I have to make compatible the java function with jQuery. The jQuery Real Person captcha validate function is written by java. I want to validate the captcha in jQuery. So, I tried to write the java function in jQuery. But the function could not return the right value.
Java:
private String rpHash(String value) {
int hash = 5381;
value = value.toUpperCase();
for(int i = 0; i < value.length(); i++) {
hash = ((hash << 5) + hash) + value.charAt(i);
}
return String.valueOf(hash);
}
jQuery:
function rpHash(value) {
hash = 5381;
value = value.toUpperCase();
for($i = 0; $i < value.length; $i++) {
hash = ((hash << 5) + hash) + value.charAt($i);
}
return hash;
}
It'd help if you'd post the expected result (from Java) and the actual result (from JavaScript). I tested it with the string "abc" and the Java function returns 193450027 while the JavaScript version returns "00177573ABC". Looking at these results, the problem is much easier to spot.
The problem is that you did not translate the loop body correctly. hash is an int, thus ((hash << 5) + hash) is also an int. However, value.charAt(i) is a char. What happens when Java needs to add an int to a char? Simple: a char is a Unicode character represented by two bytes. The char is extended to an int with the same Unicode value and is then added to the int.
However, JavaScript works very differently. hash is a number and value.charAt(i) is a string with one character. When adding a number to a string, the + is interpreted as a concatenation of two strings and thus the result is also a string. Since JavaScript is not statically typed, it is perfectly legal for the value of hash to change from number to string.
In further iterations, hash will (most likely) be a string holding both numerical and non-numerical characters. Thus, when evaluating hash << 5, the implicit conversion from a string to a number needed for the shifting operation silently fails and returns 0. The + hash as well as + value.charAt($i) become concatenations for the same reason as above.
Therefore, the correct translation would be to retrieve the character code rather than the character:
hash = ((hash << 5) + hash) + value.charCodeAt($i);
This will correctly result in the intended addition of two numbers.
On an unrelated note, you should write var $i = 0 instead of $i = 0 in your for initialization to prevent leaking $i to the global scope. The same applies to hash, which should be declared as var hash = 5381. Also, the convention is that variables prefixed with $ represent a jQuery object, which $i doesn't. You better just name it i like in Java.
Related
I am obviously new to java. I have this assignment where I am supposed to write a program which performs arithmetic operations on numbers expressed as a character string.
I don't know where to start. I have tried googling, looking through my book, big java, in the relevant sections but can't seem to find helpful information.
I found a program that have completed the same assignment but I want to learn write my own and understand how to go about.
I can show you one of the methods that he used.
I have bolded a few comments where I get confused.
public static String add(String num1, String num2) {
while (num1.length() > num2.length()) {
num2 = "0" + num2;
}
while (num1.length() < num2.length()) {
num1 = "0" + num1;
}
int carry = 0; // whats the point of this?
String result = "";
// look at the for loop bellow. I don't understand why he is converting the strings to ints this
// way? this doesn't even return the correct inputed numbers?
for (int i = 1; i <= num1.length(); i++) {
int digit1 = Character.getNumericValue(num1.charAt(num1.length() - i));
int digit2 = Character.getNumericValue(num2.charAt(num2.length() - i));
int sum = digit1 + digit2 + carry;
carry = sum / 10;
result = (sum % 10) + result;
// why is he dividing the sum with 10? If the user inputs a 5, would't the result become 0.5
// which isn't a valid int value? this line is also confusing
}
if (carry > 0) {
result = carry + result;
}
return result;
}
Any explanation or even guidance to a page where I am trying to do is explained would be very appreciated.
I found a program that have completed the same assignment but I want to learn write my own and understand how to go about.
That is the right idea. I suggest that you stop looking at the code that you found. (I'm sure that your teachers don't want you to look up the answers on the internet, and you will learn more from your homework if you don't do it.)
So how to proceed?
(I am assuming that you are supposed to code the methods to do the arithmetic, and not just convert the entire string to a primitive number or BigInteger and use them to do the arithmetic.)
Here's my suggested approach:
What you are trying to program is the equivalent of doing long addition with a pencil and paper. Like you were taught in primary school. So I suggest that you think of that pencil-and-paper procedure as an algorithm and work out how to express it as Java code. The first step is to make sure that you have the steps of this algorithm clearly in your head.
Try to break the larger problem into smaller sub-problems. One sub-problem could be how to convert a character representing a decimal digit into an integer; e.g. how to convert '1' to 1. Next sub-problem is adding two numbers in the range 0 to 9 and dealing with the "carry". A final sub-problem is converting an integer in the range 0 to 9 into the corresponding character.
Write sample Java code fragments for each sub-problem. If you have been taught about writing methods, some of the code fragments could be expressed as Java methods.
Then you assemble the solutions to the sub-problems into a solution for the entire problem. For example, adding two (positive!) numbers represented as strings involves looping over the digits, starting at the "right hand" end.
