how to compare two objects using viewstate - asp.net

how to compare two objects using viewstate.
what is the meaning of below line.
if (!((byte[])ViewState["ROW"]).SequenceEqual(obj.RowID))
{
return null
}
can anyone please help on this

ViewState["ROW"] : This part will retrieve data from the ViewState stored with key ROW
(byte[])ViewState["ROW"] : This part will cast your data stored in ViewState to byte array
SequenceEqual : is extension method from System.Linq, which checks whether two sequences are same or not
((byte[])ViewState["ROW"]).SequenceEqual(obj.RowID) : Compares sequence of ViewState["Row"] and obj.RowID
if (!((byte[])ViewState["ROW"]).SequenceEqual(obj.RowID)) : This will return null if sequences of ViewState["Row"] and obj.RowID are not same.

what is the meaning of below line.
Basically, SequenceEqual is a LINQ Enumerable extension function which desinged to determine if a source sequence (e.g. byte[]) is equals to another sequence.
Assuming you are comparing two byte arrays (sequenses) in your provided code, if they both equals in their sequence of elements, you will get true, otherwise, false would be the result.
For example, the following sequences are equals and the SequenceEqual will return true:
byte[] chars1 = {56,32,12,32,65, 87};
byte[] chars2 = {56,32,12,32,65, 87};
bool res = chars1.SequenceEqual(chars2); // Will return true

Related

Swiftui: how do you assign the value in a "String?" object to a "String" object?

Swiftui dictionaries have the feature that the value returned by using key access is always of type "optional". For example, a dictionary that has type String keys and type String values is tricky to access because each returned value is of type optional.
An obvious need is to assign x=myDictionary[key] where you are trying to get the String of the dictionary "value" into the String variable x.
Well this is tricky because the String value is always returned as an Optional String, usually identified as type String?.
So how is it possible to convert the String?-type value returned by the dictionary access into a plain String-type that can be assigned to a plain String-type variable?
I guess the problem is that there is no way to know for sure that there exists a dictionary value for the key. The key used to access the dictionary could be anything so somehow you have to deal with that.
As described in #jnpdx answer to this SO question (How do you assign a String?-type object to a String-type variable?), there are at least three ways to convert a String? to a String:
import SwiftUI
var x: Double? = 6.0
var a = 2.0
if x != nil {
a = x!
}
if let b = x {
a = x!
}
a = x ?? 0.0
Two key concepts:
Check the optional to see if it is nil
if the optional is not equal to nil, then go ahead
In the first method above, "if x != nil" explicitly checks to make sure x is not nil be fore the closure is executed.
In the second method above, "if let a = b" will execute the closure as long as b is not equal to nil.
In the third method above, the "nil-coalescing" operator ?? is employed. If x=nil, then the default value after ?? is assigned to a.
The above code will run in a playground.
Besides the three methods above, there is at least one other method using "guard let" but I am uncertain of the syntax.
I believe that the three above methods also apply to variables other than String? and String.

New to coding, Can not get this to work [duplicate]

