arrays,assembly,calculator,8086,addition

I figured it out myself, finally. Turns out I didn't need to use the DEC DI in afteradd: part. That command caused the marker to be placed twice, one after another, which fooled the procedure that was supposed to print the result.

c,formatting,integer,printf,addition

print one number before without "+" printf("%d", num); then do in a loop printf("+%d", num); according to ur code. this should work int perfectNumGen(int num) { int perfNumAlgo = pow(2, num - 1)*(pow(2, num) - 1); int factorNum = 1; printf("%d=", num); if (num % factorNum == 0) printf("%d", factorNum);...

The + in this case calls the unary magic method __pos__ rather than __add__: >>> class A(int): def __pos__(self): print '__pos__ called' return self ... >>> a = A(5) >>> +a __pos__ called 5 >>> +++a __pos__ called __pos__ called __pos__ called 5 Python only supports 4(unary arithmetic operations) of...

r,date,data.frame,lookup,addition

I thought it might be informative to let users know that I have solved this. I ended up leaving the Lookup data.frame in the melted form i.e.: df <- data.frame(Date.Time, Nest, value) I then used paste to combine the Nest and Date.Time columns paste(df$Nest, df$Date.Time, sep = "_") so that...

c,floating-point,add,bit,addition

Code is doing signed int shifts and certainly unsigned long long is desired. // unsigned long long x = (1 << (exp2 - exp1 + MANBITS)); unsigned long long x = (1LLU << (exp2 - exp1 + MANBITS)); Notes: Suggest more meaningful variable names like x_mantissa. Rounding not implemented. Rounding...

binary,overflow,addition,twos-complement,underflow

It seems like whenever you have 3 carries in a row, it drops a 1. Same with the next one, 15 + 7. It isn't doing the carry correctly. 11 <- carry 0011 = 3 +0111 = 7 1010 = 10 ...

Well, based on your requirements I'd suggest you just subtract your first random number from 100. Then you have two random numbers with that sum. Finally, Java naming convention is lower case letter first. Like, Random r = new Random(); int low = 10; int high = 100; int r1...

This construction is hardly special. You just have to understand the basic logic of what multiplication is x*y is adding x to itself y times. Multiplying x by 0 is 0 Multiplying x by 1 is x Multiplying x by N is adding x to multiplying x by N-1. Assuming...

that depends on what you are trying to do if you are just computing simple/single +/- operations then the overflow is usually ignored when you need to handle overflow/underflow for example if you need to clamp the result for some reason usually safety of the result range ... then Carry...

Parse it to integer, you are concatinating it because one element ($left_position) is a string. $new_left_position = parseInt($left_position) + 400; function ".replace()" will ofcourse always return a string so your variable is a string aswell....

c,ruby,function-pointers,addition

You may not do this if the function pointers are not in the same array of function pointers. The additive operator for pointers and integral values is valid only for pointers in an array. A single object is considered as an array with one element. Also take into account that...

C[i+1] == 1 does a comparison, not an assignment.

You can write totalScore as a computed property so that it will always be the sum of the other 3 properties. var totalScore: Int { get { return section1score + section2score + section3score } } ...

rmultinom might be handy here: x <- rmultinom(n = 1, size = 100, prob = rep(1/4, 4)) x colSums(x) Here I draw one vector, with a total size of 100, which is splitted into 4 groups....

This should work for you: <?php $val1 = "$10.00"; $val2 = "$10"; $myvalue = ltrim($val1, "$") + ltrim($val2, "$"); echo "$" . number_format((float)$myvalue, 2, ".", ""); ?> Output: $20.00 And if you have only 1 string this should work: <?php $string = "$10.00 + $10"; $values = preg_split("/(\+)/", $string); array_walk($values,...

python,variables,reference,addition,augmented-assignment

No, you cannot. You cannot use augmented assignment together with multiple targets. You can see this in the Augmented assignment statements section you linked to: augmented_assignment_stmt ::= augtarget augop (expression_list | yield_expression) augtarget ::= identifier | attributeref | subscription | slicing The augtarget rule only allows for one target. Compare...

