expression-evaluation

问题 When I work with my favorite containers, I tend to chain operations. For instance, in the well-known Erase–remove idiom: v.erase( std::remove_if(v.begin(), v.end(), is_odd), v.end() ); From what I know of the order of evaluation, v.end() (on the rhs) might be evaluated before the call to std::remove_if . This is not a problem here since std::remove* only shuffle the vector without changing its end iterator. But it could lead to really surprising constructs, like for instance (demo): #include

问题 When I work with my favorite containers, I tend to chain operations. For instance, in the well-known Erase–remove idiom: v.erase( std::remove_if(v.begin(), v.end(), is_odd), v.end() ); From what I know of the order of evaluation, v.end() (on the rhs) might be evaluated before the call to std::remove_if . This is not a problem here since std::remove* only shuffle the vector without changing its end iterator. But it could lead to really surprising constructs, like for instance (demo): #include

问题 When I work with my favorite containers, I tend to chain operations. For instance, in the well-known Erase–remove idiom: v.erase( std::remove_if(v.begin(), v.end(), is_odd), v.end() ); From what I know of the order of evaluation, v.end() (on the rhs) might be evaluated before the call to std::remove_if . This is not a problem here since std::remove* only shuffle the vector without changing its end iterator. But it could lead to really surprising constructs, like for instance (demo): #include

来源： https://stackoverflow.com/questions/57324522/library-to-evaluate-excel-other-language-expression-evaluation-in-java-without-u

问题 I have a table, inside there is a field called x, the x field contain value of '1+2', '1+3' etc, how to get these value and calculate it and save into another field? 回答1: If the case you have mentioned needs to be considered then I will suggest you use the following query: select xmlquery('3+4' returning content ).getNumberVal() from dual; If More operators are involved except division then the aforementioned query will work but if the division operator is also involved then you must have to

问题 This question already has answers here : Undefined behavior and sequence points (5 answers) Why are these constructs using pre and post-increment undefined behavior? (14 answers) Closed 5 years ago . Why does not C/C++ evaluates expression in order of left to right in these cases: Initially x=1 Evaluating x + ++x gives 4. If normal evaluation is carried out (precedence of ++ is higher than +) then the result should be 1 + 2 = 3 Similarly: x + ++x + x gives 6 x + x + ++x gives 4 Why are

问题 I can't seem to find a pointer in the right direction, I am not even sure what the terms are that I should be researching but countless hours of googling seem to be spinning me in circles, so hopefully the collective hive of intelligence of Stack Overflow can help. The problem is this, I need a way to filter data in what I can only call a compound logic tree. Currently the system implements a simple AND filtering system. For example, lets say we have a dataset of people. You add a bunch of

问题 From a data.table d such as for example require(data.table) d = data.table(a = 1:4, b = 11:14, c = 21:24, group = c(1,1,2,2)) I would like to sum all variables which names are stored in the vector varsToSum by unique values of group . varsToSum = c("a", "b") For the above d and varsToSum , the expected outcome is d[,list(a = sum(a), b = sum(b)),list(group)] group a b 1: 1 3 23 2: 2 7 27 Related posts: Select / assign to data.table variables which names are stored in a character vector How to

问题 It's easy to implement a "Calculator" to parse a string (e.g., 2 ^ 3 / 2 ) and compute the result of operations. But, is there a library already capable of doing this? 回答1: The dotMath library does this. 回答2: You are going to need some kind of math parser in order to do that. I've used C# Expression Parser using RPN by DeepEddie before, or you could make your own if the complexity of the expressions you use are of more limited scope. Don't let it scare you, it is actually quite easy to make.

问题 I am going to write an expression evaluator which only does addition and subtraction. I have a simple algorithm to do that; but, I have some implementation problems. I considered an expression as (it is a String) "(" <expression1> <operator> <expression2> ")" Here is my algorithm String evaluate( String expression ) if expression is digit return expression else if expression is "(" <expression1> <operator> <expression2> ")" cut the brackets out of it expression1 = evaluate( <expression1> )