我在Racket中创建了一个阶乘函数,需要进行单元测试以传递一个非常大的数字来捕获溢出异常。如果捕获到异常,则测试应该通过,反之亦然。这是我的代码。
#lang racket
(provide recursive_factorial)
(provide tail_factorial)
(define (recursive_factorial number)
(cond [(= 0 number) 1]
[(negative? number) (raise-argument-error 'recursive_factorial "negative?" number)]
[(* number (recursive_factorial (- number 1)))]))
(define (tail_factorial number accumulator)
(cond[( = number 0 ) accumulator]
[(negative? number) accumulator (raise-argument-error 'tail_factorial "negative?" number accumulator)]
[(tail_factorial (- number 1) (* accumulator number ))]
))
这是我尝试对其进行单元测试。
(check-not-exn (λ () (recursive_factorial(100000000)))"stack overflow")
(check-not-exn (λ () (tail_factorial(100000000)))"stack overflow")
在很多帮助下,我能够得到负面的工作条件。任何帮助表示赞赏。
答案 0 :(得分:1)
当-1
为否定时,您的函数会返回number
。所以测试不应该是:
(check-equal? (recursive_factorial -4) -1)
更新
怎么样:
#lang racket
(provide recursive_factorial)
(define (recursive_factorial number)
(cond [(= 0 number) 1]
[(negative? number) (error 'recursive_factorial
"Cannot pass a negative number")]
[else (* number (recursive_factorial (- number 1)))]))