Another approach to convert an unsigned integer to a string and write it: section .text global _start _start: mov eax, 1234567890 mov ebx, 5 add eax, ebx ; Convert EAX to ASCII and store it onto the stack sub esp, 16 ; reserve space on the stack mov ecx, 10...

Add Calculations(); to toString() public String toString() { Calculations(); return "Purchase: " + purchase + "\nDate: " + date +"\nQuantity: " + quantity +"\nCost:" + fmt.format(totalCost); } OR since the calculations method is public you can call it from the main method with myBasket[j].Calculations(); after myBasket[j] = new Basket(purchase, date,...

javascript,forms,addition,parsefloat

please compare your code with mine.i have fixed your first section which is calculating score in test and lab.please do the same for time based test. <html> <head> <script type="text/javascript" src="https://code.jquery.com/jquery-2.1.1.min.js"></script> <script type="text/javascript"> $(document).ready(function(){ }); </script> <script type="text/javascript"> var result=0; function do_addition() { var lab1 = parseFloat(document.forms["form1"]["Lab1Score"].value); var lab2 =...

If I understood correctly, you want var obj = {mybuttonid: 99}; $('button').click(function () { alert(obj[this.id]); obj[this.id]++; alert(obj[this.id]); }); var obj = { "foo": 99, "bar": 50 }; $('button').click(function () { alert(obj[this.id]); obj[this.id]++; alert(obj[this.id]); }); <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script> <button id="foo">99</button> <button id="bar">50</button> ...

How is 01234 being treated? It is treated here as octal. Any number starting with 0 is octal....

overflow,verilog,addition,signed,subtraction

{OFAdd, AddAB} <= A + B; In the example the MSB (OFAdd) is not an overflow bit. If you had access to the carry out of the final bit of an adder this could act as an overflow, but in RTL you do not have access to this only another...

This is easier if you use DateTime() function addMonth($year_month){ $date = DateTime::createFromFormat('Ym', $year_month); $date->modify('+1 month'); return $date->format('Ym'); } Alternative using DateInterval() function addMonth($year_month){ $date = DateTime::createFromFormat('Ym', $year_month); $date->add(new DateInterval('P1M')); return $date->format('Ym'); } ...

php,variables,session,addition

Your ternary statement is currently only checking if $_SESSION['vip'] is set. The problem is that it is set no matter what the user chooses ("yes" or "no") because in the part where you set the session variable, the conditional statement sets the session variable if $_POST['vip'] isn't empty. Try checking...

Summarizing the above comments: [myself]: Adding a very small floating point decimal amount to a (relatively) much larger number only appears to return an incorrect value: ## default R> options()["digits"] # $digits # [1] 7 ## a <- 2.656779e-08 b <- 1 ## R> a+b # [1] 1 ## This...

If the part is FF, change it to 00 and increment the next part. It's precisely the same way you increment a decimal number when a digit is a 9 -- you make it a 0 and increment the next digit to the left, so 19 becomes 20.

Sum should be an int int sum = 0; Your for loop should be for (int i = 0; i < tokens.length; i++) { sum += Integer.parseInt(tokens[i]); } ...

You have two stray commas: x = raw_input("Give me the first number:" ), # ^ y = raw_input("Give me the second number:" ), # ^ The comma operator creates a tuple, so both x and y now contain a tuple with one element. Remove those commas. See the Expression lists...

I think we'd need to see the "user is sent to the confirmation page" portion of the application, but if you validate and then redirect user to a new page, the $_POST global will be empty. It appears you're pushing POST into SESSION, so maybe something like this instead: <b>Total...

Ok this seems to work : <script> $(document).ready(function(){ $('#invoices').bind('DOMNodeInserted DOMNodeRemoved', function() { var sum = 0; $('.grandtotal').each(function(){ var tal = $(this).text(); if(isNaN(tal)) { sum += 0; }else{ sum += parseFloat($(this).text());}; }); $('#total_price').text('£' + sum); }); }); </script> ...

