => 123
console.log(Math.ceil('123.454')) // => 124
console.log(Number('123.123')) // => 123.123
// Unary
console.log(+'123') // => 123
console.log('002'-0) // => 2
console.log(1*'5') // => 5
console.log('7.7'*1) // => 7.7
console.log(3.3/1) // => 3.3
console.log('123.123'+0, typeof ('123.123'+0)) // => 123.1230 string
// Bitwise
console.log(~~'123') // => 123
console.log('0123'<<0) // => 123
console.log(~~'123.123') // => 123
console.log('123.123'<<0) // => 123
function parseIntSmarter(str) {
// ParseInt is bad because it returns 22 for "22thisendsintext"
// Number() is returns NaN if it ends in non-numbers, but it returns 0 for empty or whitespace strings.
return isNaN(Number(str)) ? NaN : parseInt(str, 10);
}
You can use plus.
For example:
var personAge = '24';
var personAge1 = (+personAge)
then you can see the new variable's type bytypeof personAge1 ; which is number.
Summing the multiplication of digits with their respective power of ten:
i.e: 123 = 100+20+3 = 1100 + 2+10 + 31 = 1*(10^2) + 2*(10^1) + 3*(10^0)
function atoi(array) {
// Use exp as (length - i), other option would be
// to reverse the array.
// Multiply a[i] * 10^(exp) and sum
let sum = 0;
for (let i = 0; i < array.length; i++) {
let exp = array.length - (i+1);
let value = array[i] * Math.pow(10, exp);
sum += value;
}
return sum;
}
The safest way to ensure you get a valid integer:
let integer = (parseInt(value, 10) || 0);
Examples:
// Example 1 - Invalid value:
let value = null;
let integer = (parseInt(value, 10) || 0);
// => integer = 0
// Example 2 - Valid value:
let value = "1230.42";
let integer = (parseInt(value, 10) || 0);
// => integer = 1230
// Example 3 - Invalid value:
let value = () => { return 412 };
let integer = (parseInt(value, 10) || 0);
// => integer = 0
Another option is to double XOR the value with itself:
var i = 12.34;
console.log('i = ' + i);
console.log('i ⊕ i ⊕ i = ' + (i ^ i ^ i));
This will output:
i = 12.34
i ⊕ i ⊕ i = 12
I only added one plus(+) before string and that was solution!
+"052254" // 52254
Number()
Number(" 200.12 ") // Returns 200.12
Number("200.12") // Returns 200.12
Number("200") // Returns 200
parseInt()
parseInt(" 200.12 ") // Return 200
parseInt("200.12") // Return 200
parseInt("200") // Return 200
parseInt("Text information") // Returns NaN
parseFloat()
It will return the first number
parseFloat("200 400") // Returns 200
parseFloat("200") // Returns 200
parseFloat("Text information") // Returns NaN
parseFloat("200.10") // Return 200.10
Math.floor()
Round a number to the nearest integer
Math.floor(" 200.12 ") // Return 200
Math.floor("200.12") // Return 200
Math.floor("200") // Return 200
function doSth(){
var a = document.getElementById('input').value;
document.getElementById('number').innerHTML = toNumber(a) + 1;
}
function toNumber(str){
return +str;
}
<input id="input" type="text">
<input onclick="doSth()" type="submit">
<span id="number"></span>
This (probably) isn't the best solution for parsing an integer, but if you need to "extract" one, for example:
"1a2b3c" === 123
"198some text2hello world!30" === 198230
// ...
this would work (only for integers):
var str = '3a9b0c3d2e9f8g'
function extractInteger(str) {
var result = 0;
var factor = 1
for (var i = str.length; i > 0; i--) {
if (!isNaN(str[i - 1])) {
result += parseInt(str[i - 1]) * factor
factor *= 10
}
}
return result
}
console.log(extractInteger(str))
Of course, this would also work for parsing an integer, but would be slower than other methods.
