CODEWITHSUNDEEP
java
koshish kharel
koshish kharel

Reputation: 347

Getting error in if statement

I am just trying to sort the string in descending order. input provided by the user is 10,a,1,#,15,.,6 output must be 10,a,15,#,6,.,1 I have tried.

String input = JOptionPane.showInputDialog("Enter a string:");
String[] num = input.split(",");
ArrayList<String> arr = new ArrayList<>();

 System.out.println(num);
  for ( int i = 0; i < num.length - 1; i ++ )  
   {
    for (int j = i + 1; j < num.length; j ++ )
     {
       if(Integer.parseInt(num[i]) 
       && Integer.parseInt(num[j])  
        && num[i] < num[j])  {
          String temp = num[ i ];   //swapping
           num[ i ] = num[ j ];
           num[ j ] = temp;
            }
            }           
      } 
     }

In if statement i get error. error:- The operator && is undefined for the argument type(s) int, int - The operator < is undefined for the argument type(s) java.lang.String, java.lang.String

Upvotes: 1

Views: 95

Answers (4)

bcsb1001
bcsb1001

Reputation: 2907

I'm not going to look at the rest of your code, but the bit with the error should give an error. Here is an illustration:

Integer.parseInt(num[i]) --> an int
Integer.parseInt(num[j]) --> another int

&& takes two booleans and gives a boolean, true if both evaluate to true.

However, two ints are given. What do we do with those? That's right, an error.

num[i] --> a String object (remember the declaration: String[] num = ...)
num[j] --> another String object

< is the 'less than' comparison operator, which works on ints, longs, shorts, bytes, floats and doubles.

As String is not one of those numeric types, however, an error is thrown.

I think that you want to say Integer.parseInt(num[i]) < Integer.parseInt(num[j]). This works, because you give the < operator two ints. Some ints are greater than others, not the case with Strings, so it compiles fine. But note the warning in Eran's answer; you will get a NumberFormatException if Integer.parseInt() is given an invalid input. To handle this, use a try/catch:

int numI, numJ;
try {
    numI = Integer.parseInt(num[i]);
    numJ = Integer.parseInt(num[j]);
    // rest of code
} catch (NumberFormatException e) {
    // not a number, handle here!
}

Upvotes: 0

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726579

The first thing is to make your code compile: recall that Integer.parse returns the integer, so you cannot use it in the sequence of logical expressions.

Then, you should make your code more efficient: move parsing of num[i] outside the loop, and skip the loop when parsing fails, like this:

for ( int i = 0; i < num.length - 1; i ++ ) {
    int numI;
    try { numI = Integer.parse(num[i]); } catch (NumberFormatException nfe) { continue; }
    for (int j = i + 1; j < num.length; j ++ ) 
        int numJ;
        try { numJ = Integer.parse(num[i]); } catch (NumberFormatException nfe) { continue; }
        if (numI < numJ) {
            String temp = num[ i ];   //swapping
            num[ i ] = num[ j ];
            num[ j ] = temp;
        }
    }
}

Now when parsing num[i] fails, the program skips over the nested loop, going to the next string instead.

Upvotes: 0

SMA
SMA

Reputation: 37023

You should change your if from:

Integer.parseInt(num[i]) 
   && Integer.parseInt(num[j])  
    && num[i] < num[j]

To:

Integer.parseInt(num[i]) 
   < Integer.parseInt(num[j])  
   ^^^
    && num[i].compareTo(num[j]) < 0
              ^^^^^^^^^

If you want to compare two numbers, you could compare for less than/greater than and/or equal to. You could apply && on two booleans like condition1 && condition2

Aside note, if any of the number is not parseable as integer then you might get NumberFormatException. So i would suggest you first recognize input and then use the api to convert it to a number.

Upvotes: 1

Eran
Eran

Reputation: 393821

You can replace 

   if(Integer.parseInt(num[i]) 
      && Integer.parseInt(num[j])  
      && num[i] < num[j])  {

with

   if(Integer.parseInt(num[i])  <
      Integer.parseInt(num[j])) {

However, you'll get a NumberFormatException if one of the Strings can't be parsed as an integer.

Upvotes: 2

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