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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.
I have created sample function here in c# which set the location value based on the parameter. I want to write below expression by using arm template style format.
public static Main(string name)
{
string location = string.Empty;
if(name == "uksouth")
{
location = "UKS";
}else if(name == "ukwest")
{
location = "UKE";
}else if(name == "IndiaWest")
{
location = "INDW";
}
else {
location = "INDS";
}
}
I have written this for one match condition, but i want to return value based on the user resource group.
"value": "[if(equals(resourceGroup().location,'uksouth'), 'UKS', 'EUS')]"
Unfortunately, ARM templates don't provide the equivalent of a "switch" mechanism, which is what might make this easier. However, you can nest multiple if statements. The syntax is a bit clunky, but this should be the equivalent of the code you've written:
"value": "[if(equals(resourceGroup().location,'uksouth'), 'UKS', [if(equals(resourceGroup().location,'ukwest'), 'UKE', [if(equals(resourceGroup().location,'IndiaWest'), 'INDW', 'INDS')])])]"
Here's the same code with a little formatting applied to make it more obvious what's happening here:
"value": "
[if(equals(resourceGroup().location,'uksouth'),
'UKS',
[if(equals(resourceGroup().location,'ukwest'),
'UKE',
[if(equals(resourceGroup().location,'IndiaWest'),
'INDW',
'INDS')])])]
"
You might also consider the approach described in this answer for a bit of a cleaner solution.
Try defining a variable that is an object used like a hashtable. Retrieve different properties from the object by key-name, accessing the properties as key-value pairs. I use something very similar to lookup values inside my ARM templates.
"variables" {
"locationShorten": {
"uksouth": "UKS",
"ukwest": "UKE",
"IndiaWest": "INDW",
"IndiaSouth": "INDS"
},
"locationShort": "[variables('locationShorten')[resourceGroup().location]]"}
Microsoft defines an object's properties as key-value pairs. "Each property in an object consists of key and value. The key and value are enclosed in double quotes and separated by a colon (:)."
Source: https://learn.microsoft.com/en-us/azure/azure-resource-manager/templates/data-types#objects
As for the documentation on using [] to access object properties, I can no longer find it for JSON but it is there for BICEP. "You can also use the [] syntax to access a property."
Source: https://learn.microsoft.com/en-us/azure/azure-resource-manager/bicep/data-types#objects
With arrays you can use a subscript to access Array Elements directly. You can read or write to them. With Sets I am not sure of a way to write its Elements.
For example, if I access a set element matching a condition I'm only able to read the element. It is passed by copy and I can't therefore write to the original.
For example:
columns.first(
where: {
$0.header.last == Character(String(i))
}
)?.cells.append(value: addValue)
// ERROR: Cannot use mutating member on immutable value: function call returns immutable value
You can't just change things inside a set, because of how a (hash) set works. Changing them would possibly change their hash value, making the set into an invalid state.
Therefore, you would have to take the thing you want to change out of the set, change it, then put it back.
if var thing = columns.first(
where: {
$0.header.last == Character(String(i))
}) {
columns.remove(thing)
thing.cells.append(value: addValue)
columns.insert(thing)
}
If the == operator on Column doesn't care about cells (i.e. adding cells to a column doesn't suddenly make two originally equal columns unequal and vice versa), then you could use update instead:
if var thing = columns.first(
where: {
$0.header.last == Character(String(i))
}) {
thing.cells.append(value: addValue)
columns.update(thing)
}
As you can see, it's quite a lot of work, so maybe sets aren't a suitable data structure to use in this situation. Have you considered using an array instead? :)
private var _columns: [Column]
public var columns : [Column] {
get { _columns }
set { _columns = Array(Set(newValue)) }
// or any other way to remove duplicate as described here: https://stackoverflow.com/questions/25738817/removing-duplicate-elements-from-an-array-in-swift
}
You are getting the error because columns might be a set of struct. So columns.first will give you an immutable value. If you were to use a class, you will get a mutable result from columns.first and your code will work as expected.
Otherwise, you will have to do as explained by #Sweeper in his answer.
I am looking for examples of Chapel passing by reference. This example works but it seems like bad form since I am "returning" the input. Does this waste memory? Is there an explicit way to operate on a class?
class PowerPuffGirl {
var secretIngredients: [1..0] string;
}
var bubbles = new PowerPuffGirl();
bubbles.secretIngredients.push_back("sugar");
bubbles.secretIngredients.push_back("spice");
bubbles.secretIngredients.push_back("everything nice");
writeln(bubbles.secretIngredients);
proc kickAss(b: PowerPuffGirl) {
b.secretIngredients.push_back("Chemical X");
return b;
}
bubbles = kickAss(bubbles);
writeln(bubbles.secretIngredients);
And it produces the output
sugar spice everything nice
sugar spice everything nice Chemical X
What is the most efficient way to use a function to modify Bubbles?
Whether Chapel passes an argument by reference or not can be controlled by the argument intent. For example, integers normally pass by value but we can pass one by reference:
proc increment(ref x:int) { // 'ref' here is an argument intent
x += 1;
}
var x:int = 5;
increment(x);
writeln(x); // outputs 6
The way that a type passes when you don't specify an argument is known as the default intent. Chapel passes records, domains, and arrays by reference by default; but of these only arrays are modifiable inside the function. ( Records and domains pass by const ref - meaning they are passed by reference but that the function they are passed to cannot modify them. Arrays pass by ref or const ref depending upon what the function does with them - see array default intent ).
Now, to your question specifically, class instances pass by "value" by default, but Chapel considers the "value" of a class instance to be a pointer. That means that instead of allowing a field (say) to be mutated, passing a class instance by ref just means that it could be replaced with a different class instance. There isn't currently a way to say that a class instance's fields should not be modifiable in the function (other than making them to be explicitly immutable data types).
Given all of that, I don't see any inefficiencies with the code sample you provided in the question. In particular, here:
proc kickAss(b: PowerPuffGirl) {
b.secretIngredients.push_back("Chemical X");
return b;
}
the argument accepting b will receive a copy of the pointer to the instance and the return b will return a copy of that pointer. The contents of the instance (in particular the secretIngredients array) will remain stored where it was and won't be copied in the process.
One more thing:
This example works but it seems like bad form since I am "returning" the input.
As I said, this isn't really a problem for class instances or integers. What about an array?
proc identity(A) {
return A;
}
var A:[1..100] int;
writeln(identity(A));
In this example, the return A in identity() actually does cause a copy of the array to be made. That copy wasn't created when passing the array in to identity(), since the array was passed by with a const ref intent. But, since the function returns something "by value" that was a reference, it's necessary to copy it as part of returning. See also arrays return by value by default in the language evolution document.
In any case, if one wants to return an array by reference, it's possible to do so with the ref or const ref return intent, e.g.:
proc refIdentity(ref arg) ref {
return arg;
}
var B:[1..10] int;
writeln(refIdentity(B));
Now there is no copy of the array and everything is just referring to the same B.
Note though that it's currently possible to write programs that return a reference to a variable that no longer exists. The compiler includes some checking in that area but it's not complete. Hopefully improvements in that area are coming soon.