Java:带有char数组的println给出乱码

问题描述

这就是问题所在.这段代码:

String a = "0000";System.out.println(a);char[] b = a.toCharArray();System.out.println(b);

返回

<上一页>00000000

但是这段代码:

String a = "0000";System.out.println("字符串 a:" + a);char[] b = a.toCharArray();System.out.println("char[] b:" + b);

返回

<上一页>字符串 a:0000字符 [] b: [C@56e5b723

世界上到底发生了什么?似乎应该有一个足够简单的解决方案，但我似乎无法弄清楚.

当你说

System.out.println(b);

它导致调用 print(char[] s) 然后 println()

print(char[] s) 的 JavaDoc 说:

<块引用>

打印一个字符数组.字符转换为字节根据平台的默认字符编码，而这些字节完全按照 write(int) 方法的方式写入.

所以它会逐字节打印出来.

当你说

System.out.println("char[] b: " + b);

这会导致调用 print(String)，因此您实际上正在做的是将 String 附加到 Object它在 Object 上调用 toString() - 这与默认情况下的所有 Object 以及在 Array 的情况下一样，打印引用的值(内存地址).

你可以这样做:

System.out.println("char[] b: " + new String(b));

请注意，这是错误的"，因为您不关心编码并且使用系统默认值.尽早了解编码.

Here's the problem. This code:

String a = "0000";
 System.out.println(a);
char[] b = a.toCharArray();
 System.out.println(b);

returns

0000
0000

But this code:

String a = "0000";
 System.out.println("String a: " + a);
char[] b = a.toCharArray();
 System.out.println("char[] b: " + b);

returns

String a: 0000
char[] b: [C@56e5b723

What in the world is going on? Seems there should be a simple enough solution, but I can't seem to figure it out.

When you say

System.out.println(b);

It results in a call to print(char[] s) then println()

The JavaDoc for print(char[] s) says:

Print an array of characters. The characters are converted into bytes according to the platform's default character encoding, and these bytes are written in exactly the manner of the write(int) method.

So it performs a byte-by-byte print out.

When you say

System.out.println("char[] b: " + b);

It results in a call to print(String), and so what you're actually doing is appending to a String an Object which invokes toString() on the Object -- this, as with all Object by default, and in the case of an Array, prints the value of the reference (the memory address).

You could do:

System.out.println("char[] b: " + new String(b));

Note that this is "wrong" in the sense that you're not paying any mind to encoding and are using the system default. Learn about encoding sooner rather than later.

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