Looping through all functions in a namespace in c++ - c++

My goal is to iterate over all functions of the namespace until a function returns me a valid strategy(enum). Each function can take different arguements.
enum class strategyType
{
Strategy1,
Strategy2,
Strategy3,
Strategy4,
Strategy5,
InvalidStrategy
}
namespace allFunctionStrategy
{
strategyType algorithmMl(int v1,int v2);
strategyType algorithmHeuristic(int v3,string s1);
strategyType algorithmHeuristic(string s1,string s2);
... n such function's
}
class strategyContext
{
int v1,v2,v3;
string s1,s2;
/*i want to add iterator logic here*/
}

Write a function that calls them:
void strategyContext::execute() {
if (algorithmM1(v1, v2) != strategyType::InvalidStrategy)
;
else if (algorithmHeuristic(v3, s1) != strategyType::InvalidStrategy)
;
else if (algorithmHeuristic(s1, s2) != strategyType::InvalidStrategy)
;
else
throw no_valid_strategy();
}
or, if the word "looping" was meant as a requirement, write wrappers and put the function pointers into an array:
strategyType strategyContext::wrap_algorithmM1() {
return algorithmM1(v3, s1);
}
strategyType strategyContext::wrap_algorithmHeuristic1() {
return algorithmHeuristic(v3, s1);
}
strategyType strategyContext::wrap_algorithmHeuristic2() {
return algorithmHeuristic(s1, s2);
}
typedef strategyType (strategyContext::*fptr)();
fptr pointers[] = {
&strategyContext::wrap_algorithmM1,
&strategyContext::wrap_algorithmHeuristic1,
&strategyContext::wrap_algorithmHeuristic2
};
void strategyContext::execute() {
for (auto iter = std::begin(pointers);
iter != std::end(pointers);
++iter) {
if ((this->*(*iter))() != strategyType::InvalidStrategy)
return;
}
throw no_valid_strategy();
}

Okay, first things first, you probably don't need to iterate over every function in the namespace. I'm going to show you how to use two macros to implement basic reflection, but you should probably also rethink what you're trying to do. Also, this is going to be really messy code.
First, we want every function to take the same parameters and return the same thing. For your use case, you can add all the parameters that the functions could use to a struct.
struct PossibleParams
{
//You can expand this if you want
int x, y, z;
};
struct Output {};
Next, we need a place to store the functions. Import functional and unordered_map and add the following macro.
#define ReflectableNamespace(name) namespace __REFLECTABLE__\
{ struct __NAMESPACE_ ## name ## __ { inline static std::unordered_map<std::string, std::function<Output(PossibleParams)>> funcs; }; \
struct __ ## name ## _MAP_ADDER__ \
{\
__ ## name ## _MAP_ADDER__ (const std::string& Name, std::function<Output(PossibleParams)> func) { __NAMESPACE_ ## name ## __::funcs[Name] = func; }\
};\
}\
namespace name
Here's what that macro does:
Defines an unordered_map that maps between the function name and the function in the __REFLECTABLE__ namespace.
Defines a struct that will let us add the function to this map at startup through its constructor
Begins the namespace that you're trying to make
This second macro will declare a function that can be iterated over and add it to the functions map.
#define ReflectableFunction(namespace, name) Output name (PossibleParams);\
struct __NAMESPACE_ ## namespace ## _ ## name ## _adder__ {\
inline static const __REFLECTABLE__::__ ## namespace ## _MAP_ADDER__ Adder = __REFLECTABLE__::__ ## namespace ## _MAP_ADDER__ (#name, name);\
}
This second struct declares a function that returns an Output and takes a PossibleParams as a parameter. It then creates an instance of that namespace's map adder object so that the function will be added to the namespace's map on startup. Since everything is inline, this should be able to be used in header files.
To call one of these functions, just use __REFLECTABLE__::__NAMESPACE_YourNamespace__::funcs["TheFunction"]({...});
Full working example:
struct PossibleParams
{
//Can expand this if you want
int x, y, z;
};
struct Output {};
#include <unordered_map>
#include <functional>
#define ReflectableNamespace(name) namespace __REFLECTABLE__\
{ struct __NAMESPACE_ ## name ## __ { inline static std::unordered_map<std::string, std::function<Output(PossibleParams)>> funcs; }; \
struct __ ## name ## _MAP_ADDER__ \
{\
__ ## name ## _MAP_ADDER__ (const std::string& Name, std::function<Output(PossibleParams)> func) { __NAMESPACE_ ## name ## __::funcs[Name] = func; }\
};\
}\
namespace name
#define ReflectableFunction(namespace, name) Output name (PossibleParams);\
struct __NAMESPACE_ ## namespace ## _ ## name ## _adder__ {\
inline static const __REFLECTABLE__::__ ## namespace ## _MAP_ADDER__ Adder = __REFLECTABLE__::__ ## namespace ## _MAP_ADDER__ (#name, name);\
}
#include <iostream>
ReflectableNamespace(MyNamespace)
{
ReflectableFunction(MyNamespace, func);
Output func(PossibleParams)
{
std::cout << "Hello" << std::endl;
return {};
}
}
int main()
{
std::cout << __REFLECTABLE__::__NAMESPACE_MyNamespace__::funcs.size() << std::endl;
__REFLECTABLE__::__NAMESPACE_MyNamespace__::funcs["func"]({});
return 0;
}
One last thing. If you're planning to iterate through this map a lot, it may be better to replace it with a normal map since it may be faster to iterate through. Or, if you don't care about the function's name, you could consider replacing the map with a vector. I hope this helps!