As part of your program, write a collection of test cases that you can use to automate the checking. For example:
String test1 = add("8", "3");
if (!test1.equals("11")) {
System.out.println("test1 incorrect: expected '11' go '" +
test1 + "'");
}
Hints:
You can "explode" a String to a char[] using the toCharArray method. Or you could use charAt to get characters individually.
You can convert between a char representing a digit and an int using Character methods or with some simple arithmetic.
You can use + to concatenate a string and a character, or a character and a string. Or you can use a StringBuilder.
If you need to deal with signed numbers, strip off and remember the sign, do the appropriate computation, and put it back; e.g. "-123 + -456" is "- (123 + 456)".
If you need to do long multiplication and long division, you can build them up from long addition and long subtraction.
You can convert a number in String format to a number in numeric format by “long n = Long. parseLong(String)” or “Long n = Long.valueOf(String)”. Then just add 2 long variables using a + sign. It will throw NumberFormatException if the String is not a number but a character. Throw that exception back to the caller.
The first part of the code pads both numbers to equal lengths.
e.g. "45" + "789" will be padded to "045" + "789"
The for loop evaluates one character at a time, starting from the right hand most.
iteration 1 -> (right most)
5 + 9 -> 14
when you divide an integer with another integer, you will always get an integer.
hence carry = 14/10 = 1 (note: not 1.4, but 1, because an int cannot have decimal places)
and the remainder is 14 % 10 = 4 (mod operation)
we now concatenate this remainder into "result" (which is "" empty)
result = (14%10)+ result; // value of result is now "4"
iteration 2 -> (second right most)
4+8 + (carry) = 4 + 8 + 1 = 13
same thing, there is a carry of 13/10 = 1
and the remainder is 13%10 = 3
we concatenate the remainder into result ("4")
result = (13%10) + result = 3 +"4" = "34"
iteration 3->
0 + 7 + 1 = 8
this time 8/10 will give you 0 (hence carry = 0)
and 8%10 will give a remainder of 8.
result = 8 + "34" = "834"
after all the numbers have been evaluated, the code checks if there are anymore carry. if the value is more than 0, then that value is added to the front of the result.
Is there any way that I can use a hashcode of a string in java, and recreate that string?
e.g. something like this:
String myNewstring = StringUtils.createFromHashCode("Hello World".hashCode());
if (!myNewstring.equals("Hello World"))
System.out.println("Hmm, something went wrong: " + myNewstring);
I say this, because I must turn a string into an integer value, and reconstruct that string from that integer value.
This is impossible. The hash code for String is lossy; many String values will result in the same hash code. An integer has 32 bit positions and each position has two values. There's no way to map even just the 32-character strings (for instance) (each character having lots of possibilities) into 32 bits without collisions. They just won't fit.
If you want to use arbitrary precision arithmetic (say, BigInteger), then you can just take each character as an integer and concatenate them all together. Voilà.
No. Multiple Strings can have the same hash code. In theory you could create all the Strings that have have that hash code, but it would be near infinite.
Impossible I'm afraid. Think about it, a hashcode is a long value i.e. 8 bytes. A string maybe less than this but also could be much longer, you cannot squeeze a longer string into 8 bytes without losing something.
The Java hashcode algorithm sums every 8th byte if I remember correctly so you'd lose 7 out of 8 bytes. If your strings are all very short then you could encode them as an int or a long without losing anything.
For example, "1019744689" and "123926772" both have a hashcode of -1727003481. This proves that for any integer, you might get a different result (i.e. reversehashcode(hashcode(string)) != string).
Let's assume the string consists only of letters, digits and punctuation, so there are about 70 possible characters.
log_70{2^32} = 5.22...
This means for any given integer you will find a 5- or 6-character string with this as its hash code. So, retrieving "Hello World": impossible; but "Hello" might work if you're lucky.
You could do something like this:
char[] chars = "String here".toCharArray();
int[] ints = new int[chars.length];
for (int i = 0; i < chars.length; i++) {
ints[i] = (int)chars[i];
}
Then:
char[] chars = new char[ints.length]
for (int i = 0; i < chars.length; i++) {
chars[i] = (char)ints[i];
}
String final = new String(chars);
I have not actually tested this yet... It is just "concept" code.
Hiii,I need to do the opposite of what my hash method does, I want a number to convert it to a string, unlike my other method.
I need you to do it in the same way coding as decoding
That would only be possible if there was a 1 to 1 mapping between Strings and longs. Since there are 264 possible long values, and many many more possible String values (even if you limit yourself to Strings of 64 characters, there are still K64 of them, where K is the number of possible unique characters), there cannot be a method that reverses your long hash(String c) method for all possible Strings.
I think that you are trying to implement the Vigenère cipher.
I have fixed your code for executing the two functions (hash and hash2) and I have noticed that hash("javaguay") returns the long number 2485697837967351 and then hash2(2485697837967351L) returns yaugavaj, the reverse string that you want.
A quick response could be the next, but I think you must fix your algorithm.
Add these lines to the end hash2 function:
String res2 = "";
for (int i = res.length() -1; i >=0; i--) {
res2 += res.charAt(i);
}
return res2;
I want to hash a word into fixed bit hash value say 64 bit,32 bit (binary).
