You'll want to cast to float (or use floatval()) in your condition statement. This leaves your values untouched in case you need other data from your variables, such as trailing text: $dividend = 100; $divisor = '0.0'; if ((float)$dividend && (float)$divisor) //both are non-zero { $quotient = $dividend/$divisor; } else...

math,floating-point,division,ieee-754

One practical way of correctly rounding the result of iterative division is to produce a preliminary quotient to within one ulp of the mathematical result, then use the exactly-computed residual to compute the final result. The tool of choice for the exact computation of residuals is the fused-multiply add (FMA)...

java,user-interface,division,multiplication

Here's something simpler, but it essentially does what you want out of your program. I added an ActionListener to each of the buttons to handle what I want, which was to respond to what was typed into the textbox. I just attach the ActionListener to the button, and then in...

Using np.array(range(...)) has to first create a range, which in Python 2 is a regular list. This means it will have to create a 48,000-element list just as a preliminary to creating the numpy array. Instead, try using np.arange(1, n+1) to directly create a numpy array with a certain range...

quot rounds towards zero, div rounds towards negative infinity: div (-3) 2 == (-2) quot (-3) 2 == (-1) As to the overhead of div, quot has a corresponding primitive GHC operation, while div does some extra work: quotRemInt :: Int -> Int -> (Int, Int) (I# x) `quotRemInt` (I#...

You've done the multiplication of the RHS vector by the inverse matrix incorrectly. If the variables in your code x and y match your equation, then they shouldn't be arrays. Your variable names are confusing. I don't think of the RHS as solution; that's what x and y will be....

Your for loop is never executing because you're testing if i == Dlist.Count. It should be: for (int i = 0;i < Dlist.Count; i++) Alternatively, you could do this with LINQ: public List<int> PaceCalc(List<int> Dlist, List<int> Slist) { return Dlist.Zip(Slist, (a, b) => a / b).ToList(); } ...

javascript,jquery,html,css,division

As I see you are using jQuery but in a very wrong way. I've created a JSFiddle for you. take a look at this Update 1: Edited The Code For Better Performance By Adding: $("#slider").stop() $(document).ready(function() { $("#slider").animate({ "left": $(".item:first").position().left + "px", "width": $(".item:first").width() + "px" }, 0); $(".item").hover(function() {...

algorithm,division,modulo,brainfuck,divmod

Here's what happens: # >n 0 d This line, is a comment line telling you what the memory should be like before the operation. Dividend as n, divisor as d. According to the code the next 3 cells should be empty as well, but it is ignored here, assuming you...

python,python-2.7,floating-point,division

repr shows more digits (I'm guessing just enough to reproduce the same float): >>> print result 100.0 >>> print repr(result) 99.99999999999997 >>> result 99.99999999999997 >>> print step 4.7619047619 >>> print repr(step) 4.761904761904762 >>> step 4.761904761904762 ...

Approximately everything. If you want a linear interpolation from 0 to x with n steps then you need to multiply x by k/n where k goes from 0 to n. You... don't do this. [maxVal * k / 10 for k in range(0, 11)] ...

I would perform each modulus and store the result(s) in boolean variables. Like, boolean mod3 = i % 3 == 0; boolean mod7 = i % 7 == 0; if (mod3 && mod7) { System.out.printf("%d is divisible by 3 and 7.%n", i); } else if (mod3 || mod7) { System.out.printf("%d...

python,calculator,division,subtraction

You can always use operandList[0] as the starting point if you wish, as long of course as operandList is not empty (up to you what you want to do when it is empty). if not operandList: raise ValueError('Empty operand list') # or whatever result = operandList[0] for operand in operandList[1:]...

You are missing: .chat-container { overflow: hidden; } DEMO: https://jsfiddle.net/jw727odf/1/...

This should work - it creates a list of owner 001 make and color, joins all the other owners against the list, and only includes those owners who match for all of owner 001's cars: WITH Owner001_Cars AS ( SELECT car_make, car_color FROM myTable WHERE owner_id = '001') SELECT owner_id...

It should be pretty simple. The logic may go like this: Randomly generate a number and call it divisor Multiply your divisor with a randomly generated number and call it dividend Now you've got two integer numbers (dividend & divisor) that should solve into a whole number quotient. ...

int quant = (int)(money / 0.55); if ( quant >= 1 ) printf( "You selected Mars and have enough money to buy %d bars", quant ); This will remove everything after the decimal point. If you want to force round up or down you need the functions floor and ceil...

assembly,double,mips,division,mars-simulator

The apparent success of the first division is due to an artifact of how small positive integers and positive subnormal doubles are represented. Both have leading zeros, with the binary bit pattern corresponding to the significant bits of the value in the least significant bits. The effect of treating the...

linux,bash,scripting,division,operand

To assign the output of a command to a variable, you need to use the backticks or (preferably) the $() syntax. s=$(cat /sys/bus/w1/devices/28-000006c5772c/w1_slave | grep t= | cut -d "=" -f2) will set $s to 24187 That, and removing the spaces after the = signs as suggested by chepner will...

c++,codeblocks,long-integer,division

If Population_A is in the range -16 .. 16, the integer multiply and divide will result in 0. If you want floating point division results rather than converting integer values to floats, change 6 to 6.0F and 3 to 3.0F....

assembly,optimization,bit-manipulation,division,multiplication

That method is called, "Division by Invariant Multiplication". The constants that you're seeing are actually approximates of the reciprocal. So rather than computing: N / D = Q you do something like this instead: N * (1/D) = Q where 1/D is a reciprocal that can be precomputed. Fundamentally, reciprocals...

The code you have posted is working just fine. $timesby = 360 / 36; echo($timesby); > 10 In PHP, division is done using the / operator, just as you have done above....

Divide each number of outer array (arr1) with respective number in other array(arr2) The code is shown below. You should also put check to verify that both arrays have same length. public static void main(String[] args) { int arr1[]={8,4,6,8,4}; int arr2[]={2,4,2,1,2}; for (int x =0;x <arr1.length;x++){ int result = arr1[x]...

The idea here is to use the if condition from most specific to least specific. In you case the most specific condition is a divisor of 4 and 7, followed by divisor of 4, divisor fo 7 and finally the least specific one which means everything else. If you could...

Instead of while (divisor <= dividend) { // this will not work with UINT_MAX divisor <<= 1; current <<= 1; } divisor >>= 1; current >>= 1; try unsigned k = 1 << (sizeof(unsigned) * 8 - 1); while (((divisor & k) == 0) && ((divisor << 1) <= dividend))...

If I'm understanding correctly, there are multiple ways to do this. The simplest approach, treating it sort of like a vector: Add your hours for each class, divide by the total. So from your example: teacher : andy lesson : math (4 hours),chemistry (2 hours) total hours : 30 hours...

c++,algorithm,integer,division,bigint

First of all, you can implement division in time O(n^2) and with reasonable constant, so it's not (much) slower than the naive multiplication. However, if you use Karatsuba-like algorithm, or even FFT-based multiplication algorithm, then you indeed can speedup your division algorithm using Newton-Raphson. A Newton-Raphson iteration for calculating the...

def divide(dividends, divisors): ret = dict() for key, dividend in dividends.iteritems(): if key in divisor: ret[key] = dividend/divisors[key] else: ret[key] = dividend return ret ...

You formatting with the integer specifier. Try this method instead. print('{} / {} = {}'.format(numberA, numberB, answer)) ...

Your num happens to be the most negative number that can be represented by a signed 32-bit variable (assuming 2's complement); unfortunately for you, the most positive number that can be so represented is 2147483647 - one less than -2147483648/-1.

sql-server,count,subquery,division

You can use your first query to populate a local variable and the use it in the second query like this: declare @IniDate as int = 20140610, @EndDate as int = 20150425, @Week int select @Week = COUNT(distinct sk_calendar_week) from TableA where Sk_Date between @IniDate and @EndDate ) Select sk_calendar_week,...

python,performance,division,modulus,divmod

To measure is to know: >>> import timeit >>> timeit.timeit('divmod(n, d)', 'n, d = 42, 7') 0.22105097770690918 >>> timeit.timeit('n // d, n % d', 'n, d = 42, 7') 0.14434599876403809 The divmod() function is at a disadvantage here because you need to look up the global each time. Binding it...

You're applying a where clause to the outer query (b), where there is no such column Season there. You should move it to the inner query: $result = mysql_query(" SELECT team , COUNT(*) played , SUM(win) wins , SUM(loss) lost , SUM(win)/count(*) pctWon1 , SUM(draw) draws , SUM(SelfScore) ptsfor ,...

sql,sql-server,count,division,percentage

You need conditional Aggregate to find the percentage of output start with 'RT @' SELECT ( Count(CASE WHEN tweet LIKE 'RT @%' THEN 1 END) / Cast(Count(*) AS NUMERIC(10, 2)) ) * 100 FROM temp ...

