The "common knowledge" of programming is that x += y is an equivalent shorthand notation of x = x + y. As long as x and y are of the same type (for example, both are ints), you may consider the two statements equivalent.
However, in Java, x += y is not identical to x = x + y in general.
If x and y are of different types, the behavior of the two statements differs due to the rules of the language. For example, let's have x == 0 (int) and y == 1.1 (double):
int x = 0;
x += 1.1; // just fine; hidden cast, x == 1 after assignment
x = x + 1.1; // won't compile! 'cannot convert from double to int'
+= performs an implicit cast, whereas for + you need to explicitly cast the second operand, otherwise you'd get a compiler error.
Quote from Joshua Bloch's Java Puzzlers:
Answer from jakub.g on Stack Overflow(...) compound assignment expressions automatically cast the result of the computation they perform to the type of the variable on their left-hand side. If the type of the result is identical to the type of the variable, the cast has no effect. If, however, the type of the result is wider than that of the variable, the compound assignment operator performs a silent narrowing primitive conversion [JLS 5.1.3].
The "common knowledge" of programming is that x += y is an equivalent shorthand notation of x = x + y. As long as x and y are of the same type (for example, both are ints), you may consider the two statements equivalent.
However, in Java, x += y is not identical to x = x + y in general.
If x and y are of different types, the behavior of the two statements differs due to the rules of the language. For example, let's have x == 0 (int) and y == 1.1 (double):
int x = 0;
x += 1.1; // just fine; hidden cast, x == 1 after assignment
x = x + 1.1; // won't compile! 'cannot convert from double to int'
+= performs an implicit cast, whereas for + you need to explicitly cast the second operand, otherwise you'd get a compiler error.
Quote from Joshua Bloch's Java Puzzlers:
Answer from jakub.g on Stack Overflow(...) compound assignment expressions automatically cast the result of the computation they perform to the type of the variable on their left-hand side. If the type of the result is identical to the type of the variable, the cast has no effect. If, however, the type of the result is wider than that of the variable, the compound assignment operator performs a silent narrowing primitive conversion [JLS 5.1.3].
The "common knowledge" of programming is that x += y is an equivalent shorthand notation of x = x + y. As long as x and y are of the same type (for example, both are ints), you may consider the two statements equivalent.
However, in Java, x += y is not identical to x = x + y in general.
If x and y are of different types, the behavior of the two statements differs due to the rules of the language. For example, let's have x == 0 (int) and y == 1.1 (double):
int x = 0;
x += 1.1; // just fine; hidden cast, x == 1 after assignment
x = x + 1.1; // won't compile! 'cannot convert from double to int'
+= performs an implicit cast, whereas for + you need to explicitly cast the second operand, otherwise you'd get a compiler error.
Quote from Joshua Bloch's Java Puzzlers:
(...) compound assignment expressions automatically cast the result of the computation they perform to the type of the variable on their left-hand side. If the type of the result is identical to the type of the variable, the cast has no effect. If, however, the type of the result is wider than that of the variable, the compound assignment operator performs a silent narrowing primitive conversion [JLS 5.1.3].
x += yisx = x + yx -= yisx = x - yx *= yisx = x * yx /= yisx = x / yx %= yisx = x % yx ^= yisx = x ^ yx &= yisx = x & yx |= yisx = x | y
and so on ...
Does Java have a "IN" operator or function like SQL? - Stack Overflow
How do the post increment (i++) and pre increment (++i) operators work in Java? - Stack Overflow
What does <? mean in java?
is Java a good language to start learning programming?
Send this question to the java reddit and you'll get the answer yes. Send it to the python, ruby, C#, etc reddit and you'll get the answer no.
Send it to the programming reddit and you'll get 1,000 completely different answers from 800 people.
You can start learning programming with Java. It is a fine language to use and it is similar enough (syntax wise) to C, C++, C# that you could transition to those if you need to.
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There are many collections that will let you do something similar to that. For example:
With Strings:
String s = "I can has cheezeburger?";
boolean hasCheese = s.contains("cheeze");
or with Collections:
List<String> listOfStrings = new ArrayList<String>();
boolean hasString = listOfStrings.contains(something);
However, there is no similar construct for a simple String[].
In SQL
x in ('Alice', 'Bob', 'Carol')
In Java:
Arrays.asList("Alice", "Bob", "Carol").contains(x)
++aincrements and then uses the variable.a++uses and then increments the variable.
If you have
a = 1;
and you do
System.out.println(a++); //You will see 1
//Now a is 2
System.out.println(++a); //You will see 3
codaddict explains your particular snippet.
Does this help?
a = 5;
i=++a + ++a + a++; =>
i=6 + 7 + 7; (a=8)
a = 5;
i=a++ + ++a + ++a; =>
i=5 + 7 + 8; (a=8)
The main point is that ++a increments the value and immediately returns it.
a++ also increments the value (in the background) but returns unchanged value of the variable - what looks like it is executed later.
I have come across some code recently that uses <? in a function and I do not understand what that means. I understand that the ? is used as the conditional operator and outputs the first value if true and the second if false. I just don't understand how the < effects the ? operator.
The function looks like this:
private static Pair divideAndConquer(List<? extends Point> pointsSortedByX, List<? extends Point> pointsSortedByY)
This is from the Closet-Pair problem on Rosettacode.org
Link Here: http://rosettacode.org/wiki/Closest-pair_problem#Java
Any help with this is much appreciated.