This question's answers are a community effort. Edit existing answers to improve this post. It is not currently accepting new answers or interactions.
I've been using the == operator in my program to compare all my strings so far.
However, I ran into a bug, changed one of them into .equals() instead, and it fixed the bug.
Is == bad? When should it and should it not be used? What's the difference?
== tests for reference equality (whether they are the same object).
.equals() tests for value equality (whether they contain the same data).
Objects.equals() checks for null before calling .equals() so you don't have to (available as of JDK7, also available in Guava).
Consequently, if you want to test whether two strings have the same value you will probably want to use Objects.equals().
// These two have the same value
new String("test").equals("test") // --> true
// ... but they are not the same object
new String("test") == "test" // --> false
// ... neither are these
new String("test") == new String("test") // --> false
// ... but these are because literals are interned by
// the compiler and thus refer to the same object
"test" == "test" // --> true
// ... string literals are concatenated by the compiler
// and the results are interned.
"test" == "te" + "st" // --> true
// ... but you should really just call Objects.equals()
Objects.equals("test", new String("test")) // --> true
Objects.equals(null, "test") // --> false
Objects.equals(null, null) // --> true
You almost always want to use Objects.equals(). In the rare situation where you know you're dealing with interned strings, you can use ==.
From JLS 3.10.5. String Literals:
Moreover, a string literal always refers to the same instance of class String. This is because string literals - or, more generally, strings that are the values of constant expressions (§15.28) - are "interned" so as to share unique instances, using the method String.intern.
Similar examples can also be found in JLS 3.10.5-1.
Other Methods To Consider
String.equalsIgnoreCase() value equality that ignores case. Beware, however, that this method can have unexpected results in various locale-related cases, see this question.
String.contentEquals() compares the content of the String with the content of any CharSequence (available since Java 1.5). Saves you from having to turn your StringBuffer, etc into a String before doing the equality comparison, but leaves the null checking to you.
== tests object references, .equals() tests the string values.
Sometimes it looks as if == compares values, because Java does some behind-the-scenes stuff to make sure identical in-line strings are actually the same object.
For example:
String fooString1 = new String("foo");
String fooString2 = new String("foo");
// Evaluates to false
fooString1 == fooString2;
// Evaluates to true
fooString1.equals(fooString2);
// Evaluates to true, because Java uses the same object
"bar" == "bar";
But beware of nulls!
== handles null strings fine, but calling .equals() from a null string will cause an exception:
String nullString1 = null;
String nullString2 = null;
// Evaluates to true
System.out.print(nullString1 == nullString2);
// Throws a NullPointerException
System.out.print(nullString1.equals(nullString2));
So if you know that fooString1 may be null, tell the reader that by writing
System.out.print(fooString1 != null && fooString1.equals("bar"));
The following are shorter, but it’s less obvious that it checks for null:
System.out.print("bar".equals(fooString1)); // "bar" is never null
System.out.print(Objects.equals(fooString1, "bar")); // Java 7 required
== compares Object references.
.equals() compares String values.
Sometimes == gives illusions of comparing String values, as in following cases:
String a="Test";
String b="Test";
if(a==b) ===> true
This is because when you create any String literal, the JVM first searches for that literal in the String pool, and if it finds a match, that same reference will be given to the new String. Because of this, we get:
(a==b) ===> true
String Pool
b -----------------> "test" <-----------------a
However, == fails in the following case:
String a="test";
String b=new String("test");
if (a==b) ===> false
In this case for new String("test") the statement new String will be created on the heap, and that reference will be given to b, so b will be given a reference on the heap, not in String pool.
Now a is pointing to a String in the String pool while b is pointing to a String on the heap. Because of that we get:
if(a==b) ===> false.
String Pool
"test" <-------------------- a
Heap
"test" <-------------------- b
While .equals() always compares a value of String so it gives true in both cases:
String a="Test";
String b="Test";
if(a.equals(b)) ===> true
String a="test";
String b=new String("test");
if(a.equals(b)) ===> true
So using .equals() is always better.
The == operator checks to see if the two strings are exactly the same object.
The .equals() method will check if the two strings have the same value.
Strings in Java are immutable. That means whenever you try to change/modify the string you get a new instance. You cannot change the original string. This has been done so that these string instances can be cached. A typical program contains a lot of string references and caching these instances can decrease the memory footprint and increase the performance of the program.
When using == operator for string comparison you are not comparing the contents of the string, but are actually comparing the memory address. If they are both equal it will return true and false otherwise. Whereas equals in string compares the string contents.
So the question is if all the strings are cached in the system, how come == returns false whereas equals return true? Well, this is possible. If you make a new string like String str = new String("Testing") you end up creating a new string in the cache even if the cache already contains a string having the same content. In short "MyString" == new String("MyString") will always return false.
Java also talks about the function intern() that can be used on a string to make it part of the cache so "MyString" == new String("MyString").intern() will return true.
Note: == operator is much faster than equals just because you are comparing two memory addresses, but you need to be sure that the code isn't creating new String instances in the code. Otherwise you will encounter bugs.
String a = new String("foo");
String b = new String("foo");
System.out.println(a == b); // prints false
System.out.println(a.equals(b)); // prints true
Make sure you understand why. It's because the == comparison only compares references; the equals() method does a character-by-character comparison of the contents.
