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why am i getting this "the type of the expression must be an array type but resolved to string" in JAVA [duplicate]

I am getting the "Must be an array type but it resolved to string" error in my code. It also says that i (in the code below) cannot be resolved to a variable which I don't get.
public class DNAcgcount{
public double ratio(String dna){
int count=0;
for (int i=0;i<dna.length();i++);
if (dna[i]== "c"){
count+= 1;
if (dna[i]=="g"){
count+=1;
double answer = count/dna.length();
return answer;
}
}
}
}
Could you guys please help me figure out where the problem lies? I'm new to coding in Java so I am not entirely comfortable with the format yet.
Thanks a lot,
Junaid

You cannot access a String's character using subscript (dna[i]). Use charAt instead:
dna.charAt(i) == 'c'
Also, "c" is a String, 'c' is a char.
One more thing - integer division ( e.g. int_a / int_b ) results in an int, and so you lose accuracy, instead - cast one of the ints to double:
double answer = count/(double)dna.length();

Use {} to define the scope of the loop. Also, as others already pointed out, use charAt instead of [] and use ' for characters, and use floating point division for the ratio.
for (int i = 0; i < dna.length(); i++) {
if (dna.charAt(i) == 'c') {
count += 1;
}
if (dna.charAt(i) == 'g') {
count += 1;
}
}
Or a bit shorter, use || to or the two clauses together
if (dna.charAt(i) == 'c' || dna.charAt(i) == 'g') {
count += 1;
}

I think you are currently a bit weak at brackets , this is what i understood from your code and corrected it;
public class DNAcgcount{
public double ratio(String dna){
int count=0;
for (int i=0;i<dna.length();i++){
if (dna.charAt(i)== 'c')
count+= 1;
if (dna.charAt(i)=='g')
count+=1;
}
double answer = count/(double)dna.length();
return answer;
}
}
After if we have to close the brackets when what you want in if is finished . I think you wanted count to be the number of time c or g is present in the dna.
You also did some other mistakes like you have to use 'c' and 'g' instead of "c" and "g" if you are using .charAt(i) because it will be treated like a character and then only you can compare .
You may view this link
http://docs.oracle.com/javase/tutorial/java/nutsandbolts/if.html
http://docs.oracle.com/javase/tutorial/java/nutsandbolts/for.html
and you may also have a look at works you can do with string like charAt.

It seems like that you have a few problems with the main syntax of basic java functions like loops or if-else statement. Click here for a good tutorial on these.
You must correct your for-loop and your if-statement:
for(int i=0;i<dna.length();i++){
if(...){
...;
}
if(...){
...;
}
}
Now you wont get the Cant be resolved to a variable... exception.
Second thing is the usage of your string. You have to use it like this:
for(int i=0;i<dna.length();i++){
if(dna.charAt(i) == 'c'){
count += 1;
}
if(dna.charAt(i) == 'g'){
count += 1;
}
}
Now all your exceptions should be eleminated.

Your problem is with syntax dna[i], dna is a string and you access it as it would be an array by []. Use dna.charAt(i); instead.

You using String incorrectly. Instead of accessing via [] use dna.charAt(i).
Altough logically a string is an array of characters in Java a String type is a class (which means it has attributes and methods) and not a typical array.
And if you want to compare a single character to another enclose it with '' instead of "":
if (dna.charAt(i) == 'c')
.
.

There are two errors:
count should be double or should be casted do double answer = (double)count / dna.length();
and as mentioned above you should replace dna[i] with dna.charAt(i)

How to check whether an element of a character array is empty?

I need to check if an element of a java array of characters is empty. I tried out the following code but didn't work. What's wrong?
char c[] = new char[3];
c[0] = 'd';
for (int i = 0; i < c.length; i++) {
if(c[i] == null) {
System.out.println(i + " is empty");
}
}

Let's take arrays out of the equation - an array is just a collection of variables, really. Let's consider a single variable. Suppose we had a method like this:
public boolean isEmpty(char c)
What would that do? The value of c cannot be null, because char is a primitive type. It could be U+0000 (aka '\u0000' or '\0' as character literals in Java), and that's the default value of char (for array elements and fields) but it's not the same as a null reference.
If you want a type which is like char but is a reference type, you should consider using Character - the wrapper type for char just like Integer is for int. Or if you know that you'll never use U+0000 as a valid value, you could just stick with that.
However, a better alternative would often be to design your code so that you don't need the concept of "empty or not empty". We don't know what you're trying to achieve here, but it's usually a good thing to at least consider. For arrays, the alternative is often to use an ArrayList<E> - but of course you can't have an ArrayList<char> in Java as Java generics don't allow primitive type arguments :(

An element of a primitive array can't be null. It will always have a default value if you didn't initialize it yourself. The default for char is 0.

