Syntax error while defining an array as a property of a class
You cannot use non-constant values while initializing class properties in PHP versions earlier than 5.6.
These are initialized at compile time, at which PHP will do no calculations or execute any code. (5 * (1024 * 1024))
is an expression that requires evaluation, which you cannot do there. Either replace that with the constant value 5242880
or do the calculation in __construct
.
PHP 5.6, introduced in 2014, allows "constant scalar expressions" wherein a scalar constant or class property can be initialized by an evaluated expression in the class definition rather than the constructor.
Syntax error while defining an array as a property of a class
You cannot use non-constant values while initializing class properties in PHP versions earlier than 5.6.
These are initialized at compile time, at which PHP will do no calculations or execute any code. (5 * (1024 * 1024))
is an expression that requires evaluation, which you cannot do there. Either replace that with the constant value 5242880
or do the calculation in __construct
.
PHP 5.6, introduced in 2014, allows "constant scalar expressions" wherein a scalar constant or class property can be initialized by an evaluated expression in the class definition rather than the constructor.
Syntax error when accessing static array with in class
You want to access variable so you have to add $
.
self::$config['base_url']
Read some more about that here.
Unfortunately you can't assign any variable (even static) to other static property as you can see in linked manual page.
Like any other PHP static variable, static properties may only be initialized using a literal or constant; expressions are not allowed. So while you may initialize a static property to an integer or array (for instance), you may not initialize it to another variable, to a function return value, or to an object.
Creating an array of objects gives Syntax Error
tiles[0] = new Tile("air", -1);
You can't do this in your class. You need to assign it in your constructor.
Read the difference between initialization and assignment.
Initialization of field belongs outside a method, but of course is encapsulated by a class.
Assignment belongs in a method.
or you can initialize like so:
Tiles []tiles = new Tiles[]{new Tile("air",-1)} ;
And use .equals
not ==
Why do I get a syntax error on array declaration?
Each Java application needs an entry point, so the compiler knows where to begin executing the application. In the case of Java applications, you need to wrap up your code in the main() method.
public class HelloWorld{
public static void main(String []args){
System.out.println("Hello World");
}
}
Your code should be
package sort;
public class InsertionSort {
public static void main (String[] args) {
int[] polje = { -2, 5, -14, 35, 16, 3, 25, -100 };
for(int firstUnsorted = 1; firstUnsorted < polje.length; firstUnsorted++) {
int newElement = polje[firstUnsorted];
int i;
for (i = firstUnsorted; i > 0 && polje[i - 1] > newElement; i--) {
}
}
for (int i = 0; i < polje.length; i++){
int firstUnsorted = 1;
int elemant;
}
}
}
Errors trying to declare array in class definition
You got compilation error. It is because array is defined in namespace std. Either add
using namespace std;
at the top of your file or add std::
before you use any type defined in there:
std::array< int, 5> mapArray;
The latter is preferred because you don't have to bring all symbols from the standard library just to use it's array
type.
Syntax error on token with array
You need to put the code into a main
method as it is the entry of your program.
On that, the instructions are not allowed inside the body of the class, but inside one of its methods or blocks.
Solution
public class multiples_of_13 {
public static void main(String[] args) {
int[] thirteens = new int[400];
int numFound = 0;
// candidate: the number that might be a multiple
// of 13
int candidate = 1;
System.out.println("the first 400 multiples of 13:" );
while (numFound < 400) {
if (candidate % 13 == 0) {
thirteens[numFound] = candidate;
numFound++;
}
candidate++;
}
System.out.println("First 400 multiples of 13:");
for (int i = 0; i < 400; i++) {
System.out.print(thirteens[i] + " ");
}
}
}
JavaScript dictionary syntax error when injecting array element as property name
You have to wrap
the array's item into brackets because you're creating dynamic
keys.
let surveyTypes = [1,2];
let surveyDuration = { [surveyTypes[0]]: 'first',[surveyTypes[1]]: 'second' }
console.log(surveyDuration);
Related Topics
Double Not (!!) Operator in PHP
How to Solve Time Out in PHPmyadmin
How to Pass Get Parameters to Laravel from with Get Method
Continue Processing After Closing Connection
Converting HTML to PDF (Not PDF to HTML) Using PHP
Getting Pear to Work on Xampp (Apache/MySQL Stack on Windows)
Laravel Eloquent, Group by Month/Year
Codeigniter Default Controller in a Sub Directory Not Working
How to Decode This JSON String
Loop Code for Each File in a Directory
When Should I Use Stdclass and When Should I Use an Array in PHP Oo Code
How to Get Time Zone Through Ip Address in PHP
Guzzle: Handle 400 Bad Request
PHP + SQL Server - How to Set Charset for Connection