Difficult to answer at the rate you are editing but: m_angle = math.degrees(math.atan(float(thing.position()[1]-stuff.position()[1]) / float(thing.position()[0]-stuff.position()[0]))) the bit inside the atan is equivalent to something/(a-b) m_angle = math.degrees(math.atan(float(thing.position()[1]-stuff.position()[1]) / thing.position()[0]-stuff.position()[0])) the bit inside the atan is equivalent to (something/a)-b...

Assume that there is no negative values. Only use assign, loop, increment, sub, add. Div(A, B) { temp = 1; result = -1; loop(A) { temp = 0; increment(result) A = sub(A, B) } return add(result, temp); } ...

If M is either a compile time constant or is constant within a loop then instead of using division you can calculated a reciprocal and then do multiplication and a shift. We can write x/M = (x*(2^n/M))>>n The factor 2^n/M (aka magic number) should be calculated before the loop or...

The book probably means round towards negative infinity, i.e. -∞: Directed roundings Round toward 0 – directed rounding towards zero (also known as truncation). Round toward +∞ – directed rounding towards positive infinity (also known as rounding up or ceiling). Round toward −∞ – directed rounding towards negative infinity (also...

You can also divide by keeping it as a data.frame df[] <- lapply(df, function(x) as.numeric(gsub('[.]', '', x))) res <- df[1:3,]/unlist(df[4,])[col(df)[-4,]] Or res <- t(t(df[1:3,])/unlist(df[4,])) res[,1] #[1] 1.638736e-06 3.277471e-07 7.374310e-06 df[1:3,1]/df[4,1] #[1] 1.638736e-06 3.277471e-07 7.374310e-06 Update The updated data have numeric last row 6 res <- df[1:5,]/unlist(df[6,])[col(df)[-6,]] res[,6] #[1] NaN NaN...

The usual reason: edx is supposed to be the "upper half" of a 64bit dividend. So if you have a negative number, edx can not be zero (at least the sign bit must be set, otherwise it's just not negative), and if you had a 32bit signed number, you have...

Using itertools.izip (Python2.7): import itertools [[float(aaa) / bbb for (aaa, bbb) in itertools.izip(aa, bb)] \ for (aa, bb) in itertools.izip(a, b)] ...

You can use a subquery : SELECT SUM(i.price)/(SELECT COUNT(*) FROM customers) FROM items i LEFT JOIN orders o ON i.itemID = o.itemID AND o.date BETWEEN '2013-03-01' AND '2013-03-31' WHERE o.itemID IS NOT NULL ...

I would approach this by removing all of the factors of 3, 5, and 7 from the original number, and seeing what's left. while(num % 3 == 0) { num = num / 3; } while(num % 5 == 0) { num = num / 5; } while(num % 7...

Don't know if this helps, but it really looks like you're doing the division twice, since (40.423/714861)*100 is actually 0.005595493389623997. I would definitely check that.

You are reading in int values using the %f formatter flag, which is meant for float. a and b are of type int, so you want to use %d.

As mentioned by @Borgleader in a comment, the problem was library compilation with the -fast-math compiler option in a computation library but not my application. In this case it wasn't OpenCV but rather another library in my distribution, but it was this discrepancy that caused the difference. The problem was...

You must add L at the end of 3600 or 1000000: Example: long hour = 92233720368L / (3600 * 1000000L ); Here's what's hapenning: System.out.println(3600 * 1000000); // Gives -694967296 because it exceeds the max limit of an integer size. So 92233720368L / -694967296 = -132 That's exactly what's happening...

EAX and EDX are integer registers, so DIV is an integer division. You cannot expect a rational number like 0.9. DIV gives you in EDX the remainder of an integer division. You can use the FPU floating point registers or - better - multiply testScore with 100 before DIV: #include...

100/65 is an integer division. What you need is double d = (100d / 65) % 1; This will give you 0.53846153846153855...

javascript,arrays,division,parsefloat

Due to the particular value you get, I'm going to guess that your currentAmount isn't 1000 as you claim, but rather 1,000 or 1 000 or even 1'000 in some countries. When parseFloat gets hold of that, it sees only 1 because , isn't a valid character in numbers. 1...

130 / 20 performs integer divison From / Operator (C# Reference) When you divide two integers, the result is always an integer. For example, the result of 7 / 3 is 2 That's why it always discards the fractional part and it returns 6. That's why your result will be...

php,loops,foreach,division,modulus

I got this to work. My problem was that my array was multidimensional, so I converted it to a single array. Afterwards, I used array_chunks. $chunks = array_chunk($l, 2); $i=0; foreach($chunks as $mychunk){ if($i%2== 0){ echo "<tr class=\"r0\">"; } else { echo "<tr class=\"r1\">"; } $i++; foreach($mychunk as $newchunk) {...

