c - 使用 Google Test 测试 C 函数

Question

我正在尝试使用 Google Test 来测试一些 C 代码，但我很困难。我真的不明白如何测试代码。我尝试过一些东西，但我不知道它是否可以。有人可以向我解释一下我所做的事情是否可以以及如何测试开关部分吗？

这是我写的测试:

    TEST_F(MPS_Tests, Wai_for_discharging_of_several_rails) {


        MPS__tenPowerUpTrigger enTrigger = MPS__enGetPowerUpTrigger();
        EXPECT_EQ(enPowerUpTriggerFstp, enTrigger);
        MPS__vOnEnterStateAcPu1();
        EXPECT_EQ(MPS__enGetInternalState(enAC), enStateAcPu1);
        EXPECT_EQ(MPS__u16GetSmEvent(enAC), enStateAcPu1);

这是经过测试的代码:

MPS__tenPowerUpTrigger MPS__enGetPowerUpTrigger(void)
{
    /* TODO: rework after startup trigger is considerd by design */
    return enPowerUpTriggerFstp;
}

/* Wait for discharging of several rails */
void MPS__vOnEnterStateAcPu1(void)
{
    MPS__tenPowerUpTrigger enTrigger = MPS__enGetPowerUpTrigger();
    MPS__vSetInternalState(enAC, enStateAcPu1);

    switch (enTrigger)
    {
    case enPowerUpTriggerFstp:
        MPS__vSetWait(enAC, u16Time_AC1ASU_t4);
        break;
    case enPowerUpTriggerWakeUp:
        MPS__vSetWait(enAC, MPS_TIMER_NOWAIT);
        break;
    case enPowerUpTriggerReset:
        MPS__vSetWait(enAC, u16Time_AC1CSU_t2);
        break;
    default:
        /* error handling */
        break;
    }
}

谢谢!

Answer 1

覆盖开关中所有选项的简单方法是将MPS__enGetPowerUpTrigger更改为如下所示:

MPS__tenPowerUpTrigger MPS__enGetPowerUpTrigger(void)
{
    const MPS__tenPowerUpTrigger return_values[3] = { 
      enPowerUpTriggerFstp,
      enPowerUpTriggerWakeUp,
      enPowerUpTriggerReset
    };
    static int idx = 0;

    MPS__tenPowerUpTrigger return_value = return_values[idx];
    idx += 1;
    if (idx >= 3) idx = 0;
    return return_value;
}

...然后调用 MPS__vOnEnterStateAcPu1 3 次？