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;鉴于你有Resource1
;这是一种快速简便的方法,需要Resource1
和Resource2
。现在,这些资源是宝贵的,所以我把它们包装成RAII实现ResNHolder
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()”会引用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?
答案 0 :(得分:2)
我认为您的代码会很好,但是可以考虑以下方法:
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);
}
当资源无法初始化时,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>();
}
}
答案 1 :(得分:1)
你的代码看起来很好,这是我可以想象你可能遇到的唯一问题,因为异常被认为不是很有效。如果是这样,您可以将代码更改为:
void DoTheJob(Logger& log/*, some other params */) {
Res1HolderNoThrow r1(/* arguments for creating resource #1 */);
if( r1 ) {
Res2HolderNoThrow r2(/* arguments */);
if( r2 )
DoTheJobQuicklyAndEasily(log, r1, r2);
else {
log.log("Can't obtain resource 2, that'll slowdown us a bit");
DoTheJobWithModerateSuffering(log, r1);
}
} else {
log.log("Can't obtain resource 1, using fallback");
DoTheJobTheSlowAndPainfulWay(log);
}
}
你需要另一个不抛出异常但有状态的RAII对象,并在运算符bool()或其他地方返回它。但是你的代码看起来不容易出错,除非你遇到性能问题或者需要避免异常,否则我宁愿使用它。