When you call new for a and b, each one gets a new reference that points to the "foo" in the string table. The references are different, but the content is the same.
Yea, it's bad...
== means that your two string references are exactly the same object. You may have heard that this is the case because Java keeps sort of a literal table (which it does), but that is not always the case. Some strings are loaded in different ways, constructed from other strings, etc., so you must never assume that two identical strings are stored in the same location.
Equals does the real comparison for you.
Yes, == is bad for comparing Strings (any objects really, unless you know they're canonical). == just compares object references. .equals() tests for equality. For Strings, often they'll be the same but as you've discovered, that's not guaranteed always.
Java have a String pool under which Java manages the memory allocation for the String objects. See String Pools in Java
When you check (compare) two objects using the == operator it compares the address equality into the string-pool. If the two String objects have the same address references then it returns true, otherwise false. But if you want to compare the contents of two String objects then you must override the equals method.
equals is actually the method of the Object class, but it is Overridden into the String class and a new definition is given which compares the contents of object.
Example:
stringObjectOne.equals(stringObjectTwo);
But mind it respects the case of String. If you want case insensitive compare then you must go for the equalsIgnoreCase method of the String class.
Let's See:
String one = "HELLO";
String two = "HELLO";
String three = new String("HELLO");
String four = "hello";
one == two; // TRUE
one == three; // FALSE
one == four; // FALSE
one.equals(two); // TRUE
one.equals(three); // TRUE
one.equals(four); // FALSE
one.equalsIgnoreCase(four); // TRUE
I agree with the answer from zacherates.
But what you can do is to call intern() on your non-literal strings.
From zacherates example:
// ... but they are not the same object
new String("test") == "test" ==> false
If you intern the non-literal String equality is true:
new String("test").intern() == "test" ==> true
== compares object references in Java, and that is no exception for String objects.
For comparing the actual contents of objects (including String), one must use the equals method.
If a comparison of two String objects using == turns out to be true, that is because the String objects were interned, and the Java Virtual Machine is having multiple references point to the same instance of String. One should not expect that comparing one String object containing the same contents as another String object using == to evaluate as true.
.equals() compares the data in a class (assuming the function is implemented).
== compares pointer locations (location of the object in memory).
== returns true if both objects (NOT TALKING ABOUT PRIMITIVES) point to the SAME object instance.
.equals() returns true if the two objects contain the same data equals() Versus == in Java
That may help you.
== performs a reference equality check, whether the 2 objects (strings in this case) refer to the same object in the memory.
The equals() method will check whether the contents or the states of 2 objects are the same.
Obviously == is faster, but will (might) give false results in many cases if you just want to tell if 2 Strings hold the same text.
Definitely the use of the equals() method is recommended.
Don't worry about the performance. Some things to encourage using String.equals():
Implementation of String.equals() first checks for reference equality (using ==), and if the 2 strings are the same by reference, no further calculation is performed!
If the 2 string references are not the same, String.equals() will next check the lengths of the strings. This is also a fast operation because the String class stores the length of the string, no need to count the characters or code points. If the lengths differ, no further check is performed, we know they cannot be equal.
Only if we got this far will the contents of the 2 strings be actually compared, and this will be a short-hand comparison: not all the characters will be compared, if we find a mismatching character (at the same position in the 2 strings), no further characters will be checked.
When all is said and done, even if we have a guarantee that the strings are interns, using the equals() method is still not that overhead that one might think, definitely the recommended way. If you want an efficient reference check, then use enums where it is guaranteed by the language specification and implementation that the same enum value will be the same object (by reference).
If you're like me, when I first started using Java, I wanted to use the "==" operator to test whether two String instances were equal, but for better or worse, that's not the correct way to do it in Java.
In this tutorial I'll demonstrate several different ways to correctly compare Java strings, starting with the approach I use most of the time. At the end of this Java String comparison tutorial I'll also discuss why the "==" operator doesn't work when comparing Java strings.
Option 1: Java String comparison with the equals method
Most of the time (maybe 95% of the time) I compare strings with the equals method of the Java String class, like this:
if (string1.equals(string2))
This String equals method looks at the two Java strings, and if they contain the exact same string of characters, they are considered equal.
Taking a look at a quick String comparison example with the equals method, if the following test were run, the two strings would not be considered equal because the characters are not the exactly the same (the case of the characters is different):
String string1 = "foo";
String string2 = "FOO";
if (string1.equals(string2))
{
// this line will not print because the
// java string equals method returns false:
System.out.println("The two strings are the same.")
}
But, when the two strings contain the exact same string of characters, the equals method will return true, as in this example:
String string1 = "foo";
String string2 = "foo";
// test for equality with the java string equals method
if (string1.equals(string2))
{
// this line WILL print
System.out.println("The two strings are the same.")
}
Option 2: String comparison with the equalsIgnoreCase method
In some string comparison tests you'll want to ignore whether the strings are uppercase or lowercase. When you want to test your strings for equality in this case-insensitive manner, use the equalsIgnoreCase method of the String class, like this:
