所以我想写一个可以像这样使用的缩进输出类:
Debug f;
f.open("test.txt");
f << Debug::IndS << "Start" << std::endl;
f << Debug::Ind << "test" << std::endl;
f << Debug::IndE << "End" << std::endl;
输出:
Start
test
End
因此 IndS 将打印出当前缩进量和递增缩进量,Ind 将打印出当前缩进量,IndE 将递减缩进量并打印出当前缩进量。我试过这样创建它:
class Debug : public std::ofstream {
public:
Debug();
~Debug();
private:
std::string IndentText;
int _Indent;
public:
void SetIndentText(const char* Text);
inline void Indent(int Amount);
inline void SetIndent(int Amount);
inline std::ofstream& Ind (std::ofstream& ofs);
inline std::ofstream& IndS(std::ofstream& ofs);
inline std::ofstream& IndE(std::ofstream& ofs);
};
Debug::Debug () : std::ofstream() {
IndentText = " ";
}
Debug::~Debug () {
}
void Debug::SetIndentText (const char* Text) {
IndentText = Text;
}
void Debug::Indent (int Amount) {
_Indent += Amount;
}
void Debug::SetIndent(int Amount) {
_Indent = Amount;
}
std::ofstream& Debug::Ind (std::ofstream& ofs) {
for (int i = 0;i < _Indent;i++) {
ofs << IndentText;
}
return ofs;
}
std::ofstream& Debug::IndS (std::ofstream& ofs) {
ofs << Ind;
_Indent++;
return ofs;
}
std::ofstream& Debug::IndE (std::ofstream& ofs) {
_Indent--;
ofs << Ind;
return ofs;
}
所以我认为这有几个问题:
它不编译。
no match for 'operator<<' (operand types are 'std::ofstream {aka std::basic_ofstream<char>}' and '<unresolved overloaded function type>') ofs << Ind; candidates are:
错误呜呜呜我没有覆盖所有构造函数。有没有办法做到这一点?我想我只需要重写所有的构造函数来做
IndentText = " ";
并委托(delegate)重载的构造函数
有人可以帮我解决这个问题吗?谢谢!
最佳答案
你通常不应该继承std::ostream
或者像 std::ofstream
这样的实现.而是将它们包装到另一个类中。
这是我在评论中提到的想法的简短概述
#include <iostream>
#include <fstream>
using namespace std;
class Logger {
public:
Logger(ostream& os) : os_(os), curIndentLevel_(0) {}
void increaseLevel() { ++curIndentLevel_; }
void decreaseLevel() { --curIndentLevel_; }
private:
template<typename T> friend ostream& operator<<(Logger&, T);
ostream& os_;
int curIndentLevel_;
};
template<typename T>
ostream& operator<<(Logger& log, T op) {
for(int i = 0; i < log.curIndentLevel_ * 4; ++i) {
log.os_ << ' ';
}
log.os_ << op;
return log.os_;
}
int main() {
Logger log(cout);
log.increaseLevel();
log << "Hello World!" << endl;
log.decreaseLevel();
log << "Hello World!" << endl;
return 0;
}
输出
Hello World!
Hello World!
这里有一个小变体,展示了如何使用 operator<<()
简化编码重载:
class Logger {
public:
Logger(ostream& os) : os_(os), curIndentLevel_(0) {}
Logger& increaseLevel() { ++curIndentLevel_; return *this; }
Logger& decreaseLevel() { --curIndentLevel_; return *this; }
// ... as before ...
};
int main() {
Logger log(cout);
log.increaseLevel() << "Hello World!" << endl;
log.decreaseLevel() << "Hello World!" << endl;
return 0;
}
您可以采用相同的方式提供额外的 I/O 操纵器样式的免费功能。
关于C++ 缩进输出类继承 ofstream,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/25216993/