I used the following code
long murmur_hash= MurmurHash.hash64(word);
Then murmur_hash value is converted into binary by the following function
public static String intToBinary (int n, int numOfBits) {
String binary = "";
for(int i = 0; i < numOfBits; ++i) {
n/=2;
if(n%2 == 0)
{
binary="0"+binary;
}
else
binary="1"+binary;
}
return binary;
}
Is there any direct hash method to convert into binary?
Just use this
Integer.toBinaryString(int i)
If you want to convert into a fixed binary string, that is, always get a 64-character long string with zero padding, then you have a couple of options. If you have Apache's StringUtils, you can use:
StringUtils.leftPad( Long.toBinaryString(murmurHash), Long.SIZE, "0" );
If you don't, you can write a padding method yourself:
public static String paddedBinaryFromLong( long val ) {
StringBuilder sb = new StringBuilder( Long.toBinaryString(val));
char[] zeros = new char[Long.SIZE - sb.length()];
Arrays.fill(zeros, '0');
sb.insert(0, zeros);
return sb.toString();
}
This method starts by using the Long.toBinaryString(long) method, which conveniently does the bit conversion for you. The only thing it doesn't do is pad on the left if the value is shorter than 64 characters.
The next step is to create an array of 0 characters with the missing zeros needed to pad to the left.
Finally, we insert that array of zeros at the beginning of our StringBuilder, and we have a 64-character, zero-padded bit string.
Note: there is a difference between using Long.toBinaryString(long) and Long.toString(long,radix). The difference is in negative numbers. In the first, you'll get the full, two's complement value of the number. In the second, you'll get the number with a minus sign:
System.out.println(Long.toString(-15L,2));
result:
-1111
System.out.println(Long.toBinaryString(-15L));
result:
1111111111111111111111111111111111111111111111111111111111110001
Another other way is using
Integer.toString(i, radix)
you can get string representation of the first argument i in the radix ( Binary - 2, Octal - 8, Decimal - 10, Hex - 16) specified by the second argument.
Integer.toBinaryString(data)
gives me a binary String representation of my array data.
However I would like a simple way to add leading zeros to it, since a byte array equal to zero gives me a "0" String.
I'd like a one-liner like this:
String dataStr = Integer.toBinaryString(data).equals("0") ? String.format(format, Integer.toBinaryString(data)) : Integer.toBinaryString(data);
Is String.format() the correct approach? If yes, what format String should I use?
Thanks in advance!
Edit: The data array is of dynamic length, so should the number of leading zeros.
For padding with, say, 5 leading zeroes, this will work:
String.format("%5s", Integer.toBinaryString(data)).replace(' ', '0');
You didn't specify the expected length of the string, in the sample code above I used 5, replace it with the proper value.
EDIT
I just noticed the comments. Sure you can build the pattern dynamically, but at some point you have to know the maximum expected size, depending on your problem, you'll know how to determine the value:
String formatPattern = "%" + maximumExpectedSize + "s";
This is what you asked for—padding is added only when the value is zero.
String s = (data == 0) ? String.format("%0" + len + 'd', 0) : Integer.toBinaryString(data);
If what you really want is for all binary values to be padded so that they are the same length, I use something like this:
String pad = String.format("%0" + len + 'd', 0);
String s = Integer.toBinaryString(data);
s = pad.substring(s.length()) + s;
Using String.format() directly would be the best, but it only supports decimal, hexadecimal, and octal, not binary.
You could override that function in your own class:
public static String toBinaryString(int x){
byte[] b = new byte[32]; // 32 bits per int
int pos = 0;
do{
x = x >> 1; // /2
b[31-pos++] = (byte)(x % 2);
}while(x > 0);
return Arrays.toString(b);
}
would this satisfy your needs?
String dataStr = data == 0 ? "00" + Integer.toBinaryString(data) : Integer.toBinaryString(data);
edit: noticed the comment about dynamic length:
probably some of the other answers are more suited:)
This, in concept, is almost same as #Óscar López answer, but different methods are used, so i thought i should post it. Hope this is fine.
1] Building the format string
String format = "%0" + totalDigits + "d";
2] Integer to Binary Conversion
String dataStr = Integer.toBinaryString(data);
3] Padding with Leading Zeros
dataStr = String.format(format, new Integer(dataStr));
The major difference here is the 3rd step. I believe, its actually a hack.
#erickson is right in String.format() not supporting binary, hence, i converted the binary number to an integer (not its equivalent), i.e., "100" will be converted to hundred (100), not four(4). I then used normal formatting.
Not sure about how much optimized this code is, but, i think its more easy to read, but, maybe, its just me.
EDIT
1] Buffer Over-run is possible for longer binary strings. Long can be used, but, even that has limitations.
2] BigInteger can be used, but, I'm sure, it will be the costliest at runtime compared to all the other methods.
So, it seems, unless only shorter binary strings are expected, replace() is the better method.
Seniors,
please correct me if I'm wrong.
Thanks.