String string1 = "foo";
String string2 = "FOO";
// java string compare while ignoring case
if (string1.equalsIgnoreCase(string2))
{
// this line WILL print
System.out.println("Ignoring case, the two strings are the same.")
}
Option 3: Java String comparison with the compareTo method
There is also a third, less common way to compare Java strings, and that's with the String class compareTo method. If the two strings are exactly the same, the compareTo method will return a value of 0 (zero). Here's a quick example of what this String comparison approach looks like:
String string1 = "foo bar";
String string2 = "foo bar";
// java string compare example
if (string1.compareTo(string2) == 0)
{
// this line WILL print
System.out.println("The two strings are the same.")
}
While I'm writing about this concept of equality in Java, it's important to note that the Java language includes an equals method in the base Java Object class. Whenever you're creating your own objects and you want to provide a means to see if two instances of your object are "equal", you should override (and implement) this equals method in your class (in the same way the Java language provides this equality/comparison behavior in the String equals method).
You may want to have a look at this ==, .equals(), compareTo(), and compare()
Function:
public float simpleSimilarity(String u, String v) {
String[] a = u.split(" ");
String[] b = v.split(" ");
long correct = 0;
int minLen = Math.min(a.length, b.length);
for (int i = 0; i < minLen; i++) {
String aa = a[i];
String bb = b[i];
int minWordLength = Math.min(aa.length(), bb.length());
for (int j = 0; j < minWordLength; j++) {
if (aa.charAt(j) == bb.charAt(j)) {
correct++;
}
}
}
return (float) (((double) correct) / Math.max(u.length(), v.length()));
}
Test:
String a = "This is the first string.";
String b = "this is not 1st string!";
// for exact string comparison, use .equals
boolean exact = a.equals(b);
// For similarity check, there are libraries for this
// Here I'll try a simple example I wrote
float similarity = simple_similarity(a,b);
The == operator check if the two references point to the same object or not. .equals() check for the actual string content (value).
Note that the .equals() method belongs to class Object (super class of all classes). You need to override it as per you class requirement, but for String it is already implemented, and it checks whether two strings have the same value or not.
Case 1
String s1 = "Stack Overflow";
String s2 = "Stack Overflow";
s1 == s2; //true
s1.equals(s2); //true
Reason: String literals created without null are stored in the String pool in the permgen area of heap. So both s1 and s2 point to same object in the pool.
Case 2
String s1 = new String("Stack Overflow");
String s2 = new String("Stack Overflow");
s1 == s2; //false
s1.equals(s2); //true
Reason: If you create a String object using the new keyword a separate space is allocated to it on the heap.
== compares the reference value of objects whereas the equals() method present in the java.lang.String class compares the contents of the String object (to another object).
I think that when you define a String you define an object. So you need to use .equals(). When you use primitive data types you use == but with String (and any object) you must use .equals().
If the equals() method is present in the java.lang.Object class, and it is expected to check for the equivalence of the state of objects! That means, the contents of the objects. Whereas the == operator is expected to check the actual object instances are same or not.
Example
Consider two different reference variables, str1 and str2:
str1 = new String("abc");
str2 = new String("abc");
If you use the equals()
System.out.println((str1.equals(str2))?"TRUE":"FALSE");
You will get the output as TRUE if you use ==.
System.out.println((str1==str2) ? "TRUE" : "FALSE");
Now you will get the FALSE as output, because both str1 and str2 are pointing to two different objects even though both of them share the same string content. It is because of new String() a new object is created every time.
Operator == is always meant for object reference comparison, whereas the String class .equals() method is overridden for content comparison:
String s1 = new String("abc");
String s2 = new String("abc");
System.out.println(s1 == s2); // It prints false (reference comparison)
System.out.println(s1.equals(s2)); // It prints true (content comparison)
All objects are guaranteed to have a .equals() method since Object contains a method, .equals(), that returns a boolean. It is the subclass' job to override this method if a further defining definition is required. Without it (i.e. using ==) only memory addresses are checked between two objects for equality. String overrides this .equals() method and instead of using the memory address it returns the comparison of strings at the character level for equality.
A key note is that strings are stored in one lump pool so once a string is created it is forever stored in a program at the same address. Strings do not change, they are immutable. This is why it is a bad idea to use regular string concatenation if you have a serious of amount of string processing to do. Instead you would use the StringBuilder classes provided. Remember the pointers to this string can change and if you were interested to see if two pointers were the same == would be a fine way to go. Strings themselves do not.
You can also use the compareTo() method to compare two Strings. If the compareTo result is 0, then the two strings are equal, otherwise the strings being compared are not equal.
The == compares the references and does not compare the actual strings. If you did create every string using new String(somestring).intern() then you can use the == operator to compare two strings, otherwise equals() or compareTo methods can only be used.
In Java, when the == operator is used to compare 2 objects, it checks to see if the objects refer to the same place in memory. In other words, it checks to see if the 2 object names are basically references to the same memory location.
The Java String class actually overrides the default equals() implementation in the Object class – and it overrides the method so that it checks only the values of the strings, not their locations in memory.
This means that if you call the equals() method to compare 2 String objects, then as long as the actual sequence of characters is equal, both objects are considered equal.
The == operator checks if the two strings are exactly the same object.
The .equals() method check if the two strings have the same value.

