the counter is not updating with my loop and I don't know why-CodePudding

I'm trying to practice for a techniqual test where I have to count the number of characters in a DNA sequence, but no matter what I do the counter won't update, this is really frustrating as I learnt code with ruby and it would update, but Java seems to have an issue. I know there's something wrong with my syntaxt but for the life of me I can't figure it out.

public class DNA {

  public static void main(String[] args) {
    String dna1 = "ATGCGATACGCTTGA";
    String dna2 = "ATGCGATACGTGA";
    String dna3 = "ATTAATATGTACTGA";
    String dna = dna1;
    int aCount = 0;
    int cCount = 0;
    int tCount = 0; 

    for (int i = 0; i <= dna.length(); i  ) {

      if (dna.substring(i) == "A") {
        aCount = 1;
      }
      else if (dna.substring(i) == "C") {
        cCount  ;
      } 
      else if (dna.substring(i) == "T") {
        tCount  ;
      }
      System.out.println(aCount);
    } 
  }
}

It just keeps returning zero instead of adding one to it if the conditions are meet and reassigning the value.

CodePudding user response：

Good time to learn some basic debugging!

Let's look at what's actually in that substring you're looking at. Add System.out.println(dna.substring(i)); to your loop. You'll see:

ATGCGATACGCTTGA
TGCGATACGCTTGA
GCGATACGCTTGA
CGATACGCTTGA
GATACGCTTGA
ATACGCTTGA
TACGCTTGA
ACGCTTGA
CGCTTGA
GCTTGA
CTTGA
TTGA
TGA
GA
A

So, substring doesn't mean what you thought it did - it's taking the substring starting at that index and going to the end of the string. Only the last character has a chance of matching your conditions.

Though, that last one still won't match your condition, which is understandably surprising if you're new to the language. In Java, == is "referential equality" - when applied to non-primitives, it's asserting the two things occupy the same location in memory. For strings in particular, this can give surprising and inconsistent results. Java keeps a special section of memory for strings, and tries to avoid duplicates (but doesn't try that hard.) The important takeaway is that string1.equals(string2) is the correct way to check.

It's a good idea to do some visibility and sanity checks like that, when your program isn't doing what you think it is. With a little practice you'll get a feel for what values to inspect.

CodePudding user response：

Edward Peters is right about misuse of substring that returns a String. In Java, string must be places between double quotes. A String is an object and you must use method equals to compare 2 objects:

String a = "first string";
String b = "second string";
boolean result = a.equals(b));

In your case, you should consider using charAt(int) instead. Chars must be places between simple quotes. A char is a primitive type (not an object) and you must use a double equals sign to compare two of them:

char a = '6';
char b = 't';
boolean result = (a==b);

So, your code should look like this:

public class DNA {
    
        public static void main(String[] args) {
            String dna1 = "ATGCGATACGCTTGA";
            String dna2 = "ATGCGATACGTGA";
            String dna3 = "ATTAATATGTACTGA";
            String dna = dna1;
            int aCount = 0;
            int cCount = 0;
            int tCount = 0;
    
            for (int i = 0; i < dna.length(); i  ) {
                if (dna.charAt(i) == 'A') {
                    aCount  = 1;
                } else if (dna.charAt(i) == 'C') {
                    cCount  ;
                } else if (dna.charAt(i) == 'T') {
                    tCount  ;
                }
                System.out.println(aCount);
            }
        }
    }

CodePudding user response：

substring(i) doesn't select one character but all the characters from i to the string length, then you also made a wrong comparison: == checks 'object identity', while you want to check that they are equals.

You could substitute

if (dna.substring(i) == "A")

with:

if (dna.charAt(i) == 'A')

this works because charAt(i) returns a primitive type, thus you can correctly compare it to 'A' using ==

CodePudding user response：

One of the problems, as stated, was the way you are comparing Strings. Here is a way that uses a switch statement and a iterated array of characters. I put all the strings in an array. If you only have one string, the outer loop can be eliminated.

public class DNA {

    public static void main(String[] args) {
        String dna1 = "ATGCGATACGCTTGA";
        String dna2 = "ATGCGATACGTGA";
        String dna3 = "ATTAATATGTACTGA";

        String[] dnaStrings =
            {dna1,dna2,dna3};
        int aCount = 0;
        int cCount = 0;
        int tCount = 0;
        int gCount = 0;
        for (String dnaString : dnaStrings) {
            for (char c : dnaString.toCharArray()) {
                switch (c)  {
                    case 'A' -> aCount  ;
                    case 'T' -> tCount  ;
                    case 'C' -> cCount  ;
                    case 'G' -> gCount  ;
                }
            }
        }
        System.out.println("A's = "   aCount);
        System.out.println("T's = "   tCount);
        System.out.println("C's = "   cCount);
        System.out.println("G's = "   gCount);
}

prints

A's = 14
T's = 13
C's = 6
G's = 10