problem with help menu.

[Toggo]

Hi all im new in java (1 month). I am reading a book and wanted to make my learning easy by making a help menu, but there is a problem i cant solve on my own.
p.s. sorry about the language its Latvian, so i could understand it more easy.

class Help {
	void helpon (int what) {
 
		switch(what) {
			case '1' :
				System.out.println("\"if\" veido: ");
				System.out.println("if(nosacijums)kas notiek; ");
				System.out.println("\telse kas notiek; \n");
				break;
			case '2' :
				System.out.println("\"switch\" veido : ");
				System.out.println("switch (nemainigais) { ");
				System.out.println("\tcase 'attiecigais nosaukums' : un ko dara; ");
				System.out.println("break; lai visu partrauktu\n");
				break;
			case '3' :
				System.out.println("\"for\" veido: ");
				System.out.println("for(kur sakas; kur beidzas; formula");
				System.out.println("bet ir ari protams daudz dazadu veidu, ka manipulet ar \"for\" \n");
				break;
			case '4' :
				System.out.println("\"while\" veido : ");
				System.out.println("while(nosacijums) ko dara; ");
				System.out.println("un dara tik ilgi, lidz paradas \"false\" pie izpildes \n");
				break;
			case '5' :
				System.out.println("\"do-while\" veido : ");
				System.out.println("do { nosacijums ko dara; ");
				System.out.println("\twhile cik ilgi dara\n ");
				break;
			case '6' :
				System.out.println("\"break\" veido : ");
				System.out.println("... kods (kas ir aizverts(logiski pabeigts)) break; vai break label; ");
				System.out.println("pec break atlikuso koda dalu(kas atrodas taja pasa bloka) vairs neizpilda\n");
			case '7' :
				System.out.println("\"continue\" veido : ");
				System.out.println("nosacijums continue;");
				System.out.println("pec continue izpildas tikai nodefinetas formulas vai ari neizpildas nenodefinetas\n");
			}
		System.out.println();
	}
	void showmenu() {
		System.out.println("Sis ir palidzibas logs: ");
		System.out.println("1. Lai iegutu info par \"if\" spied : 1 ");
		System.out.println("2. Lai iegutu info par \"switch\" spied : 2 ");
		System.out.println("3. Lai iegutu info par \"for\" spied : 3 ");
		System.out.println("4. Lai iegutu info par \"while\" spied : 4 ");
		System.out.println("5. Lai iegutu info par \"do-while\" spied : 5 ");
		System.out.println("6. Lai iegutu info par \"break\" spied : 6 ");
		System.out.println("7. Lai iegutu info par \"continue\" spied : 7\n");
		System.out.println("spied 'q' lai izietu");
	}
	boolean isvalid(int a) {
			if(a < '1' | a > '7'& a != 'q') return false;
			else return true;
	}
}
class HelpClassDemo {
	public static void main(String args [])
		throws java.io.IOException{
		char b;
		Help hlpobj = new Help();
 
		for( ; ; ) {
			do{
				hlpobj.showmenu();
				do{
				b = (char) System.read.in();
				} while(b == '\n' | b == '\r');
			}while( !hlpobj.isvalid(b) );
		if(b == 'q') break;
 
		System.out.println("\n");
		hlpobj.helpon(b);
		}
	}
}

[javapenguin]

In your Demo class, it appears that you have two dos but only one while.

[javapenguin]

if(a < '1' | a > '7'& a != 'q') return false;

Using ' ' means a char value.

Typically, when saying if conditions are true, they use || or &&. I did hear that there was something that they would use
| or & for, but I don't think that's what you need to do in this case.

also, you have a as an int.

That will create an error as it'll say "Cannot convert from int to char" or something like that.

[javapenguin]

Is the stuff in the switch part of the program, or what you'd like the help menu to be doing? Those are some weird print outs.

[javapenguin]

In your switch, you probably should get rid of the ' ' around your numbers, as I think that's for using a switch with char variables, which you're not doing.

[javapenguin]

You do realize that do...while will do whatever is in the do brackets, even if the while condition is way false from the start, though it'll only do it once.

[Toggo]

Thanks for all those advices. But the main problem here is in this

 b = (char) System.read.in();

part of code, cant really execute it because my jc returns an error message: input must be char or something.
p.s. those weird print outs are because i am self-learning and this way i can learn really well. Making difficult tasks (for me) to become real.