Number data type automatically loses precision in Flex

private function getPercentage(max:Number, value:Number):int
{
return Number((value*100) / max);
}
I call the above function to assign a percentage to an object.
var max:Number = findMax();
p.percentage = getPercentage(max, p.value);
Assume that p is some object with percentage defined as
public var percentage:Number;
When I put a breakpoint and check for the value returned in getPercentage it will something like 1.22343342322 but when I assign it to p.percentage it automatically becomes 1, losing the precision.
How do I handle this kind of a situation?
It says in the LiveDocs
To store a floating-point number,
include a decimal point in the number.
If you omit a decimal point, the
number will be stored as an integer.
But how do I do that? Any insight to this problem is highly appreciated.
Your method getPercentage() returns int. Change it to Number.

How To update list using lambda expression

I have two lists. lst contains ViewState Data i.e All the records of the gridview & second lstRank contains the list of integer(i.e ID) for only those records which are marked as checked (i.e Gridview contains a columns for checkbox). Now i want to update the lst bool status depending upon integer ID of lstRank. How it can be achieved by lambda expression
List<Tuple<int, string, bool>> lst = (List<Tuple<int, string,bool>>)ViewState["gvData"];
List<int> lstRank = gvDetails.Rows.OfType<GridViewRow>().Where(s => ((CheckBox)s.FindControl("chkSelect")).Checked)
.Select(s => Convert.ToInt32(((Label)s.FindControl("lblRankCD")).Text)).ToList();
Your question isn't clear, but I'm guessing you want to change lst's contents so that the boolean values are true if the int exists in lstRank, or false if not?
Obviously, tuples are immutable, so if you want to change one of the values, you would have to generate new tuple instances. It's not clear what you mean when you say you specifically want to do this with a lambda expression, but I assume you probably mean that you don't want a solution that involves an explicit loop. So how about this:
lst = lst.Select(oldValues =>
Tuple.Create(oldValues.Item1,
oldValues.Item2,
lstRank.Contains(oldValues.Item3))).ToList();
If lstRank is large, you might want to optimize by first building a HashSet out of it, since you'll be doing a lot of Contains calls.

Sorting in AdvancedDatagrid in Flex 3

I am using AdvancedDatagrid in Flex 3. One column of AdvancedDatagrid contains numbers and alphabets. When I sort this column, numbers come before alphabets (Default behavior of internal sorting of AdvancedDatagrid). But I want alphabets to come before number when I sort.
I know I will have to write the custom sort function. But can anybody give some idea on how to proceed.
Thanks in advance.
Use sortCompareFunction
The AdvancedDataGrid control uses this function to sort the elements of the data provider collection. The function signature of the callback function takes two parameters and has the following form:
mySortCompareFunction(obj1:Object, obj2:Object):int
obj1 — A data element to compare.
obj2 — Another data element to compare with obj1.
The function should return a value based on the comparison of the objects:
-1 if obj1 should appear before obj2 in ascending order.
0 if obj1 = obj2.
1 if obj1 should appear after obj2 in ascending order.
<mx:AdvancedDataGridColumn sortCompareFunction="mySort"
dataField="colData"/>
Try the following sort compare function.
public function mySort(obj1:Object, obj2:Object):int
{
var s1:String = obj1.colData;
var s2:String = obj2.colData;
var result:Number = s1.localeCompare(s2);
if(result != 0)
result = result > 0 ? 1 : -1;
if(s1.match(/^\d/))
{
if(s2.match(/^\d/))
return result;
else
return 1;
}
else if(s2.match(/^\d/))
return -1;
else
return result;
}
It checks the first character of strings and pushes the ones that start with a digit downwards in the sort order. It uses localeCompare to compare two strings if they both start with letters or digits - otherwise it says the one starting with a letter should come before the one with digit. Thus abc will precede 123 but a12 will still come before abc.
If you want a totally different sort where letters always precede numbers irrespective of their position in the string, you would have to write one from the scratch - String::charCodeAt might be a good place to start.

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