这是一个很好的方式来嵌入调试消息在我的程序? (宏) [英] Is this a good way to embed debugging message in my program? (Macros)
问题描述
在Debug.h文件中,我有以下:
#ifdef DEBUG_FLAG
#define DEBUG msg)std :: cerr<< #msg<< std :: endl
#else
#define DEBUG(msg)for(; true == false;)
#endif
在其他地方,我可以写如下
void process数据数据)
{
DEBUG(Function'process'starts);
// Blah blah
//更多blah blah ...
DEBUG(Function'process'returns);
}
编译器会优化for(; true == false;); ?
此外,这种做法是否正确?
p>这里有一个替代方法,使用编译器的死代码删除:
#define DEBUG(msg)if(!DEBUG_ENABLED){ } \
else dbglog()<< __FILE__<< :<< __LINE__<< < msg
#ifdef DEBUG_FLAG
#define DEBUG_ENABLED 1
#else
#define DEBUG_ENABLED 0
#endif
dbglog
实例是一个 ostream
如果日志行以换行符结束或不结束。如果没有,它会添加一个。
struct dbglog {
std :: ostream&
mutable bool has_endl_;
dbglog(std :: ostream& os = std :: cerr):os_(os),has_endl_(false){}
〜dbglog(){if(!has_endl_)os_< std :: endl; }
template< typename T> static bool has_endl(const T&){return false; }
static bool has_endl(char c){return(c =='\\\
'); }
static bool has_endl(std :: string s){return has_endl(* s.rbegin()); }
static bool has_endl(const char * s){return has_endl(std :: string(s)); }
template< typename T>
static bool same_manip(T&(* m)(T&),T&(* e)(T&)){return(m == e); }
const dbglog&运算符<< (std :: ostream&(* m)(std :: ostream&))const {
has_endl_ = same_manip(m,std :: endl);
os_<< m;
return * this;
}
template< typename T>
const dbglog&运算符<< (const T& v)const {
has_endl_ = has_endl(v);
os_<< v;
return * this;
}
};
现在,您可以添加一个类似这样的简单消息(注意,换行符是可选的) p>
DEBUG(A simple message);
DEBUG(一个简单的消息with newline \\\
);
DEBUG(带有endl的简单消息)< std :: endl;
或者,如果要添加更多调试信息:
DEBUG(Entering:)< __func__<< ,argc =< argc<< ,argv =< argv;
// ...
DEBUG(Leaving:)<< __func__<< std :: endl;
in the Debug.h file, I have the following:
#ifdef DEBUG_FLAG
#define DEBUG(msg) std::cerr << #msg << std::endl
#else
#define DEBUG(msg) for(;true==false;)
#endif
In other places, I may write something like
void process (Data data)
{
DEBUG("Function 'process' starts");
// Blah blah
// More blah blah...
DEBUG("Function 'process' returns");
}
Will the compiler optimize away the for(;true==false;); ?
Also, is this kind of practice okay? If not, what would be a better way?
Thanks!
Here's an alternative, that uses the compiler's dead code removal:
#define DEBUG(msg) if (!DEBUG_ENABLED) {} \
else dbglog() << __FILE__ << ":" << __LINE__ << " " << msg
#ifdef DEBUG_FLAG
#define DEBUG_ENABLED 1
#else
#define DEBUG_ENABLED 0
#endif
The dbglog
instance is a ostream
wrapper that detects if the log line ended with a newline or not. If not, it adds one.
struct dbglog {
std::ostream &os_;
mutable bool has_endl_;
dbglog (std::ostream &os = std::cerr) : os_(os), has_endl_(false) {}
~dbglog () { if (!has_endl_) os_ << std::endl; }
template <typename T> static bool has_endl (const T &) { return false; }
static bool has_endl (char c) { return (c == '\n'); }
static bool has_endl (std::string s) { return has_endl(*s.rbegin()); }
static bool has_endl (const char *s) { return has_endl(std::string(s)); }
template <typename T>
static bool same_manip (T & (*m)(T &), T & (*e)(T &)) { return (m == e); }
const dbglog & operator << (std::ostream & (*m)(std::ostream &)) const {
has_endl_ = same_manip(m, std::endl);
os_ << m;
return *this;
}
template <typename T>
const dbglog & operator << (const T &v) const {
has_endl_ = has_endl(v);
os_ << v;
return *this;
}
};
Now, you can add a simple message like this (note, the newline is optional):
DEBUG("A simple message");
DEBUG("A simple message with newline\n");
DEBUG("A simple message with endl") << std::endl;
Or, if you want to add more debugging information:
DEBUG("Entering: ") << __func__ << ", argc=" << argc << ", argv=" << argv;
//...
DEBUG("Leaving: ") << __func__ << std::endl;
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