问题描述
我正在尝试为日志记录机制编写宏.我编写了一个可变参数宏,但是它不适用于std::string
.该代码如下所示:
I am trying to write a macro for logging mechanism. I wrote a variadic macro but it does not work with std::string
. The code looks like the following:
#include <stdio.h>
#include <string>
#define LOG_NOTE(m, ...) printf(m, ##__VA_ARGS__)
int main()
{
std::string foo = "random string";
int bar = 5;
LOG_NOTE("%s %d %s", "Hello World", bar, foo);
return 0;
}
如果我按如下所示调用宏,则不会出现任何错误.
If I would call the macro like following, I would not get any error.
LOG_NOTE("%s %d %s", "Hello World", bar, "random string");
编译器输出:
推荐答案
不要.使用可变参数模板功能.
Don't. Use a variadic template function.
您遇到的实际问题是,您试图通过C API(printf
)传递C ++对象(std::string
).这是不可能的.
The actual problem you have is that you're trying to pass a C++ object (std::string
) through a C API (printf
). This is not possible.
您需要某种转换机制,例如:
You'd need some mechanism for conversion, for example:
#include <stdio.h>
#include <string>
template<class T>
decltype(auto) convert_for_log_note(T const& x)
{
return x;
}
decltype(auto) convert_for_log_note(std::string const& x)
{
return x.c_str();
}
template<class...Args>
void LOG_NOTE(const char* format, Args&&...args)
{
printf(format, convert_for_log_note(args)...);
}
int main()
{
std::string foo = "random string";
int bar = 5;
LOG_NOTE("%s %d %s\n", "Hello World", bar, foo);
return 0;
}
示例输出:
Hello World 5 random string
http://coliru.stacked-crooked.com/a/beb3431114833860
更新:
对于C ++ 11,您需要手工说明返回类型:
For C++11 you'll need to spell out the return types by hand:
#include <stdio.h>
#include <string>
template<class T>
T const& convert_for_log_note(T const& x)
{
return x;
}
const char* convert_for_log_note(std::string const& x)
{
return x.c_str();
}
template<class...Args>
void LOG_NOTE(const char* format, Args&&...args)
{
printf(format, convert_for_log_note(args)...);
}
int main()
{
std::string foo = "random string";
int bar = 5;
LOG_NOTE("%s %d %s\n", "Hello World", bar, foo);
return 0;
}
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