How come when I run this main.cpp
:
#include <iostream>
#include <typeinfo>
using namespace std;
struct Blah {};
int main() {
cout << typeid(Blah).name() << endl;
return 0;
}
By compiling it with GCC version 4.4.4:
g++ main.cpp
I get this:
4Blah
On Visual C++ 2008, I would get:
struct Blah
Is there a way to make it just print Blah
or struct Blah
?
typeid().name()
is implementation dependent. It may even return empty string for every type. That would not be very useful implementation, but it would be valid.No. The result of
std::typeinfo::name()
is unspecified. It might even return the same string for all types (or, indeed, empty strings for all types) and the implementation would still be standard-conforming. You must not rely on its result. Really, the only thing I found it useful for was debugging.Tell us what what you need it for. Often traits is what you use instead.
The string returned is implementation defined.
What gcc is doing is returning the mangled name.
You can convert the mangled name into plain text with c++filt
in 4Blah, 4 is the number of letters in your class name. For example if your class name is myEmptyClass then it would print 12myEmptyClass.
The return of
name
is implementation defined : an implementation is not even required to return different strings for different types.What you get from g++ is a decorated name, that you can "demangle" using the
c++filt
command or__cxa_demangle
.As others have said, the result here is implementation-defined, meaning that the implementation (i.e., the compiler toolchain) is free to define it how it wants, so long as it documents that somewhere.
From the C++ standard, section 18.5.1/1 [lib.type.info]: