RAII和构造函数中的异常

Question

想象一下,我有一份工作要做,可以通过三种不同的方式来完成:缓慢而痛苦,但却是安全的方式; 给你带来适度痛苦的方式Resource1; 一种快速简便的方法,需要Resource1和Resource2.现在,这些资源是宝贵的,所以我将它们包装成RAII实现ResNHolders并写下这样的东西:

void DoTheJob(Logger& log/*, some other params */) {
    try {
        Res1Holder r1(/* arguments for creating resource #1 */);
        try {
            Res2Holder r2(/* arguments */);
            DoTheJobQuicklyAndEasily(log, r1, r2);
        }
        catch (Res2InitializationException& e) {
            log.log("Can't obtain resource 2, that'll slowdown us a bit");
            DoTheJobWithModerateSuffering(log, r1);
        }
    }
    catch (Res1InitializationException& e) {
        log.log("Can't obtain resource 1, using fallback");
        DoTheJobTheSlowAndPainfulWay(log);
    }
}

"DoTheJobXxx()"引用Logger/ ResNHolder,因为它们是不可复制的.我是不是太笨拙了？有没有其他聪明的方法来构建函数？

Answer 1

我认为你的代码就很好，但这里有一个可以考虑的替代方案：

void DoTheJob(Logger &log/*,args*/)
{
    std::unique_ptr<Res1Holder> r1 = acquireRes1(/*args*/);
    if (!r1) {
        log.log("Can't acquire resource 1, using fallback");
        DoTheJobTheSlowAndPainfulWay(log);
        return;
    }
    std::unique_ptr<Res2Holder> r2 = acquireRes2(/*args*/);
    if (!r2) {
        log.log("Can't acquire resource 2, that'll slow us down a bit.");
        DoTheJobWithModerateSuffering(log,*r1);
        return;
    }
    DoTheJobQuicklyAndEasily(log,*r1,*r2);
}

当资源初始化失败时，acquireRes 函数返回 null unique_ptr：

std::unique_ptr<Res1Holder> acquireRes1()
{
  try {
    return std::unique_ptr<Res1Holder>(new Res1Holder());
  }
  catch (Res1InitializationException& e) {
    return std::unique_ptr<Res1Holder>();
  }
}

std::unique_ptr<Res2Holder> acquireRes2()
{
  try {
    return std::unique_ptr<Res2Holder>(new Res2Holder());
  }
  catch (Res2InitializationException& e) {
    return std::unique_ptr<Res2Holder>();
  }
}