2013-05-09 131 views
0

我知道你可以将Java中的字符与正常运算符进行比较,例如anysinglechar == y。但是,我有这个特殊的代码有问题:Java char比较似乎不起作用

do{ 
    System.out.print("Would you like to do this again? Y/N\n"); 
    looper = inputter.getChar(); 
    System.out.print(looper); 
    if(looper != 'Y' || looper != 'y' || looper != 'N' || looper != 'n') 
     System.out.print("No valid input. Please try again.\n"); 
}while(looper != 'Y' || looper != 'y' || looper != 'N' || looper != 'n'); 

的问题不应该是另一种方法,inputter.getChar(),但无论如何,我会放弃它:

private static BufferedReader read = new BufferedReader(new InputStreamReader(System.in)); 
public static char getChar() throws IOException{ 
    int buf= read.read(); 
    char chr = (char) buf; 
    while(!Character.isLetter(chr)){ 
     buf= read.read(); 
     chr = (char) buf; 
    } 
    return chr; 
} 

输出我得到如下:

Would you like to do this again? Y/N 
N 
NNo valid input. Please try again. 
Would you like to do this again? Y/N 
n 
nNo valid input. Please try again. 
Would you like to do this again? Y/N 

正如你所看到的,我把炭是n。然后将其正确打印出来(因此可以看到两次)。但是,这种比较似乎并不真实。

我确定我忽略了一些明显的东西。

+0

你试过吗(a.equals(b))? – cgalvao1993 2013-05-09 16:39:02

+1

@CássioGalvão'(a.equals(b))'如果它的字符串能正常工作。但使用字符“==”。 – Smit 2013-05-09 16:40:40

+1

感谢您的清理,@Leigh。 – hellerve 2013-05-09 16:47:26

回答

3

您的逻辑错误。它始终是truelooper不是'Y'它不是'y'它不是...

你想的逻辑运算符 “和”:&&

if(looper != 'Y' && looper != 'y' && looper != 'N' && looper != 'n') 

和您的while条件中的类似更改。

+0

哦,该死的。我知道这是明显的。对不起,这个愚蠢的问题。 – hellerve 2013-05-09 16:46:57