The + operator is only defined for numbers (addition) and strings (concatenation). However, it coerces its arguments to number or string. Full details in the spec, §11.6.1. Basically: Processing + does a "to primitive" on the operands, which means it then ends up with numbers, strings, booleans, nulls, or undefineds....

binary,overflow,addition,signed,twos-complement

Overflow occurs only when correct answer cannot be interpreted using given number of bits.An extra bit is required to interpret the answer right.Consider the following example of byte numbers addition: +70 in binary 01000110 +80 in binary 01010000 On addition :10010110 1 in the MSB(most significant bit) indicates that answer...

The output you've posted is the trailing lines of the output. Not the first 30-31 lines. It goes so fast that in first 30 iterations it goes beyond the integer limit and start showing 0. (Remember 2^32 ? int is 4 byte) Instead of while(true) { try while(count>0) {, you...

javascript,processing,nan,numeric,addition

JavaScript identifiers are case-sensitive, so nx and ny are not the same as nX and nY: nx = x; ny = y; ... x+=((nX-x)/delay); y+=((nY-y)/delay); ...

assembly,hex,unsigned,addition,signed

binary addition has no notion if signed vs unsigned. exactly the same the bit patterns. The meaning of those bit patterns is in the eyes of the beholder (multiply and divide this is not true). This is the beauty of twos complement. 0xF0 + 0x02 = 0xF2 = 240 +...

because you are using parseInt() which convert it into integer, instead try parseFloat: var myResult = parseFloat(myBox1) + parseFloat(myBox2) + parseFloat(myBox3); Here is demo...

The problem You defined the method: def add(c) return self + c end and attempted to use it thus: 3.add(4) #=> NoMethodError: private method `add' called for 3:Fixnum Understanding this error message This error message tells you exactly what the problem is. I think your problem is simply that you...

Here is a dynamic-programming example: def make_combo_dict(nums, max_): nums = sorted(set(nums)) prev = {0: [""]} # base state: no combinations for num in nums: nxt = defaultdict(list) for total, ways in prev.items(): # k = 0 nxt[total] = ways # k > 0 max_k = (max_ - total) // num...

c,performance,math,increment,addition

In this particular example all four expressions have the exact same externally observable result so a competent compiler should generate the exact same code for them. The compiler doesn't slavishly read the code and generate a few instructions for each statement, the compiler reasons about what the result of the...

Remove the quotes in points='(points + $pointsvar)' and for SET points='(points - $pointsvar)' As for this line: if ($conn->query($sql2) === TRUE) { You're using that in conjunction with $sql3 so that should be: if ($conn->query($sql3) === TRUE) { ...

python,loops,if-statement,while-loop,addition

number = # generate random number while number != 1: if number % 2: # if number is odd, multiply by 3, add 1 number *= 3 number += 1 else: # if number is even, divide by 2 number /= 2 You can run a bit of cheeky code...

arrays,matlab,conditional,addition

The behaviour you are looking for is called (phase) unwrapping and there's a built in function unwrap for that res = unwrap(angles / 90 * pi) / pi * 90 Note that unwrap works in radians and for jumps of +/-pi and not 2*pi as you request, hence I'm intentionally...

python,loops,while-loop,python-3.4,addition

Sorry I redid your whole code, but IT WORKS NOW! Tot=input("Enter Income:") NetInc=Tot NetXpense=0 c = 0 while (c==0): Xpense=input("Enter your expenses:") NetXpense+=int(Xpense) if (int(Xpense)==0): c=1 NetInc = int(Tot)-NetXpense print("Expenditure:",NetXpense) print("Net Income:",NetInc) Try this and let me know :)...

Try something like: INSERT INTO result (no, total, average) SELECT no, A+B+C+D, (A+B+C+D)/4 FROM marks ...

I don't see the issue, your code does follow PEMDAS. The only issue is that you're not getting 0.5 because you're using integer division. Try this: System.out.println((1 + (1 - 1)) / 2.0); System.out.println(1 + (1 - 1) / 2.0); ...

c++,multithreading,math,addition

Yes, as those operations are not atomic. You use iterator to take a value, add something to it and stores the value back - there is quite some machine code behind this. Another thread can jump in and replace your value at any time. You must ensure thread-safety of this...

