I am doing a college assignment in Java that deals with currency. For that I am advised to use ints instead of doubles and then later convert it to a dollar value when I print out the statement.
Everything works fine until I do calculations on the number 4005 (as in $40.05 represented as an int). I am pasting the part of code I am having problems with, I would appreciate if someone could tell me what I am doing wrong.
import java.io.*;
class modumess {
public static void main(String[] args) {
int money = 4005; //Amount in cents, so $40.05;
// Represent as normal currency
System.out.printf("$%d.%d", money/100, money%100);
}
}
The above code, when run, shows $40.5, instead of $40.05. What gives?
Kindly note that this is for my homework and I want to learn, so I would really appreciate an explanation about the root of the problem here rather than just a simple solution.
EDIT: Following Finbarr's answer, I have added the following to the code which seems to have fixed the problem:
if (money%100 < 10) {
format = "$%d.0%d";
}
Is this a good way to do it or am I over-complicating things here?
EDIT: I just want to make it clear that it was both Finbarr and Wes's answer that helped me, I accepted Wes's answer because it made it clearer for me on how to proceed.
A better way would be something like this for a general case:
format = "%d.%02d";
%02d gives you 0 padding for 2 digits. That way you don't need the extra if statement.
See this for more explanation of things you can do in format: http://docs.oracle.com/javase/1.5.0/docs/api/java/util/Formatter.html#syntax
The modulus operator returns the remainder after division without fractional calculation. In this case, 4005%100 returns 5 as the remainder of 4005/100 is 5.
Related
I need help calculating the angle between the 5 and the 3. I have absolutely no idea how to even model the question, let alone convert it to java. This is entirely specific, does not need to be abstracted.
int a=3; int b=4; int c=5;
is honestly as far as I can get. This would be cos(b), right? I just need a quick pointer/solution
Is this a right-triangle? So the 5 is the hypotenuse (the side opposite the right angle)?
If that's the case, you need to find any one of the following (which should all be equal to each other): arctan(4/3), ArcSin(4/5), or arccos(3/5).
For example, using the first:
Math.toDegrees(Math.atan(4.0/3.0))
So the code is rather complicated so ill try to do some neat pseudo code that covers the most important issues. I'm trying to parse a math expression. For example: 1-5*(-2)+3 = 14
The syntax that im using is:
expression = term OR term+expression OR term-expression
term = factor OR factor*term OR factor/term
factor = number OR -factor OR (expression)
I have written a piece of code which checks if an expression follows this syntax and it works well for checking the expressions but not for calculating it.
The pseudo code goes something like:
double readExpression()
number = readTerm()
if token == +
number2 = readExpression()
return number + number2
else if token == -
number2 = readExpression()
return number - number2
else
return number
...
(The code for readTerm() is identical to readExpression() in structure)
...
double readFactor()
if token == number
return number
else if token == -
number = readFactor()
return (-1)*number
else if token == (
number = readExpression()
return number
else raise exception
If I do the above calculation with this code it will give me a tree that looks like this:
So anyway, as you matematicians have figured out byt now, the expression should give 14 and not 8 as the tree suggests. I have noticed the that the problem arises when there are minus-signs in front of expressions since affect the whole right term i this problem whilst they should only affect the middle-term.
Ive been thinking like crazy for weeks and thought about solutions for this and looked at other codes and so on. Please dont toss a bunch of links on me if they are not really really simple and good since ive been browsing alot myself on tree traversals and other relevant topics.
What could i do at this stage? As I said, my program can tell if its right or wrong. So now I only need to parse a correct expression. Should I write another class for the parsing of the correct expression? Is it easier? Anyway I dont see how that code would look different than this.
Yes I would parse the equation, it just looks like you miss a key part of the order of operations/parsing. You need to include an additional check for double negatives.
The key factor here is that: In a situation with two identical operators then the left most operation is always carried out first.
First lets narrow down the issue.
This 1-5*(-2)+3 is equal to 1--10+3.
Now for our purposes lets assign a positive to the first operator because it helps illustrate a point:
1--10+3 is the same as +1--10+3
Now if we where to run +1--10+3 through a correct parser we would know that this -- is equal to + but only when used in the following situation:
+X--Y = X+Y
So now our parser has turned the original expression of 1--10+3 into 1+10+3 and we know that is equal to 14.
So all up: Yes you need a parser, but pay special attention to how +X--Y and X+Y work.
Also take a look at this answer: https://stackoverflow.com/a/26227947/1270000
I have just started taking a Computer Science class online and I am quite new to Programming(a couple of week's worth of experience). I am working on an assignment, but I do not understand what a mystery method is. I have yet to find an answer that I can wrap my head around online, in my textbook, or from my professor. Any explanation using this code as an example would also be greatly appreciated!
This is the equation where I saw it in:
public static void mystery1(int n) {
System.out.print(n + " ");
if (n > 0) {
n = n - 5;
}
if (n < 0) {
n = n + 7;
} else {
n = n * 2;
}
System.out.println(n);
}
If anybody can help, that would be amazing! Thank you!
First of all, I voted your question up because I think it's a valid question for someone who is just beginning in computer programming, and I think that some people fail to understand the significance and purpose of Stack Overflow, which is to help programmers in times of need.
Secondly, I think that the couple of users that have commented on your post are on the right track. I have personally never heard of a mystery method, so I think the goal here is for you to simply figure out what the method does. In this case, the method takes a parameter for int 'n'. This means that if, at any point in the application, the 'mystery1()' method is called, an integer will have to be passed as the variable.
Let's say that a user enters the number '9'. The method would be called by the code mystery1(9). This would then run the first part of the 'if' statement, because n is greater than 0. So, n would be equal to n - 5, or 9 - 5, which is 4. (So, n=4.)
I hope my answer was somewhat helpful to you. Take care.
Your assignment is probably to figure out what this method does. More specifically, what does it print to the screen. I'll walk you through how to figure this out.
You have a function, also called a methood, called mystery1. A function is just a named block of code that you can use throughout other pieces of code. This function takes an integer argument called n. Let's assume n=12 for this example.
The first thing that happens in your function when it is called is that n is printed out via the System.out.print method. Notice that it prints a blank space after it. Notice also at the end it prints another value of n that gets assigned within the method. So the method is going to print "12 ?" without the double quotes. The question mark is what we have to figure out. The code says if n > 0 then n = n-5. Since 12 is greater than 0, n gets the new value of 7. The next if statement says if n is less than 0, n gets assigned n+7. But it is not less than zero, it is 7 at this point, so we move to the else statement. In this statement n gets multiplied by 2 which is 14. So the last statement prints 14.
So for an input value of 12 this method prints:
12 14
I hope this helps. If not, please give more detail about your assignment and what you don't understand about my explanation.
The point of this kind of exercise is that you are given a method, but they don't tell you what it does (hence the "mystery"). You are supposed to figure out what it does on your own (like "solving the mystery"). It doesn't mean that the method is special in any way.
Say I give you a "mystery" method like this:
public static void mystery(int n) {
System.out.println(n+1);
}
You would "solve the mystery" by telling me that this method prints out the number that comes after n. Nothing else is special here.
In the example you gave, your job would be to tell me why the method prints out 0 0 when n = 0, or 6 2 when n = 6.
I think the usage of the term "mystery method" is rather misleading, as it has clearly made you (and many, many, many others) believe that something about these methods is special and something that you need to learn about. There isn't anything special about them, and there's nothing to learn.
I think a lot of people would understand this better if instructors just said "tell me what this method does" instead of trying treat students like 5 year olds by saying "Here's a mystery method (ooh, fancy and entertaining). Can you play detective and solve the mystery for me?"
float per = (num / (float)totbrwdbksint) * 100;
i m getting the value of per as say 29.475342 . i want it to round off upto two decimal places only.like 29.48 .how to achieve this?
You should do this as part of the formatting - the floating point number itself doesn't have any concept of "two decimal places".
For example, you can use a DecimalFormat with a pattern of "#0.00":
import java.text.*;
public class Test
{
public static void main(String[] args)
{
float y = 12.34567f;
NumberFormat formatter = new DecimalFormat("#0.00");
System.out.println(formatter.format(y));
}
}
As Jon implies, format for display. The most succinct way to do this is probably using the String class.
float f = 70.9999999f;
String toTwoDecPlaces = String.format("%.2f", f);
This will result in the string "71.00"
If you need to control how rounding is done you should check BigDecimal ist has several rounding modes. http://java.sun.com/j2se/1.5.0/docs/api/java/math/BigDecimal.html
You need to be careful here, this answer is not related to java, it relates to all aspects of decimals in many programming languages hence it is generic. The danger lies with rounding numbers, is this, and it has happened in my experience and know that it can be tricky to deal with:
Supposing you are dealing with prices on items, the pricing you get from a retail supplier may be different to the price the computer tells you, sure it is marginally small, but it could add up to big money.
Adding a sales tax on a price can either be positive or negative, it can impact the operating margin of the profit/loss balance sheets...
If you are in this kind of arena of development, then my advice is not to adjust by rounding up/down...it may not show up on small sales of the items, but it could show up elsewhere...an accountant would spot it...Best thing to do is to simply, truncate it,
e.g. 29.475342 -> 29.47 and leave it at that, why?, the .005 can add up to big profit/loss.
In conjunction to what is discussed here...electronic tills and registers use their own variety of handling this scenario, instead of dealing with XX.XXXXXXXXXX (like computers, which has 27/28 decimal places), it deals with XX.XX.
Its something to keep in mind...
Hope this helps,
Best regards,
Tom.
you can use the formatted print method System.out.printf to do the formatted printing if that's what you need
Sorry if my question sounds dumb. But some time small things create big problem for you and take your whole time to solve it. But thanks to stackoverflow where i can get GURU advices. :)
So here is my problem. i search for a word in a string and put 0 where that word occur.
For example : search word is DOG and i have string "never ever let dog bite you" so the string
would be 000100 . Now when I try to convert this string into INT it produce result 100 :( which is bad. I also can not use int array i can only use string as i am concatinating it, also using somewhere else too in program.
Now i am sure you are wondering why i want to convert it into INT. So here my answer. I am using 3 words from each string to make this kind of binary string. So lets say i used three search queries like ( dog, dog, ever ) so all three strings would be
000100
000100
010000
Then I want to SUM them it should produce result like this "010200" while it produce result "10200" which is wrong. :(
Thanks in advance
Of course the int representation won't retain leading zeros. But you can easily convert back to a String after summing and pad the zeros on the left yourself - just store the maximum length of any string (assuming they can have different lengths). Or if you wanted to get even fancier you could use NumberFormat, but you might find this to be overkill for your needs.
Also, be careful - you will get some unexpected results with this code if any word appears in 10 or more strings.
Looks like you might want to investigate java.util.BitSet.
You could prefix your value with a '1', that would preserve your leading 0's. You can then take that prefix into account you do your sum in the end.
That all is assuming you work through your 10 overflow issue that was mentioned in another comment.
Could you store it as a character array instead? Your using an int, which is fine, but your really not wanting an int - you want each position in the int to represent words in a string, and you turn them on or off (1 or 0). Seems like storing them in a character array would make more sense.