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

Question

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 "";
  }
}

'o' is a char - a primitive type. It cannot be converted directly to the string "o" (notice the double quotes) - you'd need to explicitly do String.valueOf('o'). — Green Cloak Guy
– Green Cloak Guy, Commented Jan 6, 2022 at 18:35
just change the type of quotes from '' to "" for all return excpression — Dmitrii B
– Dmitrii B, Commented Jan 6, 2022 at 18:45

Rainmaker · Accepted Answer · 2022-01-06 18:51:26Z

4

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 "";
  }
}

answered Jan 6, 2022 at 18:51

Rainmaker

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1 Comment

user13966876 Over a year ago

Thank you so much!!

paulsm4 · Accepted Answer · 2022-01-07 16:32:56Z

1

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 "";
  }

edited Jan 7, 2022 at 16:32

answered Jan 6, 2022 at 19:03

paulsm4

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4 Comments

user13966876 Over a year ago

Wow...thank you so much! I did not know "startsWith" and the solution you provided is so clear and really helps me think differently.

user13966876 Over a year ago

This runs, but if str.length()==0, startOz("") → "" it will show StringIndexOutOfBoundsException: String index out of range: 0

paulsm4 Over a year ago

Correct. Or you'll get an NPE if it's null. I added an additional static method, equivalent to .Net string.IsNullOrEmpty().

Oleg Cherednik Over a year ago

You get java.lang.StringIndexOutOfBoundsException for input startOz("x")

Oleg Cherednik · Accepted Answer · 2022-01-06 19:42:07Z

0

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.

StringBuilder is a correct way to build a String, because String is immutable in Java. In case you are a beginner, I removed it.
<condition> ? <true> : <false> is called a ternary operand. if is similar like if(<condition>) <true> else <false>;
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.

edited Jan 6, 2022 at 19:42

answered Jan 6, 2022 at 18:51

Oleg Cherednik

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2 Comments

user13966876 Over a year ago

Thank you so much for your help! I am new to programming, and I was wondering if you could explain a little more. What's StringBuilder? I also don't understand the part "'o' ? 'o' : '\0'". Why do we need trim() at the end? Thank you so much!

Oleg Cherednik Over a year ago

@user13966876 see my answers above

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