How to check if a char is equal to an empty space?

if (c == ' ')

char is a primitive data type, so it can be compared with ==.

Also, by using double quotes you create String constant (" "), while with single quotes it's a char constant (' ').

Since char is a primitive type, you can just write c == ' '.
You only need to call equals() for reference types like String or Character.

The code you needs depends on what you mean by "an empty space".

• If you mean the ASCII / Latin-1 / Unicode space character (0x20) aka SP, then:

  if (ch == ' ') {
      // ...
  }
  

• If you mean any of the traditional ASCII whitespace characters (SP, HT, VT, CR, NL), then:

  if (ch == ' ' || ch == '\t' || ch == '\r' || ch == '\n' || ch == '\x0b') {
      // ...
  }
  

• If you mean any Unicode whitespace character, then:

  if (Character.isWhitespace(ch)) {
      // ...
  }
  

Note that there are Unicode whitespace includes additional ASCII control codes, and some other Unicode characters in higher code planes; see the javadoc for Character.isWhitespace(char).

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What you wrote was this:

    if (Equals(ch, " ")) {
        // ...
    }

This is wrong on a number of levels. Firstly, the way that the Java compiler tries to interpret that is as a call to a method with a signature of boolean Equals(char, String).

• This is wrong because no method exists, as the compiler reported in the error message.
• Equals wouldn't normally be the name of a method anyway. The Java convention is that method names start with a lower case letter.
• Your code (as written) was trying to compare a character and a String, but char and String are not comparable and cannot be cast to a common base type.

There is such a thing as a Comparator in Java, but it is an interface not a method, and it is declared like this:

    public interface Comparator<T> {
        public int compare(T v1, T v2);
    }

In other words, the method name is compare (not Equals), it returns an integer (not a boolean), and it compares two values that can be promoted to the type given by the type parameter.

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Someone (in a deleted Answer!) said they tried this:

    if (c == " ")

That fails for two reasons:

• " " is a String literal and not a character literal, and Java does not allow direct comparison of String and char values.

• You should NEVER compare Strings or String literals using ==. The == operator on a reference type compares object identity, not object value. In the case of String it is common to have different objects with different identity and the same value. An == test will often give the wrong answer ... from the perspective of what you are trying to do here.

You could use

Character.isWhitespace(c)

or any of the other available methods in the Character class.

  if (c == ' ')

also works.