c++ - 在 if 语句中调用重载的构造函数失败

Question

所以我有一个 if 语句来决定是调用一个还是另一个构造函数。这是给我错误的代码:

detect::Wrapper detector;

if (detectorFileNameSpecified){
  detector = detect::Wrapper(detectorFileName,
             saveImages,
             learnOrDetect,
             verbosity);
}
else {
  detector = detect::Wrapper(saveImages,
             learnOrDetect,
             verbosity);
}

detector.setFind(findAlgNo);
detector.setLD(LDAlgNo);

detector.run();

g++ 给我这些错误:

main.cpp:126:15: error: use of deleted function ‘detect::Wrapper& detect::Wrapper::operator=(detect::Wrapper&&)’
      verbosity);
               ^
In file included from main.cpp:1:0:
wrapper.hpp:14:9: note: ‘detect::Wrapper& detect::Wrapper::operator=(detect::Wrapper&&)’ is implicitly deleted because the default definition would be ill-formed:
   class Wrapper{
         ^~~~~~~

但是，如果我以创建冗余代码的天真方式来做，它会完美地工作。

if (detectorFileNameSpecified){
      detect::Wrapper detector(detectorFileName,
                   saveImages,
                   learnOrDetect,
                   verbosity);

      detector.setFind(findAlgNo);
      detector.setLD(LDAlgNo);

      detector.run();
    }
    else {
      detect::Wrapper detector(saveImages,
                   learnOrDetect,
                   verbosity);

      detector.setFind(findAlgNo);
      detector.setLD(LDAlgNo);

      detector.run();
    }

我想做的就是减少代码中的冗余，而不必在每次需要不同的构造函数时都编写 .setFind() 等部分。我不确定如何解决这个问题。

Answer 1

通过创建函数:

void doJob(detect::Wrapper&& detector)
{
    detector.setFind(findAlgNo);
    detector.setLD(LDAlgNo);

    detector.run();
}

然后您可以将代码分解为:

if (detectorFileNameSpecified){
    doJob(detect::Wrapper(detectorFileName, saveImages, learnOrDetect, verbosity));
} else {
    doJob(detect::Wrapper(saveImages, learnOrDetect, verbosity));
}