Pipes work on output streams, and: C="$A / $B" | bc -l does not send "$A / $B" to the output stream, instead just sends an eof. You can do this: C=$(echo "$A / $B" | bc -l) to get the result into C....

For your case, I think you need your parent div contains both 2 dynamic size children. You can set the property overflow of the parent to achieve that effect. .parent{ overflow:hidden; } ...

algorithm,integer,division,integer-division

Here's how I'd approach it, at least initially. Every bucket has a desired amount that it needs. This is based on their float values, and all the float values sum to 1. So, go through the "objects" to be distributed one by one. To figure out which bucket gets it,...

html,css,css-float,inline,division

Added float:left; to the .stats div-children. .stats { display: inline-block; margin-right: 6px; float: right; padding: 0; font-family: "Roboto" sans-serif; color: black; } .stats div{ float:left; } .separator { background: rgba(47, 187, 255, 0); height: 20px; width: 1px; vertical-align: middle; border-left: solid; border-width: 1px; border-color: rgba(0, 124, 184, 0.7); float: left;...

ruby,variables,division,literals

Ruby's parser is interpreting /1000 as the beginning of a literal regexp, since a is a token which may be a method. That is, imagine that a is a method: def a(arg); end Time.at(a /1000) Ruby will interpret this as "a invoked with an incomplete regexp as the argument". To...

The two answers you show are the same, one is in scientific notation, the other is in standard notation. For example: 0.00000000851 == 8.51e-9 ...

Take a look at this 0.30000000000000004.com Your language isn't broken, it's doing floating point math. Computers can only natively store integers, so they need some way of representing decimal numbers. This representation comes with some degree of inaccuracy. That's why, more often than not, .1 + .2 != .3. ...

I think try except (as in Cyber's answer) is usually the best way (and more pythonic: better to ask forgiveness than to ask permission!), but here's another: def safe_div(x,y): if y == 0: return 0 return x / y One argument in favor of doing it this way, though, is...

matlab,plot,division,figure,subplot

I have come this problem many times and haven't yet figured out a decent way to solve it. However what you can do is: A) Include a Label (help label) in the subplot you want. Alternatively use a "edit locked" edit text field. B) Yes in a way. Check out...

python,debugging,pandas,dataframes,division

You are doing chained assignment which you shouldn't do as it sometimes doesn't work which is what you are observing: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy moreover you should always use .loc if possible, if we compare the performance between using a mask and without a mask we can see that for a 25000 row...

c++,matrix,vector,division,eigen

After the suggestion of Avi, I put my comment as an answer so that the question is closed. Fixed it. Apparently I was thinking too complicated . A simple RowVector3d adsf = difference * normTransform.inverse(); did the trick...

java,loops,while-loop,logic,division

if you divide an integer by another integer you will always have an integer. you need the module (remainder) operation int num1 = 15; int num2 = 16; int num3 = 5; System.out.println("num1/num3 = "+num1/num3); System.out.println("num2/num3 = "+num2/num3); System.out.println("num1%num3 = "+num1%num3); System.out.println("num2%num3 = "+num2%num3); if the module is not 0...

does it mean that, when I divide one floating-point number by another and the the dividend is divisible by the divider without a reminder (like 10000.0 is divisible by 10.0), it's possible that I get a number with .99999999... at the end No. IEEE 754 division is correctly rounded....

When you divide two ints you perform integer division, which, in this case will result in 22/64 = 0. Only once this is done are you creating a float. And the float representation of 0 is 0.0. If you want to perform floating point division, you should cast before dividing:...

actionscript-3,division,flash-cs6,modulus

I think the way you've framed the question may not actually fit well with your problem. If you want to know each time your score passes a 100 increment, you could simply do this: // represents current hundreds (1=100, 2=200, etc) private var currentLevel:int = 0; // each time score...

abs expects an integer, and you are providing a float. It should work if you apply an explicit type cast: sum = c / (pow(abs((int)(x1 - x2)), 2)); ...

You can simply perform long division by hand, which is O(N) on the number of digits -- it's hard to see how you could do better than that. The only problem with long division is that it would not terminate ever if the fraction is a repeating decimal, but you...

ruby,division,ruby-2.0,integer-division

Integer 0 is exactly zero; there is no error. And since division by zero is mathematically undefined, it makes sense that integer division by 0 raises an error. On the other hand, float 0.0 is not necessarily representing exactly zero. It might originate from a number whose absolute value is...