Scheme if 语句。为什么这个显示语句不是运行？

Question

我正在运行宁以下代码：

(define (myignore x)
    0
)
(define (myinterpreter mylist)
  (evaluate mylist)
)

(define (evaluate mylist)

  (if (eqv? (car mylist) 'prog)
    (evaluate (cdr mylist))
    (display (integer? (car mylist)))
    (if (integer? (car mylist))
      (display "YESSS")
    )
  )

  (if (eqv? (car mylist) 'myignore)
    (myignore (cdr mylist))
  )

)

(myinterpreter '(prog 5))

我想知道为什么包含 (display ("YESSS")) 的行不运行尽管 (display (integer? (car mylist))) 在 [= 行之前等于 true 17=]s?

Answer 1

if的语法是：

(if predicate
    consequent
    optional-alternative)

在您的代码中，consequent 是 (evaluate (cdr mylist))，optional-alternative 是 (display (integer? (car mylist)))。之后的所有内容都将被忽略。

如果要执行多个表达式，则必须将它们包装在begin。

(define (evaluate mylist)

  (if (eqv? (car mylist) 'prog)
    (evaluate (cdr mylist))
    (begin 
      (display (integer? (car mylist)))
      (if (integer? (car mylist))
        (display "YESSS")
      )
    )
  )

  (if (eqv? (car mylist) 'myignore)
    (myignore (cdr mylist))
  )

)

Scheme if 语句。为什么这个显示语句不是运行？

Scheme if statement. Why does this display statement not run?

lisp

scheme

functional-programming