Develop Reference

Random Math Question Generator not following while loop

I'm a new java coder getting into it doing a project. I coded it how i believe the system would execute it and yet it doesn't seem to be following the While loops requirements. I want it to generate random number, do a random operation, then ask the user for an answer. The answer must be not decimal and the random numbers must be below 10 to make the questions easier as its for a lower target audience. I'm kind of stuck now on this piece. Apologies if this doesn't make sense as i say it is a first attempt for me.
import java.util.Random;
import java.lang.Math;
import java.util.Scanner;
public class RandomisedQuestions{
public static void QuestionGenerator(){
Random r = new Random();
Scanner s = new Scanner(System.in);
int intA = 0;
int intB = 0;
char operator ='?';
double value = 1.2;
for (int i = 0; i < 3; i++) {
intA = (int)(10.0 * Math.random());//the (int) forces the number to be an int
intB = (int)(10.0 * Math.random());
if (intA <= 0 && intB <= 0){
intA = (int)(10.0 * Math.random());//the (int) forces the number to be an int
intB = (int)(10.0 * Math.random());
System.out.println(intA + intB);
}
while ((value % 1) !=0 && value > 1){//Runs while value is not whole
switch (r.nextInt(4)){
case 0: operator = '+';
value = intA+intB;
break;
case 1: operator = '-';
value = intA-intB;;
break;
case 2: operator = '*';
value = intA*intB;;
break;
case 3: operator = '/';
value = intA/intB;;
break;
default: operator = '?';
}
//System.out.println(operator);
}
System.out.println(intA +""+ operator +""+ intB);
System.out.println("Enter the answer");
int uGuess = s.nextInt();
if (uGuess == value){
System.out.println("Correct");
}
else{
System.out.println("Incorrect");
}
}
}
}

It's better to use ThreadLocalRandom.nextInt to generate your numbers:
// At the start of your program initialize the generator:
ThreadLocalRandom r = ThreadLocalRandom.current();
// Later use it:
do {
intA = ThreadLocalRandom.nextInt(1, 10);
intB = ThreadLocalRandom.nextInt(1, 10);
switch (r.nextInt(4)) {
case 0: operator = '+';
value = intA + intB;
break;
case 1: operator = '-';
value = intA - intB;
break;
case 2: operator = '*';
value = intA * intB;
break;
case 3: operator = '/';
value = (double)intA / intB;
break;
default: operator = '?';
}
} while (value != (int)value || value <= 1);
Also note the conversion to double in division case, otherwise the division will be performed for integer types.

Count the number of occurrences of all the digits in the string

My task is write a program to take string as input(only numbers) and for each digit starting from 0 to 9, print the count of their occurrences in the string.
I have completed it. I have declared 10 integers with zero. Each integer will count the corresponding integers. But in the last when I am printing the result it is giving me the result as 48+count
Count represents the number of count of values occurrences.
For the correct result I need to subtract 48. I am unable to understand why I am getting value.
class TestClass {
public static void main(String args[] ) throws Exception {
Scanner sc = new Scanner(System.in);
int a='0',b='0',c='0',d='0',e='0',f='0',g='0',h='0',i='0',j='0';
String s=sc.next();
OUTER:
for (int k = 0; k<s.length(); k++) {
char ch=s.charAt(k);
switch (ch) {
case '0':
a++;
break;
case '1':
b++;
break;
case '2':
c++;
break;
case '3':
d++;
break;
case '4':
e++;
break;
case '5':
f++;
break;
case '6':
g++;
break;
case '7':
h++;
break;
case '8':
i++;
break;
case '9':
j++;
break;
case ' ':
break OUTER;
default:
break;
}
}
System.out.println("0 "+(a-48));
System.out.println("1 "+(b-48));
System.out.println("2 "+(c-48));
System.out.println("3 "+(d-48));
System.out.println("4 "+(e-48));
System.out.println("5 "+(f-48));
System.out.println("6 "+(g-48));
System.out.println("7 "+(h-48));
System.out.println("8 "+(i-48));
System.out.println("9 "+(j-48));
}
}
please anyone explain me what I can do for removing this extra value in this program.
thanks

Instead of
int a = '0'
use
int a = 0
'0' is equal to 48 in ASCII and it is a character, not a number. So by int a = '0', you actually initialize a to 48

I would suggest to use an array instead. it'll be easier to process.
String str = sc.next();
char[] input = str.toCharArray();
int[] count = new int[10]; // stores the count, int array initialized to 0 by default
for(int i = 0 ; i < input.length; i++){
// get index value by substracting ASCII value
int c = input[i] - 48; // 48 being ASCII Value of '0'
count[c]++;
}
// print the count array
System.out.println(Arrays.toString(count));
count[0] has no of 0's
count[1] has no of 1's
.....

