Java: Is there any short combination to convert array of primitive to List & receive “printable” version?

int[] arrc = new int[] {1, 2, 3};
System.out.println(new ArrayList(Arrays.asList(arrc)));

prints address, but i desire to use toString as in ArrayList.

打印地址，但我希望在ArrayList中使用toString。

Is it possible ?

可能吗？

6 个解决方案

#1

Try:

尝试：

import java.util.Arrays;

// ...

int[] arrc = new int[] {1, 2, 3};
System.out.println(Arrays.toString(arrc));

Note that the asList(...) does not take an array of primitive int's but takes Objects instead, that's why you see an address-like String appear.

请注意，asList（...）不接受原始int的数组，而是接受Object，这就是为什么你会看到类似地址的String出现的原因。

So, doing:

所以，做：

int[] array = {1, 2, 3};
List list = Arrays.asList(array);

results in the same as doing:

结果与做：

int[] array = {1, 2, 3};
List list = new ArrayList();
list.add(array);

Both result in a List that has one element in it: an array of primitive ints.

两者都导致List中包含一个元素：一个原始int数组。

(only that Arrays.asList(...) returns a List that cannot be modified...)

（只有Arrays.asList（...）返回一个无法修改的List ...）

#2

If you just want to print the array:

如果您只想打印数组：

Arrays.toString( arrc )

If you want to turn an int[] into a List, Arrays.asList does not work, unfortunately (it only works with Object[]):

如果你想把一个int []变成一个List，不幸的是Arrays.asList不起作用（它只适用于Object []）：

List<Integer> list = new ArrayList<Integer>(arrc.length);
for (int a: arrc)
   list.add(a);

System.out.println(list); // prints nicely now

#3

use Arrays.toString(arrc)

使用Arrays.toString（arrc）

#4

Use Apache Commons Lang as your main library after SDK

在SDK之后使用Apache Commons Lang作为主库

System.out.println("An object: " + ReflectionToStringBuilder.toString(anObject));

#5

using Dollar should be simple:

使用Dollar应该很简单：

int[] ary = { 1, 2, 3, 4};
String string = $(ary).toString(); // wrapping Arrays.toString()

alternatively you can convert the int array to List then use the toString method :

或者你可以将int数组转换为List然后使用toString方法：

List<Integer> list = $(ary).toList();

please note that list is an ArrayList: you can even specify which concrete class should be used simply passing a List implementation (it will work with any List implementation with a no-args constructor):

请注意，list是一个ArrayList：您甚至可以指定应该使用哪个具体类，只需传递List实现（它将与任何带有no-args构造函数的List实现一起使用）：

List<Integer> linkedList = $(ary).toList(LinkedList.class);

#6

Using Ints.asList from Guava:

使用Guava的Ints.asList：

import java.util.List;
import com.google.common.primitives.Ints;

// ...

int[] arrc = {1, 2, 3, 4};
List<Integer> list = Ints.asList(arrc);
System.out.println(list);
// Output: "[1, 2, 3, 4]"

#1