Converting this bufferedreader to string

Question

I am trying to convert this buffered reader to a string but for some reasong the if statement will not recognize the first letter as a string

BufferedReader userInputMessage = new BufferedReader(new InputStreamReader(System.in));
message = userInputMessage.readLine();

firstLetter = message.substring(0,1);

System.out.println("FirstLetter: " + firstLetter + "/");
message = message.substring(1);
System.out.println("Message: " + message + "/");

if(!firstLetter.equals("0") || !firstLetter.equals("1")){
    System.out.println("First letter not valid!");
}

re "Converting this bufferedreader to string" -- A BufferedReader is not a String, nor can it be "converted" into a String. — Hovercraft Full Of Eels
– Hovercraft Full Of Eels, Commented Oct 20, 2014 at 21:06
Also, if your code is causing an error, please print the complete error message, and also indicate which line is causing the error to occur. — Hovercraft Full Of Eels
– Hovercraft Full Of Eels, Commented Oct 20, 2014 at 21:07

Hovercraft Full Of Eels · Accepted Answer · 2014-10-20 21:08:41Z

2

This will always be true:

if(!firstLetter.equals("0") || !firstLetter.equals("1")){

think about it. If first letter is 0, then the second part is true, if it's 1, the first part will be true, if it's anything else, both will be true. You want to use && perhaps:

if(!firstLetter.equals("0") && !firstLetter.equals("1")){

answered Oct 20, 2014 at 21:08

Hovercraft Full Of Eels

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4 Comments

christopher Over a year ago

Simple but sometimes difficult to spot. Good catch! + 1

Richard Barley Over a year ago

Thanks! it is sometimes difficult to spot

Joop Eggen Over a year ago

!case1 || !case2 should as pattern immediately ring an alarm.

Hovercraft Full Of Eels Over a year ago

@JoopEggen: yep. A simple mental Venn diagram will show a big problem if case1 and case2 have an empty union set.

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