Anyone have suggestions on how to make this Java switch case more efficient/readable?

It is a pretty standard switch case used within a program that replicates an electric circuit. The main thing I am looking for is easier readability of the code and brevity, without disregarding efficiency.
edit: Didn't realize that I was unclear on the purpose of offset, the way offset works is that it will offset the input char by a number of characters equal to offset which is an integer. So for example if source is 'a' and offset is 2 this will return the value within the paths array at index 2.
char passCurrent(char source)
{
source += offset;
switch(source)
{
case 'a':
return this.paths[0];
case 'b':
return this.paths[1];
case 'c':
return this.paths[2];
case 'd':
return this.paths[3];
case 'e':
return this.paths[4];
case 'f':
return this.paths[5];
case 'g':
return this.paths[6];
case 'h':
return this.paths[7];
case 'i':
return this.paths[8];
case 'j':
return this.paths[9];
case 'k':
return this.paths[10];
case 'l':
return this.paths[11];
case 'm':
return this.paths[12];
case 'n':
return this.paths[13];
case 'o':
return this.paths[14];
case 'p':
return this.paths[15];
case 'q':
return this.paths[16];
case 'r':
return this.paths[17];
case 's':
return this.paths[18];
case 't':
return this.paths[19];
case 'u':
return this.paths[20];
case 'v':
return this.paths[21];
case 'w':
return this.paths[22];
case 'x':
return this.paths[23];
case 'y':
return this.paths[24];
case 'z':
return this.paths[25];
}
return '/';
}

Eliminate the switch and use subtraction after checking the range. Something like,
if (source >= 'a' && source <= 'z') {
return this.paths[source - 'a'];
}
return '/';
And, we can shorten that further with a ternary like
return (source >= 'a' && source <= 'z') ?
this.paths[source - 'a'] : '/';

A char is just a number; the meaning of the number has to do with how characters are laid out in Unicode (the numbers from 0 to 127 were defined a long time ago by ASCII, which got subsumed into Unicode).
Thus, if source is 'a', it actually has the integer value 97. The letters from a to z are all consecutive in Unicode, so b is 98, c is 99, etc.
That means that if you want 0 for a, 1 for b, and so on, you can get it by simple subtraction. Thus:
if (source >= 'a' && source <= 'z') {
return this.paths[source - 97];
}
or, equivalently (and more readably):
if (source >= 'a' && source <= 'z') {
return this.paths[source - 'a'];
}
since 'a' is just another way to write 97.

Coverting roman numeral exceptions to decimal

For some class homework I need to create a program that converts roman numerals to decimal form. I can convert just fine as long as there are no exception characters such as IV or IX. How do I check for these exceptions? My attempt was to translate both the current character and the next one into decimal, then to compare them and if the next one (going right to left) is smaller to then subtract it. The problem is that I get out of bounds errors from this.
My current code is this:
Scanner keyboard = new Scanner(System.in);
String roman;
int decimal = 0;
int number = 0;
System.out.print("Enter a Roman Numeral to convert to decimal form: ");
roman = keyboard.next();
roman = roman.toUpperCase();
for (int count = roman.length()-1; count >= 0; count--)
{
char numeral = roman.charAt(count);
switch (numeral){
case 'I':
decimal = 1;
break;
case 'V':
decimal = 5;
break;
case 'X':
decimal = 10;
break;
case 'L':
decimal = 50;
break;
case 'C':
decimal = 100;
break;
case 'D':
decimal = 500;
break;
case 'M':
decimal = 1000;
break;
default:
System.out.println("Error: Invalid character detected.");
break;
}
number = number + decimal;
}
System.out.println("The decimal equivalent is: " + number);
System.out.println("Later!");
I'm still a beginner and most of the information I see on this kind of problem uses advanced solutions that I simply don't understand. I know I need to compare the characters but I'm not sure how to do this in a way that won't eventually go out of bounds.
EDIT: Solved! After posting the question I was struck by insight and solved the problem myself. This code works but I would appreciate any insights into how to improve it!
Scanner keyboard = new Scanner(System.in);
String roman;
int decimal = 0;
int number = 0;
int last = 0;
System.out.println("This program converts Roman Numerals to decimal form.");
System.out.println("Note: Roman Numerals are I, V, X, L, C, D and M.");
System.out.println("All letters entered will be treated as capitalized.");
System.out.print("Enter a Roman Numeral to convert to decimal form: ");
roman = keyboard.next();
roman = roman.toUpperCase();
for (int count = roman.length()-1; count >= 0; count--)
{
char numeral = roman.charAt(count);
switch (numeral){
case 'I':
decimal = 1;
break;
case 'V':
decimal = 5;
break;
case 'X':
decimal = 10;
break;
case 'L':
decimal = 50;
break;
case 'C':
decimal = 100;
break;
case 'D':
decimal = 500;
break;
case 'M':
decimal = 1000;
break;
default:
System.out.println("Error: Invalid character detected.");
System.exit(0);
break;
}
if (decimal >= last){
number = number + decimal;
}
else {
number = number - decimal;
}
last = decimal;
}
System.out.println("The decimal equivalent is: " + number);
System.out.println("Later!");

