Number generated randomly ? and problem here with the 2nd if statement?

Question

it is show that error in the 2nd If statement why???

import java.security.SecureRandom;

public class Test 
{
   public static void main(String[] args)
   {
      System.out.println("T : " + Tfun());
   }

public static String Tfun()
{
    SecureRandom r = new SecureRandom();
    int T = 1+r.nextInt(10);


    if(T >= 1 && T <= 5)
        Integer.toString(T);
        return "Fast_Plod";

    if(T >= 6 && T <= 7)
        Integer.toString(T);
        return "Slow_Plod";
  }
}

how to fix that? what is the problem exactly?

Place braces around both lines of each `if` block. Unlike Python, indenting means nothing to the Java compiler. — rgettman, Oct 31 '18 at 18:12
This is a perfect example of why I always use brackets, even if only one line of code follows. — Michael Bruesch, Oct 31 '18 at 18:15

GBlodgett · Accepted Answer · 2018-10-31T18:20:20.090

1

In this block (And the other if block):

 if(T >= 1 && T <= 5)
    Integer.toString(T);
    return "Fast_Plod";

You forgot brackets, so the return is always executed, resulting in the rest of the code in the method being dead code (code that can never be reached). Indentation means nothing in java, so you have to specify that the return statement is within the if block by surrounding it, and all the other lines that should be in the if block, with brackets. Also the lines Integer.toString(T); do nothing. Lastly, T can be a number greater than 7, so you must specify a return for this too.

To fix this simply add brackets:

if(T >= 1 && T <= 5) {
    return "Fast_Plod";
}
if(T >= 6 && T <= 7) {
    return "Slow_Plod";
}
else {
    return "";  //return what you want to return if it's > 7
}

edited Oct 31 '18 at 18:20

answered Oct 31 '18 at 18:12

GBlodgett

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i check with brackets also it didnt work – Shaheer Ahmad Oct 31 '18 at 18:16
@ShaheerAhmad What error are you getting? – GBlodgett Oct 31 '18 at 18:18
1

@ShaheerAhmad Because it does not handle the case when `T` is greater than 7. – 001 Oct 31 '18 at 18:18
I want to return T (int type) as a String from that function – Shaheer Ahmad Oct 31 '18 at 18:24
@ShaheerAhmad Then why do you have the `if` statements at all? Why not just `return Integer.toString(T);`? – GBlodgett Oct 31 '18 at 18:25
because i generated random number and if that number is in that condition so it would return it. – Shaheer Ahmad Oct 31 '18 at 18:29
@ShaheerAhmad You want to return the String representation of `T` or a `String` based on the value of `T`? (i.e. if `T` is within a range return "Hello", or within a different range "Goodbye") – GBlodgett Oct 31 '18 at 18:39

score 0 · Answer 2 · answered Oct 31 '18 at 18:15

To add on to GBlodgett's answer you can get away with not putting brackets around the block to be executed in an if statement if it is only one statement (only one ;). Java will execute the first statement after the if and then carry on as if the job is done.

score 0 · Answer 3 · answered Oct 31 '18 at 18:23

import java.security.SecureRandom;

public class Test {

    public static void main(String[] args) {
        System.out.println("T : " + Tfun());
    }

    public static String Tfun() {
        SecureRandom r = new SecureRandom();
        int T = 1+r.nextInt(10);

        if(T >= 1 && T <= 5) {
            Integer.toString(T);
            return "Fast_Plod";
        }else if(T >= 6 && T <= 7) {
            Integer.toString(T);
            return "Slow_Plod";
        }else {
            return "your return value.";
        }
    }
}

Number generated randomly ? and problem here with the 2nd if statement?

3 Answers3