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varargs java

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coder answered on January 16, 2022 Popularity 10/10 Helpfulness 3/10

answer varargs java

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public static void printNumberOfArguments(int... numbers) {
    System.out.println(numbers.length);
}
printNumberOfArguments(1);
printNumberOfArguments(1, 2);
printNumberOfArguments(1, 2, 3);
printNumberOfArguments(new int[] { }); // no arguments here
printNumberOfArguments(new int[] { 1, 2 });

Popularity 10/10 Helpfulness 3/10 Language java

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Tags: java

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Contributed on Jan 16 2022

coder

0 Answers Avg Quality 2/10

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public class SomeClass{
    public void someMethod(int ... vars){
    // logic
}
}

var obj1 = new SomeClass();
var obj2 = new SomeClass();

obj1.someMethod(10, 20, 30, 40); // Works
obj2.someMethod(new int [] {10, 20, 30, 40}); // works

Popularity 8/10 Helpfulness 4/10 Language java

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Tags: java

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Contributed on Nov 09 2023

Innocent Iguana

0 Answers Avg Quality 2/10

varargs java

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varargs:
It's possible to pass an arbitrary number of the same type arguments to 
a method using the special syntax named varargs (variable-length arguments).
 (Varargs can be used when we are unsure about the number of 
 arguments to be passed in a method)
Use varargs for any method (or constructor) that needs an array of
T (whatever type T may be) as input
One good example is String.format.
The format string can accept any number of parameters.

Popularity 10/10 Helpfulness 1/10 Language java

Source: Grepper

Tags: java

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Contributed on Jan 16 2022

coder

0 Answers Avg Quality 2/10

varargs java

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