Use Character class instead primitive char.
Character c[] = new Character[3];
c[0] = 'd';
for(int i = 0; i < c.length; i++){
if(c[i] == null){
System.out.println(i + " is empty");
}
}

You cannot use null as char is a primitive type. null only works for objects. use \0 as it's the primitive version of null.

primitive char's default value is 0. you can check it with 0
char c[] = new char[3];
c[0] = 'd';
for(int i = 0; i < c.length; i++){
if(c[i] == 0){
System.out.println(i + " is empty");
}
}
even, char=0 is also a character

Is your code compiling?
You should be seeing an error message at this code line if(c[i] == null)
And from error message, compiler is clearly revealing that "the operator == is undefined for argument type(s) char,null".
This should suffice that element of a primitive array (character array in this case)can't be null.

Replace null with '\0' with the quotes

Java check all characters in string are present in a given string array

I am attempting to create a method that checks every character in userInput to see if they are present in operatorsAndOperands. The issue is that tempbool is always false for all values.
import java.util.*;
public class stringCalculator
{
private String userInput = null;
String[] operatorsAndOperands = {"1","2","3","4","5","6","7","8","9","0","+","-","*","/"};
public stringCalculator(String newInput)
{
userInput = newInput;
}
public boolean checkInput()
{
boolean ifExists = true;
for(int i=0; i<userInput.length(); i++)
{
char currentChar = userInput.charAt(i);
boolean tempbool = Arrays.asList(operatorsAndOperands).contains(currentChar);
if (tempbool == false)
{
ifExists = false;
}
}
return ifExists;
}
}

This is because you have an array of string objects (which you later convert to a list of string objects), but you are checking a char for presence in that array.
Efficiency is also pretty poor here - converting a fixed array to a list on each iteration takes a lot of unnecessary CPU cycles.
A simple solution to this problem is to put all characters in a string, and then check each incoming character against that string:
if ("0123456789+-*/".indexOf(currentChar) >= 0) {
... // Good character
}
Another solution would be making a regex that allows only your characters to be specified, like this:
if (expr.replaceAll("[0-9+/*-]*", "").length() == 0) {
... // Expr contains only valid characters
}

Why don't you declare
String[] operatorsAndOperands = {"1","2","3","4","5","6","7","8","9","0","+","-","*","/"};
as a String, instead of an array of String. Then you can just use the contains method to check the characters against the valid operators.

Declare: char[] operatorsAndOperands; instead of: String[] operatorsAndOperands.
Or add this: String.valueOf(charToCompare) as the "contains" argument.

As has been pointed out, the issue is that you're checking for a char in a list of String objects, so you'll never find it.
You can make this check easier, though, by using a regular expression:
Pattern operatorsAndOperands = Pattern.compile("[0-9+\\-*/]");

Why will this java string routine not print the answer?

I have been working on the Project Euler problem 4. I am new to java, and believe I have found the answer (906609 = 993 * 913, by using Excel!).
When I print the line commented out, I can that my string manipulations have worked. I've researched a few ways to compare strings in case I had not understoof something, but this routine doesn't give me a result.
Please help me identify why it is not printing the answer?
James
public class pall{
public static void main(String[] args){
int i;
int j;
long k;
String stringProd;
for(i=994;i>992; i--){
for (j=914;j>912; j--){
k=(i*j);
stringProd=String.valueOf(k);
int len=stringProd.length();
char[] forwards=new char[len];
char[] back = new char[len];
for(int l=0; l<len; l++){
forwards[l]=stringProd.charAt(l);
}
for(int m=0; m<len;m++){
back[m]=forwards[len-1-m];
}
//System.out.println(forwards);
//System.out.println(back);
if(forwards.toString().equals(back.toString())){
System.out.println(k);}
}
}
}
}

