[Chapter 3] Problem with do..while loop

Truck35 · November 20th, 2012, 08:59 PM

I am having a problem getting my program to exit my do...while loop. I am also having a problem with my if statement. This program is my variation of the program in excercise 4 chapter 3. The code is below:

Code:

/*This program will ask the user for three numbers.  Two numbers are taken to be multiplied and  the third number is to tell the program how much times to muliply the number. */

function timesTable (start,secondNum,numEnd){

  var firstNum;
  for (firstNum = start; firstNum <= numEnd; firstNum++) {

  document.write (firstNum + " * " + secondNum + " = " + firstNum *   
  secondNum + "<br />" );

  }
}

do{

var a = parseInt (prompt("Enter the number you want to start your times table:", "")); 
var b = parseInt (prompt("Enter the number were you want your times table to end", ""));
var c = parseInt (prompt("Enter the first number you want to muliply by", ""));

   if (isNaN(a||b||c)==true){
     alert("You have entered an invalid key");
     alert("Please try again");
   }
  else 
     timesTable(a,c,b);
}

while (a||b||c != -1);

jmcpeak · November 21st, 2012, 11:35 AM

Howdy, Truck35.

The problem boils down to the conditions in the if and while statements. Let's look at the if statement:

Code:

// original
if (isNaN(a||b||c)==true)

It's understandable why you wrote it this way. It reflects how we think as normal people: "If a, b, or c is not a number..." But programmers have to think in a different way: "If a is not a number, or b is not a number, or c is not a number...". That translates to the following in code:

Code:

// better
if (isNaN(a) == true || isNaN(b) == true || isNaN(c) == true)

But this code still needs work because it compares a boolean value with another boolean value--which is unnecessary in JavaScript (and most other languages). So the above could be rewritten as:

Code:

// correct
if (isNaN(a) || isNaN(b) || isNaN(c))

I'll let you apply this concept to the while statement. I'm here if you need help.