You can use a += a * b which will do a * b and then + a, same result as: a * b + a ...

math,arithmetic-expressions,bc

The key here is to be sure to use printf with the formatting spec of "%.3f" and printf will take care of doing the rounding as you wish, as long as "scale=4" for bc. Here's a script that works: echo -e "please enter math to calculate: \c" read x printf...

java,double,arithmetic-expressions

You must Typecast the ballsPlayed to double because ballsPlayed is an integer. It returns the integer part of (ballsPlayed % 6) / 10. Thus you got 0 except 0.4. Try this, double o = (double)(( ballsPlayed / 6 ) + ((double) ( ballsPlayed % 6) / 10 )); Here you...

language-agnostic,arithmetic-expressions

Yes. Here is a quick example: http://my.safaribooksonline.com/book/hardware/9788131732465/instruction-set-and-instruction-timing-of-8086/app_c those are the microcode and the timing of the operation of a massively old architecture, the 8086. it is a fairly simple point to start. of relevant note, they are measured in cycles, or clocks, and everything move at the speed of the...

c,while-loop,arithmetic-expressions

In C integer is used as a boolean: 0 is false, everything else is true. As @JonathanLeffler noted (see his comment below), in C99 and C11 there is a standart boolean datatype, but it also expands to integer constants (0 and 1). Link. When you write an expression like 5...

python,floating-point,arithmetic-expressions

== is reflexive for all numbers, zero, -zero, ininity, and -infinity, but not for nan. You can get inf, -inf, and nan in native Python just by arithmetic operations on literals, like below. These behave correctly, as in IEEE 754 and without math domain exception: >>> 1e1000 == 1e1000 True...

haskell,parsec,arithmetic-expressions,mathematical-expressions

You can get at the error messages and source position of an error using the errorMessages, messageString, errorPos and sourceColumn functions from Text.Parsec.Error. Here is an example taken from this blog post. It demonstrates the use of <?> and using the above mentioned functions to customize error processing: import Text.ParserCombinators.Parsec...

performance,swift,language-agnostic,arithmetic-expressions

Performance There's necessarily a cost to that kind of verification. At the lowest level, the operation which causes the overflow takes a single CPU operation. Checking that there was or not and overflow during this operation requires at least one more operation (e.g. using jump operations (branching) which consider the...

r,matrix,dataframes,arithmetic-expressions

The problem has to do with the way you are indexing. df["colname"] extracts the column colname as a data frame. If you subtract two data frames, R does this column-wise and row-wise, so both data frames must have the same number of columns. However, you can extract colname as a...

bash,shell,posix,ksh,arithmetic-expressions

It's worse than you think. The value of the variable is recursively treated as an arithmetic expression: $ foo='bar+bar' $ echo $((foo)) 10 The bash manual section on Shell Arithmetic says: The value of a variable is evaluated as an arithmetic expression when it is referenced, or when a variable...

algorithm,artificial-intelligence,genetic-algorithm,arithmetic-expressions

Long processing time is most probably because of small probability of an event that two chromosomes have different types (operator or digit) of characters on the same position. Note that for 1 digit numbers two chromosomes will always have the same type on the same position. The solution is to...

I think you have not understood the principle of this exercise, and the principle of abstract class and polymorphism. An expression has a calculate() method, which returns the value of this expression. This method should be abstract, because depending on the actual subclass of Expression, the calculate() method won't do...

bash,math,arithmetic-expressions

As you have mentioned that it's not necessary to do this in bash, I would recommend using awk: awk -v t1="$time1" -v t2="$time2" 'BEGIN{print (t2-t1)/t1 * 100}' Normally, awk is designed to process files but you can use the BEGIN block to perform calculations without needing to pass any files...

java,math,nan,arithmetic-expressions

You are getting the unexpected ArithmeticException because of the way that your numeric literals 3 and 0 are treated. They are interpreted as integers, despite the fact that you assign a to be a float. Java will interpret 3 as an integer, 3f as a float and 3.0 or 3d...

int,lldb,arithmetic-expressions

po stands for print object, so it may be creating a pointer. Try e instead: e minutes * 60 + seconds

I believe the snippet of code you're going to need to add to each of your string-to-number fields is CONVERT(SUBSTRING_INDEX(field,'-',-1),UNSIGNED INTEGER). For instance: CAST(REPLACE(REPLACE(IFNULL(closeout_sl_dataform.PO1_FE_HRS_CLOSEOUT_SL,0),',',''),'$','') AS DECIMAL(10,2)) + CAST(REPLACE(REPLACE(IFNULL(closeout_sl_dataform.PO2_FE_HRS_CLOSEOUT_SL,0),',',''),'$','') AS DECIMAL(10,2)) + CAST(REPLACE(REPLACE(IFNULL(closeout_sl_dataform.PO3_FE_HRS_CLOSEOUT_SL,0),',',''),'$','') AS DECIMAL(10,2)) AS TOTAL_FE_HOURS ...

bash,math,arithmetic-expressions

Ok, finally... all the code was correct... the only problem was that I am working on Windows, using Cygwin. I made both 1.txt with Notepad... if (( "$time1" < 60 )) && (( "$time2" < 60 )); then Is working perfecly. The problem was the \r In the TXT files....

exception,db2,overflow,arithmetic-expressions

Obviously, the issue here is your exceeding the maximum range representable by a 64-bit floating point number format, which is about +1.79769E+308. If you do need to work with larger numbers, consider performing calculations in DECFLOAT(34) instead (if your DB2 version supports this data type)....

bash,post-increment,arithmetic-expressions,side-effects

It gets incremented in the subshell created by the $(command substitution). When that process exits, the modified value is lost. Here are similar ways of seeing the same effect: i=0 bash -c 'let i++' # Subprocess ( let i++ ) # Explicit subshell let i++ & wait # Backgrounded process...

After multiply all the terms,please follow the order of operations.Right steps: Steps: 1) 1 * 8 = 8 2) 2 * 3 = 6 3)3-8=-5 4)-5+6=1

objective-c,operators,arithmetic-expressions

The % does not perform division. For unsigned numbers, the remainder of dividing a smaller number by a larger number is always the smaller number itself. Recall that Dividend = Divisor * Quotient + Remainder When Divisor is greater than Dividend, Quotient is zero. Hence Dividend = Remainder ...

scala,parsing,arithmetic-expressions

I think you should add it to the factor alternatives: def factor: Parser[Any] = ident | floatingPointNumber | "("~expr~")" ...

c++,function,arithmetic-expressions,truncated

Set fixed: // Output data // cout << fixed; cout << "hat size: " << setprecision(2) << hat(weight, height) << endl; cout << "jacket size: " << setprecision(2) << jacket(weight, height, age) << endl; cout << "waist size: " << setprecision(2) << waist(weight, age) << endl; ...

objective-c,arithmetic-expressions

Because addition doesn't take precedence over subtraction. Both have the same precedence, and are associated left to right., so 5 - 1 + 4 is equivalent to (5 - 1) + 4. (The order of evaluation is unspecified, which matters only if the operands have side effects.)...

java,jasper-reports,arithmetic-expressions

${Budget}.subtract($F{Actual}) will return a BigDecimal reference, so you cannot use division symbol / because this is only supported for primitives (and their class wrappers when unboxed). So, you should use BigDecimal#divide instead. ${Budget}.subtract($F{Actual}).divide($F{Budget}) From your question edit: Caused by: java.lang.ArithmeticException: Division by zero Your $F{Budget} variable should not have a...