You could also parse integers with this method and return NaN if the string isn't a number, but I don't see why you'd want to since this relies on parseInt internally and parseInt is probably faster.
var str = '3a9b0c3d2e9f8g'
function extractInteger(str) {
var result = 0;
var factor = 1
for (var i = str.length; i > 0; i--) {
if (isNaN(str[i - 1])) return NaN
result += parseInt(str[i - 1]) * factor
factor *= 10
}
return result
}
console.log(extractInteger(str))
Question:
Write an efficient algorithm that searches for a value in an m x n matrix.
This matrix has the following properties:
Integers in each row are sorted from left to right.
The first integer of each row is greater than the last integer of the previous row.
class Solution {
public:
/**
* #param matrix, a list of lists of integers
* #param target, an integer
* #return a boolean, indicate whether matrix contains target
*/
bool searchMatrix(vector<vector<int> > &matrix, int target) {
bool flag = false;
int mbegin = 0, mend = matrix.size() - 1;
// target in matrix?
while (mbegin <= mend) {
int mmid = (mbegin + mend) / 2;
vector<int> sub = matrix[mmid];
int begin = 0, end = sub.size() - 1;
// target in sub?
if (sub[0] <= target && target <= sub[end]) {
while(begin <= end) {
int mid = (begin + end) / 2;
if (sub[mid] == target) {
flag = true;
return flag;
}
else if (sub[mid] < target) {
begin = mid + 1;
}
else {
end = mid - 1;
}
}
}
else if (sub[end] < target) {
mbegin = mmid + 1;
}
else{
mend = mmid - 1;
}
}
return flag;
}
};
input
[[1,5,9,13,18,25,28,33,39,43,44,51,55,56,63,68,73,77,80],[100,117,141,164,181,199,218,241,265,285,310,326,341,354,370,380,397,408,420],[433,453,475,494,506,518,536,550,568,585,609,626,651,662,676,698,716,729,746],[766,791,809,829,844,869,894,916,930,947,967,981,992,1012,1036,1059,1083,1099,1116],[1137,1150,1163,1177,1199,1215,1231,1243,1268,1283,1304,1326,1344,1362,1387,1408,1428,1438,1461],[1485,1499,1521,1546,1566,1584,1606,1622,1637,1656,1681,1705,1726,1744,1758,1779,1802,1827,1842]]
1084
Expected
false
but there are nothing showed up! why?
Taking into account the two properties, here are the optimizations possible for your algorithm:
You can skip searching rows that begin with a number greater than the one you seek.
In each row, when you reach a number greater than the one you seek, you can stop searching.
You have to look in all rows that can't be excluded by these criteria. The specification doesn't if you must find all occurences. It also doesn't tell if the value can be missing, if it is guaranteed to occur once or if it can occur multiple times.
With int mmid = (mbegin + mend) / 2;, it looks as if you are trying to do a binary search.
I am writing a piece of code in c# to retreive number of tablets for a given dosage. For example, if a Dosage is 20 mg of DrugA (if DrugA comes in 10mg, 5mg and 2mg tablets) then the code would return (2). If Dosage is 15 then the code would return (1 & 1). If a dosage is 3 then Invalid Dosage message is returned. The code must use the highest denominations first i.e. 10mg tablets and then 5mg tablets for the remainder and so on. I am using recursive function (GetDispenseBreakdownForSingleDosage) to acheive the above functionality. My code is working fine for most of the scenarios that I tested. The one scenario that it is incorrectly returning Invalid Dosage is for 8mg dosage. The code should return (4) since 2mg tablets is a valid option. I have given my code below. My questions are:
1) Is there a better way of acheiving my objective than using my code.
2) What changes should I make to avoid the trap of 8mg as invalid dosage. It is returning it invalid because code divides 8 with 5 during second recursive call and remainder becomes 3, on third recursive call 3 is not divisible by 2 so code returns invalid dosage.
