How to use typedef and define - c++

Question

I write this code to print on the output the current line and the file name:
#ifdef _DEBUG
char buffer[10000];
std::string const full_path = __FILE__;
auto const basename = full_path.substr(full_path.rfind('\\'));
sprintf(buffer, "Line:%d %s \n\n", __LINE__, basename.c_str());
OutputDebugStringA(buffer);
#endif
I don't know much about typedef and define, is possible to convert this part of the code below, into a typedef or define?
I mean a way to call it in just a single line.
std::string const full_path = __FILE__;
auto const basename = full_path.substr(full_path.rfind('\\'));

Answer 1

I mean a way to call it in just a single line.
std::string const full_path = __FILE__;
auto const basename = full_path.substr(full_path.rfind('\\'));
Generally, that's what functions are for. The only problem is the use of the macro that depends on what file it's in (__FILE__). Luckily, C++20 allows us to not need a macro:
#include <source_location>
std::string
current_source_file_base(const std::source_location& location
= std::source_location::current())
{
return std::filesystem::path(location.file_name()).filename();
}
Now, you can call that in a reusable fashion:
std::string source_file = current_source_file_base();
Alternative that doesn't require std::string and thus won't allocate dynamic memory... but may give partial filenames if they contain escaped directory separators. This is a problem with your original code as well. The limitation could be fixed by more complex logic, or by using std::filesystem::path_view which is proposed but not part of the standard. (Please don't use escaped directory separators in source file names).
constexpr std::string_view
current_source_file_base(const std::source_location& location
= std::source_location::current())
{
char separators[] = {
std::filesystem::path::preferred_separator,
'/',
'\0',
};
std::string_view path = location.file_name();
std::size_t pos = path.find_last_of(separators);
return pos == std::string_view::npos
? path
: path.substr(pos + 1);
}
All that said, I recommend to not remove the directories from the path because that will make it impossible to distinguish dir1/file.cpp from dir2/file.cpp.
Prior to C++20, you would need to use a function-like macro.

Answer 2

You can remove the std::string construction this way:
char buffer[1000];
snprintf(buffer, sizeof buffer, "Line:%d %s\n\n", __LINE__,
strrchr(__FILE__, '\\') ? strrchr(__FILE__, '\\') + 1 : __FILE__);
OutputDebugStringA(buffer);
If must make it a one liner:
char buffer[1000]; snprintf(buffer, sizeof buffer, "Line:%d %s\n\n", __LINE__, strrchr(__FILE__, '\\') ? strrchr(__FILE__, '\\') + 1 : __FILE__); OutputDebugStringA(buffer);
Here is a more general approach:
#ifdef _DEBUG
#define DEBUG_TRACE(s) do { \
char buffer[1000]; \
snprintf(buffer, sizeof buffer, "Line:%d %s %s\n\n", __LINE__, \
strrchr(__FILE__, '\\') ? strrchr(__FILE__, '\\') + 1 : __FILE__, s); \
OutputDebugStringA(buffer); \
} while (0)
#else
#define DEBUG_TRACE()
#endif
Put the above definitions is a header file and add traces in your code:
int test(int a) {
int res = 0;
DEBUG_TRACE("entering test");
if (a % 3 == a % 5)
res = 1;
DEBUG_TRACE("leaving test");
return res;
}

Answer 3

A platform agnostic way to get the basename of a file would be:
#include <filesystem>
auto basename = std::filesystem::path(__FILE__).filename();
or, if you prefer it as a std::string object instead of a std::filesystem::path object:
auto basename = std::filesystem::path(__FILE__).filename().string();