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Author: Mr.Ehab
2) Function decimalToOctal has the following script:
function decimalToOctal (number) {
    array2.splice(0);
    array3.splice(0);
    for (i = 0; i <= number; i ++) {
        binary = Math.pow (8, i);
        if (binary > number) {
            arrayelement = i - 1;
            break;
        }
        array3 [i] = binary;
    }
    for (j = arrayelement; j >= 0; j --) {
        if (j == arrayelement) {
            binaryelement = int (number / array3 [j]);
            binaryremainder = int (number % array3 [j]);
            array2[ 0 ] = binaryelement;
        } else {
            binaryremainder1 = binaryremainder;
            binaryremainder = int (binaryremainder % array3 [j]);
            binaryelement = int (binaryremainder1 / array3 [j]);
            array2 [arrayelement - j] = binaryelement;
        }
    }
    return array2.join ("");
}

Notice that this script is the same as above but with minor differences. We use powers of 8 instead of 16. No characters are involved.

3) Function binaryToDecimal has the following script:
function binaryToDecimal (number) {
    array7.splice(0);
    array6.splice(0);
    for (i = 0; i <= length (number) - 1; i ++) {
    array6 [i] = Math.floor (number / Math.pow (10, length (number) - i - 1)) -              Math.floor (number / Math.pow (10, length (number) - i)) * 10;
    }
    decimal1 = 0;
    for (j = 0; j < array6.length; j ++) {
        array7 [j] = Math.pow (2, array6.length - j - 1);
        decimal = array7 [j] * array6 [j];
        decimal1 += decimal;
    }
    return decimal1;
}

Notice the first for structure. Length (number) specifies the number of digits in the number. Let's trace this for structure to know what it does. Suppose I entered the binary number 11011 in the input variable, then: Length of the number is 5.

 for (i = 0; i <= 4; i ++)
Values of elements of array6:
For i = 0, array6 [0] = floor(11011/10000) - floor(11011/100000)*10  = 1 
For i = 1, array6 [1] = floor(11011/1000)  - floor(11011/10000) *10  = 1
For i = 2, array6 [2] = floor(11011/100)   - floor(11011/1000)  *10  = 0
For i = 3, array6 [3] = floor(11011/10)    - floor(11011/100)   *10  = 1
For i = 4, array6 [4] = floor(11011/1)     - floor(11011/10)    *10  = 1

Cool, ha?! As you notice, this for structure initializes array6 such that its elements are the digits of the binary number.

Notice the second for structure. Let's have a trace:

for (j = 0; j < 5; j ++)
Output:
For j = 0
Array7 [0] = 2*2*2*2 = 16
Decimal    = 16*1    = 16
Decimal1   = 0 + 16  = 16
For j = 1
Array7 [0] = 2*2*2  = 8
Decimal    = 8*1    = 8
Decimal1   = 16 + 8 = 24
.........
Once execution is finished, decimal1 will be the required decimal number!
» Level Intermediate

Added: 2002-03-04
Rating: 6.08 Votes: 40
(10 being the highest)
» Author
Student studying Computer Engineering and Graphic Design.
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