Ruby：rspec 没有捕获异常？

Question

我无法理解 rspec 如何处理异常。在这种情况下，我想确保异常被正确捕获并以格式化消息的形式返回。

考虑以下几点：

class Sample

  def calculate(x, y)
    begin
      add(x,y)
    rescue Exception => e
      return "Exception occured: #{e}"
    end
  end
  
  def add(x, y)
    x + y
  end
end

有这样的规范：

describe "sample" do
  it "adds numbers" do
    test = Sample.new

    allow(test)
      .to receive(:calculate)
      .with(1, 2) { Exception.new('foo') }

    expect(test.calculate(1,2)). to eq 'Exception occurred: foo'
  end
end

我希望异常被 rspec 正确捕获，因为它是由方法 calculate 本身，但是，我在执行规范时收到以下内容：

Failures:

  1) sample adds numbers
     Failure/Error: expect(test.calculate(1,2)). to eq 'Exception occurred: foo'
     
       expected: "Exception occurred: foo"
            got: #<Exception: foo>
     
       (compared using ==)
     
       Diff:
       @@ -1,2 +1,2 @@
       -"Exception occurred: foo"
       +#<Exception: foo>

但是通过irb测试救援时，处理得当：

irb(main):001:0> test = Sample.new()
=> #<Sample:0x000055d0f864db50>
irb(main):002:0> test.calculate("i", 2)
=> "Exception occured: no implicit conversion of Integer into String"
irb(main):003:0> 

我对 rspec 很陌生，所以欢迎任何反馈：D

Answer 1

你需要存根 add 方法而不是 calculate 并且你需要引发异常而不是返回异常。

describe "sample" do
  it "adds numbers" do
    test = Sample.new

    allow(test)
      .to receive(:add)
      .with(1, 2).and_raise(Exception.new('foo'))

    expect(test.calculate(1,2)). to eq 'Exception occurred: foo'
  end
end