r5rs

问题 For example: if I want the function equal? recognize my own type or record, can I add a new behavior of equal? ? without erasing or overwriting the old one? Or for example if I want to make the function "+" accept also string? 回答1: Rather than using import , a better solution is to keep track of the original function by let -binding it. It's also better to check that the type of the argument is a string, rather than that it is not a number. Using both of these approaches means that it's

问题 For example: if I want the function equal? recognize my own type or record, can I add a new behavior of equal? ? without erasing or overwriting the old one? Or for example if I want to make the function "+" accept also string? 回答1: Rather than using import , a better solution is to keep track of the original function by let -binding it. It's also better to check that the type of the argument is a string, rather than that it is not a number. Using both of these approaches means that it's

问题 What is a simple way to output text to file in a R5RS compliant version of Scheme? I use MIT's MEEP (which uses Scheme for scripting) and I want to output text to file. I have found the following other answers on Stackoverflow: File I/O operations - Scheme How to append to a file using Scheme Append string to existing textfile [sic] in IronScheme But, they weren't exactly what I was looking for. 回答1: The answers by Charlie Martin, Ben Rudgers, and Vijay Mathew were very helpful, but I would

问题 I can use SHA256 in Scheme using external libraries (Java, C or system dependent) or using a specific Scheme implementation (like Chicken e.g.), but I wonder if there is a "pure" scheme implementation. 回答1: I wrote an implementation today. Alas, R5RS has neither bytevectors nor binary I/O, so this uses the R7RS APIs for bytevectors and binary I/O. It should be easy to bridge those APIs to your Scheme implementation's native APIs (for example, I actually tested my implementation on Racket and

问题 consider following rules in the parser: expression : IDENTIFIER | (...) | procedure_call // e.g. (foo 1 2 3) | macro_use // e.g. (xyz (some datum)) ; procedure_call : '(' expression expression* ')' ; macro_use : '(' IDENTIFIER datum* ')' ; and // Note that any string that parses as an <expression> will also parse as a <datum>. datum : simple_datum | compound_datum ; simple_datum : BOOLEAN | NUMBER | CHARACTER | STRING | IDENTIFIER ; compound_datum : list | vector ; list : '(' (datum+ ( '.'

问题 Okay. So I'm wondering how to create a function that will turn a random number into its english word component. Such as (1001 -> '(one thousand one) or 0 -> '(zero) and (number-name factorial 20 -> ’(two quintillion four hundred thirty two quadrillion nine hundred two trillion eight billion one hundred seventy six million six hundred forty thousand)) I worked with a previous user on stackoverflow to get something that turned a long number into 3 part digits (1,341,100 is one million, 341

问题 In R5RS Scheme how do you display multiple parameters, with a single call? my implementation below works, but adds extra parentheses and spaces. #!/usr/bin/env racket #lang r5rs (define (display-all . rest) (display rest)) (display-all "I " "have " "a " "lovely " "bunch " "of " "coconuts\n") results in owner@K53TA:~$ ./Template.ss (I have a lovely bunch of coconuts ) 回答1: Simplest: (define (display-all . vs) (for-each display vs)) Note the use of for-each instead of map - for-each is the same

问题 I've been self-teaching myself Scheme R5RS for the past few months and have just started learning about mutable functions. I've did a couple of functions like this, but seem to find my mistake for this one. (define (lst-functions) (let ((lst '())) (define (sum lst) (cond ((null? lst) 0) (else (+ (car lst) (sum (cdr lst)))))) (define (length? lst) (cond ((null? lst) 0) (else (+ 1 (length? (cdr lst)))))) (define (average) (/ (sum lst) (length? lst))) (define (insert x) (set! lst (cons x lst)))

问题 SCHEME/Racket/R5RS Attempting to make a recursive procedure that pairs 2 lists of the same size. Just cant get the recursive call right. This is what I have and I am stuck. (define (pairs list1 list2) (if (or (null? list1) (null? list2)) '() (cons (car list1) (car list2)) )) Test Case: (pairs '(1 2 3) '(a b c)) Desired Output: ((1 . a) (2 . b) (3 . c)) Current Output: (1 . a) 回答1: You just have to cons the current result to the recursive call of the procedure, and that's it! (define (pairs

问题 I've written some code but it's not working because the add1 function I used in Scheme is not working with R5RS. What can replace add1 in R5RS? 回答1: The procedure add1 is very simple, you can implement it yourself: (define (add1 x) (+ x 1)) 回答2: late answer but an alternative is (making use of lambdas) .. (define add1 (lambda (x) (+ x 1))) 来源： https://stackoverflow.com/questions/13292094/add1-function-from-scheme-to-r5rs