Why?! 1

[creso]

Hi,

Consider the following program:

try {
int choice = 0;
while (choice != 5) {
System.out.println(" 1 : push a onto b "
System.out.println(" 2 : push a over b "
System.out.println(" 3 : stack a onto b "
System.out.println(" 4 : stack a over b "
System.out.println(" 5 : quit "
choice = System.in.read();
System.out.println("choice :" + choice);
}
} catch (Exception e) {
System.out.println(" Error : " + e.getMessage());
}

The program accepts a number from the user, prints the entered number, and displays the menu again. Instead of the entered number being printed 3 new numbers are getting printed (with the menu)!! For example, if I enter 3, I get 51, (menu), 13, (menu), 10 !! Can anyone tell me why this is happening?

Regards.

 

[wushutwist]

You are reading in the character from the user and performing an implicit cast from a char to an int. This just gives you the unicode value of the char not the int value. In other words, if someone enters a four you are getting a '4' not a 4. Understand?

You also should wrap your input stream with a BufferedReader and take in the input line by line instead of a character at a time.

 

[creso]

Hi,

Thanks for reply... I actually had the following line of code:

choice = Integer.parseInt(System.in.read());

I thought, the above would convert "character" to "int". But, the output was as follows:

1 : push a onto b
2 : push a over b
3 : stack a onto b
4 : stack a over b
5 : quit
1
choice :49
1 : push a onto b
2 : push a over b
3 : stack a onto b
4 : stack a over b
5 : quit
choice :13
1 : push a onto b
2 : push a over b
3 : stack a onto b
4 : stack a over b
5 : quit
choice :10
1 : push a onto b
2 : push a over b
3 : stack a onto b
4 : stack a over b
5 : quit

My doubt is, how is that the loop is automatically printing the "menu options" 3 times?! (I assume 13 & 10 are carriage return & new line)

Your suggestion of using InputStreamReader, BufferedReader & readLine worked out... thanks... (I still have the above doubt !!)

Regards