以下name最终打印结果是什么 String name="123"; name += 1; System.out.println( name) ; A 123 B 1231 c 编译错误,String不可变 D 124
时间: 2024-03-31 09:32:18 浏览: 6
The final output will be "1231" (Option B).
In the given code, the variable "name" is initially assigned the value "123". Then, the "+=" operator is used to append the value "1" to the existing value of "name". This results in the string "1231". Finally, the value of "name" is printed using the "System.out.println()" method, which outputs "1231" to the console.
相关问题
String s0 = "" + sr; System.out.println(s0);
This code will result in a compilation error because "sr" is not a valid variable name and it is not being assigned any value. The correct code should be:
String s0 = "";
System.out.println(s0);
This will create an empty string variable "s0" and print it to the console, which will output a blank line.
public static void main(String[] args) throws Exception { Scanner scanner = new Scanner(System.in); String[] name= new String[5]; String[] name1={}; for (int i=0;i<name.length;i++ ){ System.out.println("请输入"); String next = scanner.next(); name1[i]=next; System.out.println(name1); } System.out.println(name1);}那行不对
这段代码存在两个问题:
1. `name1` 的长度为 0,使用 `name1[i]` 会抛出 `ArrayIndexOutOfBoundsException` 异常。可以使用 `new String[5]` 来初始化 `name1` 的长度。
2. `System.out.println(name1)` 会输出 `name1` 数组的地址,而不是数组中的元素。可以使用 `Arrays.toString()` 方法将数组转换为字符串输出。
修改后的代码如下:
```java
import java.util.Scanner;
import java.util.Arrays;
public class Main {
public static void main(String[] args) throws Exception {
Scanner scanner = new Scanner(System.in);
String[] name = new String[5];
String[] name1 = new String[5];
for (int i = 0; i < name.length; i++) {
System.out.println("请输入");
String next = scanner.next();
name1[i] = next;
System.out.println(Arrays.toString(name1));
}
System.out.println(Arrays.toString(name1));
}
}
```