我已经阅读了有关该主题的多个问题和答案,但不幸的是,他们都没有帮助。我想在两个A和B类中使用相同的调试文件输出,其中A的实例创建B的实例。我有类似的东西:
class A {
public:
A() : debug("debug.txt") { };
private:
std::ofstream debug;
}
class B {
public:
B(std::ofstream &ofs) : debug(ofs) { };
private:
std::ofstream &debug;
}
并使用
创建一个实例B *b = new B(debugUnderlying);
效果很好。但是,我现在想要一个额外的构造函数,以便能够在没有ofstream的情况下使用它。然后该对象将打开一个新文件。我明白了,因为我有一个参考,我需要在初始化列表中初始化它。我尝试了很多东西:
B() : debug() { debug.open("debug2.txt"); };
error: invalid initialization of non-const reference of type ‘std::ofstream& {aka std::basic_ofstream<char>&}’ from an rvalue of type ‘const char*’
或
B() : debug("debug2.txt") { };
error: value-initialization of reference type ‘std::ofstream& {aka std::basic_ofstream<char>&}’
或(很清楚,因为我有一个临时对象)
error: invalid initialization of non-const reference of type ‘std::ofstream& {aka std::basic_ofstream<char>&}’ from an rvalue of type ‘std::ofstream {aka std::basic_ofstream<char>}’
我该怎么做?谢谢你的任何建议!
答案 0 :(得分:1)
您可以存储指针和标志所有权:
class B {
public:
B() : stream(new ...), owner(true) {}
B(std::ostream& s) : stream(&s), owner(false) {}
~B() { if(owner) delete stream; }
private:
std::ostream* stream;
bool owner;
};
注意:我用ostream替换了ofstream。
答案 1 :(得分:0)
修复错误:
B():debug(*(new std::ofstream("debug.txt"))){}
但是:如果你这样做,我想你会忘记删除......
<强>所以:强>
最好使用单例来包装调试对象
class Debug{
std::ofstream debug;
//private ctor, you can't create an object for this class outside
Debug(const std::string& filename):debug(filename){}
public:
//only way tou use this class is to call this method
static Debug* Instance(){
//the one and only Debug object, created at first use:
static Debug theObj("debug.txt");
return &theObj;
}
//write method
std::ostream& outstream(){
return debug;
}
};
//just for simplicity of use:
#define dbgout Debug::Instance()->outstream()
您还可以像这样定义宏:
#ifdef DEBUG
#define dbgout Debug::Instance()->outstream()
#else
// a release version won't create the debug file...
#define dbgout //
#endif
现在您可以在代码中的任何位置使用它:
dbgout<<" put your " << "stream " <<" here ";