You are comparing the string representation of your array. toString() doesn't give you what you think. For example, the below code makes it clear:
char[] arr1 = {'a', 'b'};
char[] arr2 = {'a', 'b'};
System.out.println(arr1.toString() + " : " + arr2.toString());
this code prints:
[C#16f0472 : [C#18d107f
So, the string representation of both the arrays are different, even though the contents are equal. This is because arrays don't override toString() method. It inherits the Object#toString() method.
The toString method for class Object returns a string consisting of
the name of the class of which the object is an instance, the at-sign
character #, and the unsigned hexadecimal representation of the hash
code of the object. In other words, this method returns a string equal
to the value of:
getClass().getName() + '#' + Integer.toHexString(hashCode())
So, in the above output, [C is the output of char[].class.getName(), and 18d107f is the hashcode.
You can't also compare the arrays using forward.equals(back), as arrays in Java don't override equals() or hashCode() either. Any options? Yes, for comparing arrays you can use Arrays#equals(char[], char[]) method:
if (Arrays.equals(forward, back)) {
System.out.println(k);
}
Also, to get your char arrays, you don't need those loops. You can use String#toCharArray() method. And also to get the reverse of the String, you can wrap the string in a StringBuilder instance, and use it's reverse() method:
char[] forwards = stringProd.toCharArray();
char[] back = new StringBuilder(stringPod).reverse().toString().toCharArray();
And now that you have found out an easy way to reverse a string, then how about using String#equals() method directly, and resist creating those character arrays?
String stringPod = String.valueOf(k);
String reverseStringPod = new StringBuilder(stringPod).reverse().toString()
if (stringPod.equals(reverseStringPod)) {
System.out.println(k);
}
Finally, since it is about project euler, which is about speed and mostly mathematics. You should consider avoiding String utilities, and do it with general division and modulus arithmetic, to get each individual digits, from beginning and end, and compare them.

To convert a string to char[] use
char[] forward = stringProd.toCharArray();
To convert a char[] to String, use String(char[]) constructor:
String backStr = new String(back); // Not the same as back.toString()
However, this is not the most performant solution, for several reasons:
You do not need to construct a back array to check if a string is a palindrome - you can walk the string from both ends, comparing the characters as you go, until you either find a difference or your indexes meet in the middle.
Rather than constructing a new array in a loop, you could reuse the same array - in case you do want to continue with an array, you could allocate it once for the maximum length of the product k, and use it in all iterations of your loop.
You do not need to convert a number to string in order to check if it is a palindrome - you can get its digits by repeatedly taking the remainder of division by ten, and then dividing by ten to go to the next digit.
Here is an illustration of the last point:
boolean isPalindrome(int n) {
int[] digits = new int[10];
if (n < 0) n = -n;
int len = 0;
while (n != 0) {
digits[len++] = n % 10;
n /= 10;
}
// Start two indexes from the opposite sides
int left = 0, right = len-1;
// Loop until they meet in the middle
while (left < right) {
if (digits[left++] != digits[right--]) {
return false;
}
}
return true;
}

StringBuffer Append Space (" ") Appends "null" Instead

Basically what I'm trying to do is take a String, and replace each letter in the alphabet inside, but preserving any spaces and not converting them to a "null" string, which is the main reason I am opening this question.
If I use the function below and pass the string "a b", instead of getting "ALPHA BETA" I get "ALPHAnullBETA".
I've tried all possible ways of checking if the individual char that is currently iterated through is a space, but nothing seems to work. All these scenarios give false as if it's a regular character.
public String charConvert(String s) {
Map<String, String> t = new HashMap<String, String>(); // Associative array
t.put("a", "ALPHA");
t.put("b", "BETA");
t.put("c", "GAMA");
// So on...
StringBuffer sb = new StringBuffer(0);
s = s.toLowerCase(); // This is my full string
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
String st = String.valueOf(c);
if (st.compareTo(" ") == 1) {
// This is the problematic condition
// The script should just append a space in this case, but nothing seems to invoke this scenario
} else {
sb.append(st);
}
}
s = sb.toString();
return s;
}

compareTo() will return 0 if the strings are equal. It returns a positive number of the first string is "greater than" the second.
But really there's no need to be comparing Strings. You can do something like this instead:
char c = s.charAt(i);
if(c == ' ') {
// do something
} else {
sb.append(c);
}
Or even better for your use case:
String st = s.substring(i,i+1);
if(t.contains(st)) {
sb.append(t.get(st));
} else {
sb.append(st);
}
To get even cleaner code, your Map should from Character to String instead of <String,String>.

String.compareTo() returns 0 if the strings are equal, not 1. Read about it here
Note that for this case you don't need to convert the char to a string, you could do
if(c == ' ')

use
Character.isWhitespace(c)
that solves the issue. Best practice.

First, of all, what is s in this example? It's hard to follow the code. Then, your compareTo seems off:
if (st.compareTo(" ") == 1)
Should be
if (st.compareTo(" ") == 0)
since 0 means "equal" (read up on compareTo)

From the compareTo documentation: The result is a negative integer if this String object lexicographically precedes the argument string. The result is a positive integer if this String object lexicographically follows the argument string. The result is zero if the strings are equal;
You have the wrong condition in if (st.compareTo(" ") == 1) {

The compareTo method of a String returns -1 if the source string precedes the test string, 0 for equality, and 1 if the source string follows. Your code checks for 1, and it should be checking for 0.