Well, a decimal is ##.##, and an integer is ##, so you should probably change demical to num.
Set the number as you already do, and add a check after it for the exception character. But first ensure that it exists:
if(this character is not the first && the previous character is an exception)
adjust the number as necessary
This will avoid an out-of-bounds exceptions.

Convert arithmetic string into double in Java

I have a program where the user inputs 6 doubles, and the program outputs every combination of operators that can go in-between the doubles as 1024 separate strings. Here are the first two results if the user inputed 14,17,200,1,5, and 118:
"14.0+17.0+200.0+1.0+5.0+118.0"
"14.0+17.0+200.0+1.0+5.0-118.0"
What I want to do is perform the arithmetic according to the order of operations. Each double is stored as a variable a through f and each operator in-between these variables is stored as a char a_b through e_f. So:
double a, b, c, d, e, f;
char a_b, b_c, c_d, d_e, e_f;
My first thought was to write the code like this:
public double operateGroup() {
value = 0;
switch (a_b) {
case '+':
value += a + b;
break;
case '-':
value += a - b;
break;
case '*':
value += a * b;
break;
case '/':
value += a / b;
break;
default:
break;
}
switch (b_c) {
case '+':
value += c;
break;
case '-':
value += -c;
break;
case '*':
value *= c;
break;
case '/':
value /= c;
break;
default:
break;
}
switch (c_d) {
case '+':
value += d;
break;
case '-':
value += -d;
break;
case '*':
value *= d;
break;
case '/':
value /= d;
break;
default:
break;
}
switch (d_e) {
case '+':
value += e;
break;
case '-':
value += -e;
break;
case '*':
value *= e;
break;
case '/':
value /= e;
break;
default:
break;
}
switch (e_f) {
case '+':
value += f;
break;
case '-':
value += -f;
break;
case '*':
value *= f;
break;
case '/':
value /= f;
break;
default:
break;
}
return value;
}
But this doesn't work because it is the same as doing (a O b) O c) O d) O e) where O is any arbitrary operator. Any tips?

Since there are no parentheses, a trivial approach will work:
Go through the list once to process multiplications and divisions
When an operator between X and Y is * or /, replace X by X*Y or X/Y, and remove Y; also remove the operator
Now go through the list again, this time processing additions and subtractions in sequence.
To implement this approach, define two lists - the list of N Doubles, and N-1 operators, and implement the calculation as follows:
ArrayList<Double> vals = ...
ArrayList<Integer> ops = ... // 1=+, 2=-, 3=*, 4=/
for (int i = 0 ; i < ops.Count ; i++) {
int op = ops.get(i);
if (op == 3 || op == 4) {
if (op == 3) {
vals.set(i, vals.get(i) * vals.get(i+1));
} else {
vals.set(i, vals.get(i) / vals.get(i+1));
}
ops.remove(i);
vals.remove(i+1);
i--;
}
}
double res = vals.get(0);
for (int i = 0 ; i != ops.Count ; i++) {
if (op == 1) {
res += vals.get(i);
} else {
res -= vals.get(i);
}
}

If you need the operators' and operands' information, you should build a Parse Tree (this has been asked before).
If you are only interested in the result, you can evaluate the String directly:
import javax.script.ScriptEngineManager;
import javax.script.ScriptEngine;
public class Eval {
public static void main(String[] args) throws Exception {
ScriptEngineManager s = new ScriptEngineManager();
ScriptEngine engine = s.getEngineByName("JavaScript");
String exp = "14.0+17.0+200.0+1.0+5.0-118.0";
System.out.println(engine.eval(exp));
}
}
Output:
119.0

I would say you should parse it into a tree and then walk the tree to evaluate. Numbers are leaf nodes and operators are parents.