CODEWITHSUNDEEP
java
user13966876
user13966876

Reputation: 57

A question about char cannot be converted to java.lang.String

I am working on a practice question:

Given a string, return a string made of the first 2 chars (if present), however, include the first char only if it is 'o' and include the second only if it is 'z', so "ozymandias" yields "oz".

startOz("ozymandias") → "oz"
startOz("bzoo") → "z"
startOz("oxx") → "o"

Below is my solution, but I don't know why it shows "incompatible types: char cannot be converted to java.lang.String". I was wondering if anyone can help me with my code. Thank you so much!

public String startOz(String str) {
  if (str.length()==1 && str.charAt(0)=='o'){
    return 'o';
  }
  if (str.length()<1){
    return "";
  }
  if (str.length()>=2 && str.charAt(0) =='o' && str.charAt(1) !='z'){
    return 'o';
  }
  if (str.length()>=2 && str.charAt(0) !='o' && str.charAt(1) =='z'){
    return 'z';
  }
  if (str.length()>=2 && str.charAt(0) =='o' && str.charAt(1) =='z'){
    return "oz";
  } else {
    return "";
  }
}

Upvotes: 0

Views: 158

Answers (3)

paulsm4
paulsm4

Reputation: 121819

Rainmaker et al are correct: the problem is that your method should return a String (a Java object), not a char (a Java primitive). One good solution to the compile error is to return "o" (double quotes).

But perhaps you might find this a simpler solution:

public static boolean isNullOrEmpty(String s) {
  return s == null || s.isEmpty();
}

public String startOz(String str) {
  if (isNullOrEmpty(str)) {
    return "";
  } else if (str.startsWith("oz")) {
    return "oz";
  } else if (str.charAt(0) == 'o') {
    return "o";
  } else if (str.length() > 1 && str.charAt(1) == 'z') {
    return "z";
  } else {
    return "";
  }

Upvotes: 1

Oleg Cherednik
Oleg Cherednik

Reputation: 18255

You cannot retreive a char instead of Stirng. Do use String.valueOf(ch) instead.

By the way, just be a simple man! You can do the same with much less code.

public static String startOz(String str) {
    String res = "";

    if (str != null) {
        if (str.length() > 0 && str.charAt(0) == 'o')
            res += 'o';
        if (str.length() > 1 && str.charAt(1) == 'z')
            res += 'z';
    }

    return res;
}

Answer to comment.

  1. StringBuilder is a correct way to build a String, because String is immutable in Java. In case you are a beginner, I removed it.
  2. <condition> ? <true> : <false> is called a ternary operand. if is similar like if(<condition>) <true> else <false>;
  3. And finally I used trim() to remove \0 from the beginning and the end of the string. A also removed it.

You can see my updated result.

Upvotes: 0

Rainmaker
Rainmaker

Reputation: 118

When you type a string in single quotation marks, it tells the compiler to consider whatever follows as a single character. Double quotation marks indicate a string. The function cannot return a char when it is expecting a String.

public String startOz(String str) {
  if (str.length()==1 && str.charAt(0)=='o'){
    return "o";
  }
  if (str.length()<1){
    return "";
  }
  if (str.length()>=2 && str.charAt(0) =='o' && str.charAt(1) !='z'){
    return "o";
  }
  if (str.length()>=2 && str.charAt(0) !='o' && str.charAt(1) =='z'){
    return "z";
  }
  if (str.length()>=2 && str.charAt(0) =='o' && str.charAt(1) =='z'){
    return "oz";
  } else {
    return "";
  }
}

Upvotes: 4

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