本文介绍了在Java中将小数转换为二进制的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我想写一个将数字转换为二进制的代码,这就是我写的。它给我几个Eclipse中的错误,我不明白。
这是什么问题?任何其他建议?我想学习和听到任何意见,修复它。谢谢。
I'm trying to write a code that converts a number to binary, and this is what I wrote. It gives me couple of errors in Eclipse, which I don't understand.What's wrong with that? Any other suggestions? I'd like to learn and hear for any comments for fixing it. Thank you.
public class NumberConverte {
public static void main(String[] args) {
int i = Integer.parseInt(args);
public static void Binary(int int1){
System.out.println(int1 + "in binary is");
do {
System.out.println(i mod 2);
} while (int1>0);
}
}
}
错误消息:
- 方法
parseInt(String)
在类型Integer
不适用于参数(String []
) - 此行有多个标记
- 令牌上的语法错误
(
,
)
,;预期 - 令牌上的语法错误
-
void
是变量的无效类型Binary
- The method
parseInt(String)
in the typeInteger
is not applicable for the arguments (String[]
) - Multiple markers at this line
- Syntax error on token "
(
", ; expected - Syntax error on token "
)
", ; expected void
is an invalid type for the variableBinary
- Syntax error on token "
- 令牌mod的语法错误,无效AssignmentOperator
- 令牌mod的语法错误,无效的AssignmentOperator。
推荐答案
应该做的!
顺便说一下,纠正你的语法,如果你使用Eclipse我相信他抱怨很多错误。
And by the way, correct your syntax, if you're using Eclipse I'm sure he's complaining about a lot of error.
工作代码:
public class NumberConverter {
public static void main(String[] args) {
int i = Integer.parseInt(args[0]);
toBinary(i);
}
public static void toBinary(int int1){
System.out.println(int1 + " in binary is");
System.out.println(Integer.toBinaryString(int1));
}
}
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