[javapenguin]

Hi all im new in java (1 month). I am reading a book and wanted to make my learning easy by making a help menu, but there is a problem i cant solve on my own.
p.s. sorry about the language its Latvian, so i could understand it more easy.

class Help {
	void helpon (int what) {
 
		switch(what) {
			case '1' :
				System.out.println("\"if\" veido: ");
				System.out.println("if(nosacijums)kas notiek; ");
				System.out.println("\telse kas notiek; \n");
				break;
			case '2' :
				System.out.println("\"switch\" veido : ");
				System.out.println("switch (nemainigais) { ");
				System.out.println("\tcase 'attiecigais nosaukums' : un ko dara; ");
				System.out.println("break; lai visu partrauktu\n");
				break;
			case '3' :
				System.out.println("\"for\" veido: ");
				System.out.println("for(kur sakas; kur beidzas; formula");
				System.out.println("bet ir ari protams daudz dazadu veidu, ka manipulet ar \"for\" \n");
				break;
			case '4' :
				System.out.println("\"while\" veido : ");
				System.out.println("while(nosacijums) ko dara; ");
				System.out.println("un dara tik ilgi, lidz paradas \"false\" pie izpildes \n");
				break;
			case '5' :
				System.out.println("\"do-while\" veido : ");
				System.out.println("do { nosacijums ko dara; ");
				System.out.println("\twhile cik ilgi dara\n ");
				break;
			case '6' :
				System.out.println("\"break\" veido : ");
				System.out.println("... kods (kas ir aizverts(logiski pabeigts)) break; vai break label; ");
				System.out.println("pec break atlikuso koda dalu(kas atrodas taja pasa bloka) vairs neizpilda\n");
			case '7' :
				System.out.println("\"continue\" veido : ");
				System.out.println("nosacijums continue;");
				System.out.println("pec continue izpildas tikai nodefinetas formulas vai ari neizpildas nenodefinetas\n");
			}
		System.out.println();
	}
	void showmenu() {
		System.out.println("Sis ir palidzibas logs: ");
		System.out.println("1. Lai iegutu info par \"if\" spied : 1 ");
		System.out.println("2. Lai iegutu info par \"switch\" spied : 2 ");
		System.out.println("3. Lai iegutu info par \"for\" spied : 3 ");
		System.out.println("4. Lai iegutu info par \"while\" spied : 4 ");
		System.out.println("5. Lai iegutu info par \"do-while\" spied : 5 ");
		System.out.println("6. Lai iegutu info par \"break\" spied : 6 ");
		System.out.println("7. Lai iegutu info par \"continue\" spied : 7\n");
		System.out.println("spied 'q' lai izietu");
	}
	boolean isvalid(int a) {
			if(a < '1' | a > '7'& a != 'q') return false;
			else return true;
	}
}
class HelpClassDemo {
	public static void main(String args [])
		throws java.io.IOException{
		char b;
		Help hlpobj = new Help();
 
		for( ; ; ) {
			do{
				hlpobj.showmenu();
				do{
				b = (char) System.read.in();
				} while(b == '\n' | b == '\r');
			}while( !hlpobj.isvalid(b) );
		if(b == 'q') break;
 
		System.out.println("\n");
		hlpobj.helpon(b);
		}
	}
}

Usually what's read in is a String.

You could always make what's read in a String and have

b = Stringthat'sreadin.charAt(0);

[javapenguin]

Also, I don't think the class System has a field called read.

It has out, in, and err.

Out is for output (System.out.println())
In is for input(what you're reading in)
Err is for error messages

There's no read.

Nor is there a method called read.

When you read something in, at least for primitive and String types,

you use a Scanner and have it read from a file.

Scanner inFile = new Scanner("inFile.txt");

char b;

b= inFile.nextLine().charAt(0);

To read in a bunch of stuff till you run out

while (inFile.hasNext())
{
read in stuff and do stuff with it
}

[javapenguin]

Also, I think that you need a constructor for your Help class.

Simply add

public Help()
{
 
}

[javapenguin]

if(b == 'q') break;

1.) break and continue statements are discouraged outside of switch structures.

2.) This comes after you do...while loops so what's it going to break anyway?

3.) Couldn't you add
while(.... && p !='q') so that it'll only go through once if p == 'q'

?

[copeg]

1.) break and continue statements are discouraged outside of switch structures.

Discouraged by who, and why?

This comes after you do...while loops so what's it going to break anyway?

The infinite loop