英文:
Why my program doesn't print the wanted outcome?
问题
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
System.out.print("Enter: ");
Scanner scan = new Scanner(System.in);
String input = scan.next();
int whitespace = 0;
int punctuation = 0;
int consonants = 0;
for (int i = 0; i < input.length(); i++) {
char ch = input.charAt(i);
if (ch == '.' || ch == ',' || ch == '!' || ch == ';' || ch == '?')
punctuation++;
if ((ch >= 'a' && ch <= 'z') || (ch >= 'A' && ch <= 'Z'))
consonants++;
else if (ch == 'y' || ch == 'Y')
consonants--;
else
whitespace++;
}
System.out.println("Consonants: " + consonants);
System.out.println("Punctuation: " + punctuation);
System.out.println("White spaces: " + whitespace);
}
}
英文:
- Write a Java program that takes an input string and determines the number of consonants, vowels (‘y’ excluded), punctuation (‘.’, ‘,’ , ‘;’ , ‘!’ , ‘?’), and whitespace characters (‘\n’, ‘\t’, ‘ ’). Print the results to the console with reasonable clarity in output.
import java.util.Scanner;
public class a {
public static void main(String[] args) {
System.out.print("Enter: ");
Scanner scan = new Scanner(System.in);
String input = scan.next();
int whitespace = 0;
int punctuation = 0;
int consonants = 0;
for (int i = 0; i < input.length(); i++){
char ch= input.charAt(i);
if (i == '.' || i == ',' || i == '!' || i == ';' || i =='?')
punctuation++;
if ((ch >= 'a' && ch <= 'z') || (ch >= 'A' && ch <= 'Z'))
consonants++;
else if (i == 'y' || i == 'Y')
consonants--;
else
whitespace++;
}
System.out.println("Consonants: " + consonants);
System.out.println("Punctuation: " + punctuation);
System.out.println("White spaces: " + whitespace);
}
}
Enter: yuuh.
Consonants: 4
Punctuation: 0
White spaces: 1
答案1
得分: 1
else if (i == 'y' || i == 'Y')
consonants--;
你正在使用 i 来检查字符，而不是使用 ch。
<details>
<summary>英文:</summary>
else if ( i== 'y' || i== 'Y')
consonants--;
You are using i to check the character instead of ch.
</details>
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