java - java中特殊字符算作空格

Question

我正在 MacOS 上工作，使用 JDK 8。 我想计算给定字符串中特殊字符的数量，但在给定代码中，特殊字符被计为空格。我该怎么办？

public static void main(String args[])
{
 String str;
 int lc=0,uc=0,d=0,s=0,spc=0;
 Scanner sc=new Scanner(System.in);
 System.out.print("Enter the string:");
 str=sc.nextLine();

 for(int i=0;i<str.length();i++)
   {
    if(str.charAt(i)>='a' && str.charAt(i)<='z')
    {
    lc++;
    }
    else if(str.charAt(i)>='A' && str.charAt(i)<='Z')
    {
    uc++;
    }
    else if(str.charAt(i)>='0' && str.charAt(i)<='9')
    {
    d++;
    }
    else if(str.charAt(i)>=32)
    {
    s++;
    }
    else if(str.charAt(i)>=33 && str.charAt(i)<=47 || str.charAt(i)==64)
    {
    spc++;
    }
   }
System.out.println("number of small characters in "+str+" are:"+lc);
System.out.println("number of CAPITAL characters in "+str+" are:"+uc);
System.out.println("number of digits in "+str+" are:"+d);
System.out.println("number of Spaces in "+str+" are:"+s);
System.out.println("number of Special characters in "+str+" are:"+spc);
}

这是我得到的输出:

Enter the string:abc23@#$% 

number of small characters in abc23@#$% are:3

number of CAPITAL characters in abc23@#$% are:0

number of digits in abc23@#$% are:2

number of Spaces in abc23@#$% are:4

number of Special characters in abc23@#$% are:0

它应该显示 4 个特殊字符而不是 4 个空格。

Answer 1

将 str.charAt(i)>=32 更改为 str.charAt(i)==32

"abc23@#$% " 的输出将是:

number of small characters in abc23@#$%  are:3
number of CAPITAL characters in abc23@#$%  are:0
number of digits in abc23@#$%  are:2
number of Spaces in abc23@#$%  are:1
number of Special characters in abc23@#$%  are:4