My code is given below:
public string GetDispenseBreakdown(PrescriptionsBLL Prescription, double[] IndexAndNonIndexDosageForBreakdown)
{
int[] NoOfTablets = new int[Prescription.SelectedDrug.PrescriptionsDrugWeights.Count];
for (int Index = 1; Index <= IndexAndNonIndexDosageForBreakdown.Length; Index++)
{
GetDispenseBreakdownForSingleDosage(Prescription, ref NoOfTablets, IndexAndNonIndexDosageForBreakdown[(Index - 1)], Prescription.SelectedDrug.PrescriptionsDrugWeights[0].Weight, 1);//assuming that index 0 will always contain the highest weight i.e. if a drug has 2, 5, 10 as drug weights then index 0 should always contain 10 as we are sorting by Desc
}
return ConvertNumberOfTabletsIntoString(NoOfTablets);
}
public void GetDispenseBreakdownForSingleDosage(PrescriptionsBLL Prescription, ref int[] NoOfTablets, double Dosage, double Weight, int WeightCount)
{
int LoopIteration;
string TempLoopIteration = (Dosage / Weight).ToString();
if (TempLoopIteration.Contains("."))
LoopIteration = (int)Math.Floor(Dosage / Weight);
else
LoopIteration = int.Parse(TempLoopIteration);
double TempDosage = Weight * LoopIteration;
int WeightTablets = LoopIteration;
double RemainingDosage = Math.Round((Dosage - TempDosage), 2);
NoOfTablets[(WeightCount - 1)] = NoOfTablets[(WeightCount - 1)] + WeightTablets;
if (WeightCount == Prescription.SelectedDrug.PrescriptionsDrugWeights.Count && RemainingDosage > 0.0)
{
NoOfTablets[0] = -99999;//Invalid Dosage
return;
}
if (LoopIteration == 0 && Dosage > 0.0 && WeightCount == Prescription.SelectedDrug.PrescriptionsDrugWeights.Count)
{
NoOfTablets[0] = -99999;//Invalid Dosage
return;
}
if (WeightCount == Prescription.SelectedDrug.PrescriptionsDrugWeights.Count)
return;
GetDispenseBreakdownForSingleDosage(Prescription, ref NoOfTablets, RemainingDosage, Prescription.SelectedDrug.PrescriptionsDrugWeights[WeightCount].Weight, ++WeightCount);
}
public bool IsDosageValid(int[] NoOfTablets)
{
if (NoOfTablets[0] == -99999)
return false;
else
return true;
}
public string ConvertNumberOfTabletsIntoString(int[] NoOfTablets)
{
if (!IsDosageValid(NoOfTablets))
return "Dosage is Invalid";
string DispenseBreakDown = "(";
int ItemsAdded = 0;
for (int Count = 0; Count < NoOfTablets.Length; Count++)
{
if (NoOfTablets[Count] != 0)
{
if (ItemsAdded > 0)
DispenseBreakDown += " & " + NoOfTablets[Count];
else
DispenseBreakDown += NoOfTablets[Count];
ItemsAdded = ItemsAdded + 1;
}
}
DispenseBreakDown += ")";
return DispenseBreakDown;
}
This sounds like a version of the same logic required for coin change.
This site goes through that logic:
http://www.geeksforgeeks.org/dynamic-programming-set-7-coin-change/
You will also need to make a few adjustments:
You'll need to get back the possible results and accept the one that has highest number of larger pills.
You'll need to handle the possibility of no "correct change".
Here is a simple recursive method. Pass it the desired dosage and an empty list:
// Test if 2 floats are "equal", the difference between them
// is less than some predefined value (epsilon)
bool floatIsEqual(float f1, float f2)
{
float epsilon = 0.001f;
return Math.Abs(f1 - f2) <= epsilon;
}
static bool CalcDose(float desired, List<float> list)
{
// Order of array is important. Larger values will be attempted first
float[] sizes = new float[] { 8, 2, .4f, .2f };
// This path isn't working, return
if (desired < sizes[sizes.Length - 1])
{
return false;
}
// Try all combos
for (int i = 0; i < sizes.Length; i++)
{
if (floatIsEqual(desired, sizes[i]))
{
// Final step: perfect match
list.Add(sizes[i]);
return true;
}
if (sizes[i] <= desired)
{
// Attempt recursive call
if (true == CalcDose( desired - sizes[i], list))
{
// Success
list.Add(sizes[i]);
return true;
}
else break;
}
}
return false;
}