evaluation

How to evaluate functions in GDB?

由 余生长醉 提交于 2019-11-27 12:49:09
问题 I wonder why evaluate function doesn't work in gdb? In my source file I include, when debugging in gdb, these examples are wrong evaluations. (gdb) p pow(3,2) $10 = 1 (gdb) p pow(3,3) $11 = 1 (gdb) p sqrt(9) $12 = 0 回答1: My guess is that the compiler and linker does some magic with those particular functions. Most likely to increase performance. If you absolutely need pow() to be available in gdb then you can create your own wrapper function: double mypow(double a, double b) { return pow(a,b)

What is the relationship between unboxed types and strictness?

由 孤人 提交于 2019-11-27 09:52:13
问题 Unboxed types, like Int# , and strict functions, like f (!x) = ... , are something different, but I see conceptual similarity - they disallow thunks/laziness in some way. If Haskell was a strict language like Ocaml, every function would be strict and every type unboxed. What is the relationship between unboxed types and enforcing strictness? 回答1: Unboxed vs Boxed Data To support parametric polymorphism and laziness, by default Haskell data types are represented uniformly as a pointer to a

Math Expression Evaluation

由 谁说胖子不能爱 提交于 2019-11-27 08:43:44
What is the best way to implement a python program that will take a string and will output its result according to operator precedence (for example: "4+3*5" will output 19). I've googled for ways to solve this problem but they were all too complex, and I'm looking for a (relatively) simple one. clarification: I need something slightly more advanced than eval() - I want to be able to add other operators (for example a maximum operator - 4$2 = 4) or, also I am more interested in this academically than professionaly - I want to know how to do this. If you're "academically interested", you want to

Built-in functions not working with evaluated strings, why?

由 风格不统一 提交于 2019-11-27 08:13:48
问题 It seems that evaluated color strings are not working with some built-in LESS functions. I have tried using e() and ~"" and any combination of both. I might find a workaround for my particular case, I’m just asking if this is this expected behaviour, or if there is a fault in my reasoning? Any insight appreciated. For example here, the color is created from an evaluated string; note the 'missing' # in the hex value that gets added later : .broken-mixin(@hexcode: '9719e1') { @color: e("#@

Order of evaluation of expression

由 ≯℡__Kan透↙ 提交于 2019-11-27 07:09:05
问题 I've just read that order of evaluation and precedence of operators are different but related concepts in C++. But I'm still unclear how those are different but related?. int x = c + a * b; // 31 int y = (c + a) * b; // 36 What does the above statements has to with order of evaluation. e.g. when I say (c + a) am I changing the order of evaluation of expression by changing its precedence? 回答1: The important part about order of evaluation is whether any of the components have side effects.

Why does 1+++2 = 3?

由 若如初见. 提交于 2019-11-27 05:14:15
How does Python evaluate the expression 1+++2 ? How many ever + I put in between, it is printing 3 as the answer. Please can anyone explain this behavior And for 1--2 it is printing 3 and for 1---2 it is printing -1 Greg Hewgill Your expression is the same as: 1+(+(+2)) Any numeric expression can be preceded by - to make it negative, or + to do nothing (the option is present for symmetry). With negative signs: 1-(-(2)) = 1-(-2) = 1+2 = 3 and 1-(-(-2)) = 1-(2) = -1 I see you clarified your question to say that you come from a C background. In Python, there are no increment operators like ++ and

Forcing a constant expression to be evaluated during compile-time?

由 跟風遠走 提交于 2019-11-27 04:34:19
问题 A few days ago I asked by which criteria the compiler decides whether or not, to compute a constexpr function during compile time. When does a constexpr function get evaluated at compile time? As it turns out, a constexpr is only evaluated during compile-time, if all parameters are constant expressions and the variable you are assigning it to is are constant expression as well. template<typename base_t, typename expo_t> constexpr base_t POW(base_t base, expo_t expo) { return (expo != 0 )?

SQL UPDATE order of evaluation

由 眉间皱痕 提交于 2019-11-27 01:55:41
What is the order of evaluation in the following query: UPDATE tbl SET q = q + 1, p = q; That is, will "tbl"."p" be set to q or q + 1 ? Is order of evaluation here governed by SQL standard? Thanks. UPDATE After considering Migs' answer , I ran some tests on all DBs I could find. While I don't know what the standard says, implementations vary. Given CREATE TABLE tbl (p INT NOT NULL, q INT NOT NULL); INSERT INTO tbl VALUES (1, 5); -- p := 1, q := 5 UPDATE tbl SET q = q + 1, p = q; I found the values of "p" and "q" were: database p q -----------------+---+--- Firebird 2.1.3 | 6 | 6 -- But see

Does Go compiler's evaluation differ for constant expression and other expression

由 雨燕双飞 提交于 2019-11-26 23:06:50
Why does below code fail to compile? package main import ( "fmt" "unsafe" ) var x int = 1 const ( ONE int = 1 MIN_INT int = ONE << (unsafe.Sizeof(x)*8 - 1) ) func main() { fmt.Println(MIN_INT) } I get an error main.go:12: constant 2147483648 overflows int Above statement is correct. Yes, 2147483648 overflows int (In 32 bit architecture). But the shift operation should result in a negative value ie -2147483648. But the same code works, If I change the constants into variables and I get the expected output. package main import ( "fmt" "unsafe" ) var x int = 1 var ( ONE int = 1 MIN_INT int = ONE

How to write an R function that evaluates an expression within a data-frame

由 ぃ、小莉子 提交于 2019-11-26 22:59:49
问题 Puzzle for the R cognoscenti: Say we have a data-frame: df <- data.frame( a = 1:5, b = 1:5 ) I know we can do things like with(df, a) to get a vector of results. But how do I write a function that takes an expression (such as a or a > 3 ) and does the same thing inside. I.e. I want to write a function fn that takes a data-frame and an expression as arguments and returns the result of evaluating the expression "within" the data-frame as an environment. Never mind that this sounds contrived (I
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