int的错误无法解除引用？

I am getting an error with this constructor, and i have no idea how to fix? I am a beginner at java. This is from an example exercise that i was trying to learn:

我在使用此构造函数时出错，我不知道如何修复？我是java的初学者。这是我试图学习的一个示例练习：

/** 
 * Create an array of size n and store a copy of the contents of the
 * input argument
 * @param intArray array of elements to copy 
*/
public IntArray11(int[] intArray)
{
    int i = 0;
    String [] Array = new String[intArray.length];
    for(i=0; i<intArray.length; ++i)
    {
    Array[i] = intArray[i].toString();
    }
}

3 个解决方案

#1

int is not an object in java (it's a primitive), so you cannot invoke methods on it.

int不是java中的对象（它是原语），因此您无法在其上调用方法。

One simple way to solve it is using Integer.toString(intArray[i])

解决它的一种简单方法是使用Integer.toString（intArray [i]）

#2

I would write it more like this

我会写得更像这样

public String[] convertToStrings(int... ints) {
    String[] ret = new String[ints.length];
    for(int i = 0; i < intArray.length; ++i)
        ret[i] = "" + ints[i];
    return ret;
}

Or in Java 8 you might write

或者在Java 8中你可以写

public List<String> convertToStrings(int... ints) {
    return IntStream.of(ints).mapToObj(Integer::toString).collect(toList());
}

This uses;

这用;

Java coding conventions,
Java编码约定，
limited scope for the variable i,
变量i的范围有限，
we do something with the String[],
我们用String []做一些事情，
give the method a meaningful name,
给这个方法一个有意义的名字，
try to use consist formatting.
尝试使用编组格式。

If we were worried about efficiency it is likely we could do away with the method entirely.

如果我们担心效率，我们很可能完全取消该方法。

String.valueOf(int) is not faster than Integer.toString(int). From the code in String you can see that the String implementation just calls Integer.toString

String.valueOf（int）不比Integer.toString（int）快。从String中的代码可以看出String实现只调用了Integer.toString

/**
 * Returns the string representation of the {@code int} argument.
 * <p>
 * The representation is exactly the one returned by the
 * {@code Integer.toString} method of one argument.
 *
 * @param   i   an {@code int}.
 * @return  a string representation of the {@code int} argument.
 * @see     java.lang.Integer#toString(int, int)
 */
public static String valueOf(int i) {
    return Integer.toString(i);
}

#3

You code tries to call the toString() method of an int value. In Java, int is a primitive type and has no methods. Change the line:

您的代码尝试调用int值的toString（）方法。在Java中，int是基本类型，没有方法。换行：

Array[i] = intArray[i].toString();

至

Array[i] = String.valueOf(intArray[i]);

and the code should run. By the way, you should use lowerCamelCase for variables and fields.

并且代码应该运行。顺便说一下，你应该使用lowerCamelCase来表示变量和字段。

Edit: For what it's worth, String.valueOf(int) is a bit faster than Integer.toString(int) on my system (Java 1.7).

编辑：对于它的价值，String.valueOf（int）比我的系统上的Integer.toString（int）快一点（Java 1.7）。

#1