如何使用开玩笑介绍异步函数？

Question

我要测试的功能接收到异步函数作为参数。在我的参数验证中，我验证它是否存在，是一个函数，并且是这样的异步函数：

exports.consumeRecords = async (consumerFunction, records) => {

    if (!consumerFunction) {
        throw new Error(`Missing required argument, consumerFunction, in consumeRecords function`)
    }
    if (typeof consumerFunction !== 'function') {
        throw new Error(`Argument, consumerFunction, is not of type 'function' in consumeRecords function.  Type found: ${typeof consumerFunction}`)
    }
    if (args.consumerFunction.then instanceof Promise) {
        throw new Error(`Argument, consumerFunction, is not a async function in consumeRecords function.`)
    }
    await consumerFunction(records)
// ... blah blah ..
}

我对上述代码的测试看起来像这样：

describe('FUNCTION: consumeRecords', () => {
    let logSpy, mockAsyncConsumerFunction
    beforeEach(()=> {
        mockAsyncConsumerFunction = async (recordBody) => `${recordBody} - resolved consumer`
        logSpy = jest.spyOn(console, 'log').mockImplementation(() => null)
    })
        test('throws error when missing consumerFunction argument', async () => {
            await expect(async () => await utils.consumeRecords({})).rejects.toThrow(`Missing required argument, consumerFunction,`)
        })
        test('throws error when consumerFunction is not a function', async () => {
            await expect(async () => await utils.consumeRecords({ consumerFunction: 123 })).rejects.toThrow(`Argument, consumerFunction, is not of type 'function'`)
        })
        test('throws error when consumerFunction is not async function', async () => {
            let syncFunction = () => console.log()
            await expect(async () => await utils.consumeRecords({ consumerFunction: syncFunction })).rejects.toThrow(`Argument, consumerFunction, is not a async function`)
        })
    afterEach(() => {
        jest.clearAllMocks()
    })
    afterAll(() => {
        jest.restoreAllMocks()
    })
})

现在，我想测试是否通过在模拟上监视模拟，因此，我正在尝试在测试的顶部执行此操作：

mockAsyncConsumerFunction = async (recordBody) => `${recordBody} - resolved consumer`
mockAsyncConsumerFunctionSpy = jest.fn(mockAsyncConsumerFunction)

然后使用.mocks对象在jest.fn上使用标准断言，如下：

test('calls consumer function correctly', async () => {
    await utils.consumeRecords({ consumerFunction: mockAsyncConsumerFunctionSpy, records: [{ body: JSON.stringify({fake:111}) },{ body: JSON.stringify({fake:222}) }], taskNameString: "Apply image overlays" })
    expect(mockAsyncConsumerFunctionSpy).toHaveBeenCalledWith({fake:111})
})

不幸，不幸的是，做完之后这是我的测试失败，因为它不再被视为异步函数，因此我的输入验证失败了，给了我：

FUNCTION: consumeRecords › calls consumer function correct number of
 times

    Argument, consumerFunction, is not a async function in consumeRecords function.

如何在Jest中制作一个模拟/间谍功能，读起来是异步函数？