Question

RSpec.describe "Fizz Buzz Tests" do
    it "should return `Fizz` for number divisible by 3" do
        expect(fizzBuzz(3)).to eq "Fizz"
    end

    it "should return `Buzz` for number divisible by 5" do
        expect(fizzBuzz(5)).to eq "Buzz"
    end

    it "should return `FizzBuzz` for number divisible by 3 and 5" do
        expect(fizzBuzz(15)).to eq "FizzBuzz"
    end

    it "should return `FizzBuzz` for number divisible by 3 and 5" do
        expect(fizzBuzz(90)).to eq "FizzBuzz"
    end

    it "should return 7 since it is indivisible by 3 and 5" do
        expect(fizzBuzz(7)).to eq 7
    end
end

这是我的代码：

def fizz_buzz (number)
    if number%3==0&&number%5==0
      puts ("FizzBuzz")
    elsif number%3==0
      puts ("Fizz")
    elsif number%5==0
      puts ("Buzz")
    else
      puts (number)
  end
 end

 fizz_buzz(3)
 fizz_buzz(5)
 fizz_buzz(15)
 fizz_buzz(90)
 fizz_buzz(7)

请帮忙。

Answer 1

fizzBuzz方法应该返回值，而不是打印它们。

require 'rspec'

def fizzBuzz(number)
  if number % 3 == 0 && number % 5 == 0
    "FizzBuzz"
  elsif number % 3 == 0
    "Fizz"
  elsif number % 5 == 0
    "Buzz"
  else
    number
  end
end

RSpec.describe "Fizz Buzz Tests" do
  it "should return `Fizz` for number divisible by 3" do
    expect(fizzBuzz(3)).to eq "Fizz"
  end

  it "should return `Buzz` for number divisible by 5" do
    expect(fizzBuzz(5)).to eq "Buzz"
  end

  it "should return `FizzBuzz` for number divisible by 3 and 5" do
    expect(fizzBuzz(15)).to eq "FizzBuzz"
  end

  it "should return `FizzBuzz` for number divisible by 3 and 5" do
    expect(fizzBuzz(90)).to eq "FizzBuzz"
  end

  it "should return 7 since it is indivisible by 3 and 5" do
    expect(fizzBuzz(7)).to eq 7
  end
end

我不能在rspec中通过这个测试。我的代码出了什么问题？

1 个答案: