let

GHCi “let” — what does it do?

限于喜欢 提交于 2019-11-27 04:23:21
问题 I'd appreciate is someone could point to docs on what "let" does in GHCi, or failing that, explain it convincingly :-). So far as I can tell, "let" (without "in") is not part of the Haskell language per se, and on the other hand, it doesn't appear to be a GHCI command either, as it's not prefixed by colon. 回答1: While programming in GHCi, you're like programming in the IO monad with do syntax, so for example you can directly execute an IO action, or use monadic bind syntax like r <- someIOFun

What does `let 5 = 10` do? Is it not an assignment operation?

被刻印的时光 ゝ 提交于 2019-11-27 04:23:09
问题 If I say let 5 = 10 , why does 5 + 1 return 6 instead of 11 ? 回答1: When you say let 5 = 10 it's not a redefinition of 5, it's a pattern matching, the same which occurs when you say foo 5 = undefined ... foo 10 ... The pattern simply fails if it's ever matched. In let-expressions the match is lazy. This means the match is only being done when a variable bound by it is evaluated. This allows us to write things like let foo = undefined in 10 In your expression, no variable is bound, so the

Confused by the difference between let and let* in Scheme

混江龙づ霸主 提交于 2019-11-27 01:49:13
Can anyone explain the difference simply? I don't think I understand the concept from the textbooks/sites I have consulted. Óscar López If you use let , you can't reference other bindings which appear in the same let expression. For example, this won't work: (let ((x 10) (y (+ x 6))) ; error! unbound identifier: x y) But if you use let* , it is possible to refer to previous bindings which appear in the same let* expression: (let* ((x 10) (y (+ x 6))) ; works fine y) => 16 It's all here in the documentation. Let is parallel, (kind of; see below) let* is sequential. Let translates as ((lambda(a

Why was the name 'let' chosen for block-scoped variable declarations in JavaScript?

六月ゝ 毕业季﹏ 提交于 2019-11-26 23:45:42
问题 I understand why var takes that name - it is variable, const - it is a constant, but what is the meaning behind the name for let , which scopes to the current block? Let it be? 回答1: Let is a mathematical statement that was adopted by early programming languages like Scheme and Basic. Variables are considered low level entities not suitable for higher levels of abstraction, thus the desire of many language designers to introduce similar but more powerful concepts like in Clojure, F#, Scala,

What causes the different behaviors between “var” and “let” when assign them a returned value of a function which throws an error

佐手、 提交于 2019-11-26 23:38:26
问题 Please find the code in the image below. 1. Assign the returned value of a function, which throws an error actually, to the variable 'withLet' that declared by using keyword 'let'. 2. call 'withLet', an error occured: 'withLet is not defined'. 3. try to assert 'withLet' using 'let', an error shows that 'withLet' has already been declared. But the paradox is not exist for 'var' (Please find in the following image). I'm curious about what caused the different behaviors between these two

Why let and var bindings behave differently using setTimeout function? [duplicate]

自古美人都是妖i 提交于 2019-11-26 21:59:05
This question already has an answer here: What's the difference between using “let” and “var”? 32 answers Explanation of `let` and block scoping with for loops 3 answers This code logs 6 , 6 times: (function timer() { for (var i=0; i<=5; i++) { setTimeout(function clog() {console.log(i)}, i*1000); } })(); But this code... (function timer() { for (let i=0; i<=5; i++) { setTimeout(function clog() {console.log(i)}, i*1000); } })(); ... logs the following result: 0 1 2 3 4 5 Why? Is it because let binds to the inner scope each item differently and var keeps the latest value of i ? With var you

Haskell: Where vs. Let

狂风中的少年 提交于 2019-11-26 18:47:27
问题 I am new to Haskell and I am very confused by Where vs. Let . They both seem to provide a similar purpose. I have read a few comparisons between Where vs. Let but I am having trouble discerning when to use each. Could someone please provide some context or perhaps a few examples that demonstrate when to use one over the other? Where vs. Let A where clause can only be defined at the level of a function definition. Usually, that is identical to the scope of let definition. The only difference

v8 JavaScript performance implications of const, let, and var?

若如初见. 提交于 2019-11-26 18:07:20
问题 Regardless of functional differences, does using the new keywords 'let' and 'const' have any generalized or specific impact on performance relative to 'var'? After running the program: function timeit(f, N, S) { var start, timeTaken; var stats = {min: 1e50, max: 0, N: 0, sum: 0, sqsum: 0}; var i; for (i = 0; i < S; ++i) { start = Date.now(); f(N); timeTaken = Date.now() - start; stats.min = Math.min(timeTaken, stats.min); stats.max = Math.max(timeTaken, stats.max); stats.sum += timeTaken;

In Haskell, when do we use in with let?

扶醉桌前 提交于 2019-11-26 15:16:32
问题 In the following code, the last phrase I can put an in in front. Will it change anything? Another question: If I decide to put in in front of the last phrase, do I need to indent it? I tried without indenting and hugs complains Last generator in do {...} must be an expression import Data.Char groupsOf _ [] = [] groupsOf n xs = take n xs : groupsOf n ( tail xs ) problem_8 x = maximum . map product . groupsOf 5 $ x main = do t <- readFile "p8.log" let digits = map digitToInt $concat $ lines t

Chrome console already declared variables throw undefined reference errors for let

做~自己de王妃 提交于 2019-11-26 11:29:57
问题 Recently I ran into this weird thing in chrome console. Here I am intentionally assigning an undefined thing to a in order to throw an error. let a = werwr // Uncaught ReferenceError: werwr is not defined Then when I tried to assign something legit to a, this happened: let a = \"legit string\" // Uncaught SyntaxError: Identifier \'a\' has already been declared so I can\'t use \"let\" because a has already been declared. So I tried to reassign something else to the \"already declared a\" a = \