25 / 2 = 12r1 (12 with a remainder of 1) 12 / 2 = 6r0 (6 with a remainder of 0) 6 / 2 = 3r0 (3 with a remainder of 0) 3 / 2 = 1r1 (1 with a remainder of 0) 1 / 2 = 0r1 (0...

You can do this easily with Numpy broadcasting: a = np.array([0, 40, 80, 120]) b = np.array([0, 1, 2, 3]) c = a[:, None] + b[None, :] The statement a[:, None] is equivalent to a.reshape(1, -1), making a a column vector. Similarly, b[None, :] makes b a row vector. Numpy's...

java,class,netbeans,static,addition

Consider this instead: public class Money { private int m; public Money(int m) { this.m = m; } public int getDollars() { return m / 100; } public int getCents() { return m % 100; } public int get() { return m; } public Money add(Money other) { return new...

javascript,string,floating-point,addition

The Problem is: floatNumber.toPrecision(2) This creates a string from your number. The toPrecision-function is used to create a representation of your number with a fixed number of digits....

Unfortunately there is no ideal solution to this. We must use approximation type comparison. For example, instead of checking: if val1 == val2 we must try something like: if val1 > (val2 - .0005) && val1 < (val2 + .0005) ...

bash,awk,grep,compare,addition

Without sample input and expected output we're just guessing but MAYBE this is the right starting point for your script (untested, of course, since no in/out provided): #!/bin/bash awk -F, ' NR==FNR { file = $1 ".pdb" ARGV[ARGC] = file file2col2s[file] = (col1to2s[file] ? file2col2s[file] FS : "") $2 next...

TL;DR - You cannot add two pointers. Pointer(s) is(are) memory address(es). Just think for a moment, even if you would have been allowed to add two pointers, what is the significance of the result? It will produce mostly an invalid value, isn't it? OTOH, you can add an int to...

javascript,function,onchange,addition

HTML: <input type="checkbox" id="extra-bathroom" name="extra-bathroom" value= "1" style="display: inline;" onchange="updatePrice();">Extra bathroom? <input id="total-price" type="button" value= 0 style="width: 100%; margin-top: 5px;"> Javascript: window.updatePrice = function () { if (document.getElementById('extra-bathroom').checked) { document.getElementById('total-price').setAttribute("value", parseInt(document.getElementById("total-price").value) + 42); } else {...

to accumulate several frames into a 'sum' frame, you need one with a larger depth, else you will overflow (or saturate) it. Mat acc(height,width,CV_32FC3,Scalar::all(0)); cv::accumulate(frame,acc); cv::accumulate(frame,acc); cv::accumulate(frame,acc); acc /= 3; Mat mean; acc.convertTo(mean, CV_8UC3); ...

I don't think the reference is a very good answer. The better answer -- which applies in your case -- is: select coalesce(a, 0) + coalesce(b, 0) + . . . You can add as many variables as you like with this approach. Note that this returns 0, not NULL...

Try to use number_format like this: $a = 100; $b = number_format(91.51, 0, ".", "." ); $c = number_format(8.49, 0, ".", "." ); $d = $a - $b - $c; echo $d; ...

Numbers with a leading 0 are octal numbers in the same way numbers with leading 0x are hexadecimal numbers: 01 = 1 02 = 2 ... 07 = 7 010 = 8 011 = 9 Normally 09 would cause a syntax error because 9 is not an octal digit (in...

c,floating-point,printf,addition

According to IEEE 754 encoding, many numbers will have small changes to allow them to be stored.Also, the number of significant digits can change slightly since it is a binary representation, not a decimal one. Single precision (float) gives you 23 bits of significand, 8 bits of exponent, and 1...

#include <stdio.h> #include <ctype.h> int main(void){ char string[1000]; char digits[3] = {0}; int i, j, x, sum = 0; printf("Enter the string containing both digits and alphabet\n"); scanf("%999s", string); for(j = i = 0; string[i]; ++i){ if(isdigit(string[i])){ digits[j++] = string[i]; if(j==2){ sscanf(digits, "%d", &x); sum += x; j = 0;...

you need parseInt() var total = parseInt(numberOnly) + 20; because numberOnly is a string, not a number, so its adding a number to a string which results in that 14020 as a string...