我想更好地学习C ++(目前我的C ++仅限于它的C子集*咳嗽* ...),因此我决定尝试使用C ++ -ify"我从C到C ++的一个有用的日志记录功能,我认为最好用代码解释:
#include <stdarg.h>
#include <stdio.h>
enum msg_type {
LOG_DBG,
LOG_INF,
LOG_WRN,
LOG_ERR
};
int printf_log(enum msg_type mt, const char *fmt_string, ...)
{
va_list args;
va_start(args, fmt_string);
switch (mt) {
case LOG_DBG:
printf("[D] ");
break;
case LOG_INF:
printf("[I] ");
break;
case LOG_WRN:
printf("[W] ");
break;
case LOG_ERR:
printf("[E] ");
break;
default:
break;
}
int res = vprintf(fmt_string, args);
va_end(args);
return res;
}
int main()
{
int i = 0;
printf_log(LOG_DBG, "i is %d\n", i);
i++;
printf_log(LOG_INF, "i is %d\n", i);
i++;
printf_log(LOG_WRN, "i is %d\n", i);
}
这应输出:
[D] i is 0
[I] i is 1
[W] i is 2
我对C ++版本的想法就是:
#include <log.h>
int main()
{
int i = 0;
log::log(log::dbg)<<"i is " << i << "\n";
i++;
log::log(log::inf)<< "i is " << i << "\n";
i++;
log::log(log::wrn)<< "i is " << i << "\n";
}
具有相同的结果。
我特别想避免对参数进行任何解析(可能除了log::inf
或类似),但只是让它们直接传递到cout
。
但我并不知道从哪里开始,所有与此有关的事情要么想要更多的C ++知识,要么希望你实现整个自己的streambuf
或类似的。
我的想法基本相同,所以我尽了最大努力,这是我在互联网上使用不同资源制作的代码:
#include <iostream>
using std::cout;
class Log {
public:
enum msg_type {
dbg =1,
inf,
wrn,
err
};
Log(enum msg_type mt)
{
switch (mt) {
case dbg:
cout << "[D] ";
break;
case inf:
cout << "[I] ";
break;
case wrn:
cout << "[W] ";
break;
case err:
cout << "[E] ";
break;
default:
break;
}
}
template<typename T>
const Log& operator<<(const T& t)
{
std::cout << t;
return *this;
}
};
int main()
{
int i = 0;
Log(Log::dbg)<< "i is " << i++ << "\n";
Log(Log::inf)<< "i is " << i++ << "\n";
Log(Log::inf)<< "i is " << i++ << "\n";
}
显然它不起作用,但我不知道错误信息会告诉我什么。
G ++:
main.cc: In function ‘int main()’:
main.cc:46:34: error: passing ‘const Log’ as ‘this’ argument discards qualifiers [-fpermissive]
Log(Log::dbg)<< "i is " << i++ << "\n";
^
main.cc:35:14: note: in call to ‘const Log& Log::operator<<(const T&) [with T = int]’
const Log& operator<<(const T& t)
^
main.cc:46:40: error: passing ‘const Log’ as ‘this’ argument discards qualifiers [-fpermissive]
Log(Log::dbg)<< "i is " << i++ << "\n";
^
main.cc:35:14: note: in call to ‘const Log& Log::operator<<(const T&) [with T = char [2]]’
const Log& operator<<(const T& t)
^
锵:
main.cc:46:30: error: invalid operands to binary expression ('const Log' and 'int')
Log(Log::dbg)<< "i is " << i++ << "\n";
~~~~~~~~~~~~~~~~~~~~~~~ ^ ~~~
main.cc:35:14: note: candidate function not viable: 'this' argument has type
'const Log', but method is not marked const
const Log& operator<<(const T& t)
^
1 error generated.
最简单的方法当然是做一个宏来替换我记录的任何事件&#34; fake-ostream&#34;用例如。 cout << [I] <<
但这远不那么灵活。
有没有人对如何正确地做到这一点有任何想法,我可以提示一下这些资源吗?
感谢任何帮助!
答案 0 :(得分:0)
const
对象是常量,因此您只能调用常量成员函数。
你说通过在声明之后添加class Log {
public:
...
template<typename T>
const Log& operator<<(const T& t) const
// ^^^^^
// Note const keyword here
{
...
}
...
};
关键字,成员函数是不变的。
所以你的课应该是这样的(缩写):
cin