expression

问题 A function produces a result, can be used as argument of another function parameter. So, is a function call like: f(1,2,3) be considered as an "expression"? Thanks. 回答1: The C++ standard (N3376, §5.1) specifies an expression as: An expression is a sequence of operators and operands that specifies a computation. An expression can result in a value and can cause side effects. Further in the same section (§5.2.2): A function call is a postfix expression followed by parentheses containing a

问题 A function produces a result, can be used as argument of another function parameter. So, is a function call like: f(1,2,3) be considered as an "expression"? Thanks. 回答1: The C++ standard (N3376, §5.1) specifies an expression as: An expression is a sequence of operators and operands that specifies a computation. An expression can result in a value and can cause side effects. Further in the same section (§5.2.2): A function call is a postfix expression followed by parentheses containing a

问题 I study C language from "C Primer Plus" book by Stephen Prata and it came to the point : "A full expression is one that’s not a subexpression of a larger expression.Examples of full expressions include the expression in an expression statement and the expression serving as a test condition for a while loop" I can't understand clearly what is the exact definition of full expressions and why the book considers test conditions are full expressions. Could any one explain clearly what is meant by

问题 I've just read the page on Expressions in the PHP docs, and right at the top it says: The simplest yet most accurate way to define an expression is "anything that has a value". That simple definition includes all functions and most language constructs, however there a few language constructs that explicitly state they do not return a value. Here is a list of language constructs that do return a value: empty eval include include_once isset list require require_once print Here are the

问题 Suppose the user inputs an infix expression as a string? What could be the easiest ( By easiest I mean the shortes t) way to evaluate the result of that expression using C language? Probable ways are converting it to a postfix then by using stacks.But its rather a long process. Is there any way of using functions such as atoi() or eval() that could make the job easier? 回答1: Certainly the most instructive way (and possibly even the easiest, once you know how) is to learn how to write your own

问题 I need to convert month number to month name. I have date time as the date type - 2009-01-01 00:00:00.000 I also have 4-byte integer data type - 1 how do I convert this 1 to "January" for example? 回答1: i think you are in the data flow: it is really easy to get MOnth Name in a script component from Date: add a varchar column to your dataflow Mark your date column for read access enter the following script Row.[NewColumnName] = Row.[Your Date Column].ToString("MMMM"); Result: Here is a good

问题 (warning: this was posted also on https://forums.fsharp.org/t/asyncresult-and-handling-rollback/928) Trying to implement 2PC Transaction Like workflow in F# (see http://learnmongodbthehardway.com/article/transactions/) and hitting an issue with computation expressions (asyncResult for example) and rollback. If you have the following pseudocode: let rollbackWorkflow parX parY = … here calling rollbackService1 and rollbackService2 let executeWorkflow par1 par2 par3 = asyncResult { let! result1

问题 I am trying to get return the correct "Where" extension method using reflection, in order to build a custom Expression. I have tried several ways but the closest I get, throws an exception: "An unhandled exception of type 'System.Reflection.AmbiguousMatchException' occurred in mscorlib.dll" I know this is because there are two Where methods defined in the Enumrable class - but how can I return the Where method which using only just a predicate of Func<T, bool>. What I have at the moment is:

问题 I am using the DrRacket environment to try out the Scheme language. I defined sum+1 as follows: (define sum+1 '(+ x y 1)) I was wondering why the following expression does not evaluate: (let ([x 1] [y 2]) (eval sum+1)) whereas doing this returns the correct value: (define x 1) (define y 2) (eval sum+1) 回答1: eval does not work with lexical variables at all unless the lexical variable was created in the same expression: #!r7rs (import (scheme base) (scheme eval)) (define env (environment '

问题 The way structures allocate memory is that :- struct mys { int a , b, c ; }; As we can see in structs, when I declare a struct variable say, struct mys var1 , var1 takes the sum of all the basic datatypes inside it. (12 bytes here assuming the word length is 4 bytes) printf("%d",sizeof(var1)) ; output is 4 . In enum, we have, enum myvar{ id1 , id2 , id3 }; and whenever I declare an enum variable and print its size, it only prints the size of integer(4 bytes). And the id1 ,id2, id3 would get 0