Related

How do i find the struct name from the instance? [duplicate]

Is it possible to get the object name too?
#include<cstdio>
class one {
public:
int no_of_students;
one() { no_of_students = 0; }
void new_admission() { no_of_students++; }
};
int main() {
one A;
for(int i = 0; i < 99; i++) {
A.new_admission();
}
cout<<"class"<<[classname]<<" "<<[objectname]<<"has "
<<A.no_of_students<<" students";
}
where I can fetch the names, something like
[classname] = A.classname() = one
[objectname] = A.objectname() = A
Does C++ provide any mechanism to achieve this?
You can display the name of a variable by using the preprocessor. For instance
#include <iostream>
#define quote(x) #x
class one {};
int main(){
one A;
std::cout<<typeid(A).name()<<"\t"<< quote(A) <<"\n";
return 0;
}
outputs
3one A
on my machine. The # changes a token into a string, after preprocessing the line is
std::cout<<typeid(A).name()<<"\t"<< "A" <<"\n";
Of course if you do something like
void foo(one B){
std::cout<<typeid(B).name()<<"\t"<< quote(B) <<"\n";
}
int main(){
one A;
foo(A);
return 0;
}
you will get
3one B
as the compiler doesn't keep track of all of the variable's names.
As it happens in gcc the result of typeid().name() is the mangled class name, to get the demangled version use
#include <iostream>
#include <cxxabi.h>
#define quote(x) #x
template <typename foo,typename bar> class one{ };
int main(){
one<int,one<double, int> > A;
int status;
char * demangled = abi::__cxa_demangle(typeid(A).name(),0,0,&status);
std::cout<<demangled<<"\t"<< quote(A) <<"\n";
free(demangled);
return 0;
}
which gives me
one<int, one<double, int> > A
Other compilers may use different naming schemes.
use typeid(class).name
// illustratory code assuming all includes/namespaces etc
#include <iostream>
#include <typeinfo>
using namespace std;
struct A{};
int main(){
cout << typeid(A).name();
}
It is important to remember that this
gives an implementation defined names.
As far as I know, there is no way to get the name of the object at run time reliably e.g. 'A' in your code.
EDIT 2:
#include <typeinfo>
#include <iostream>
#include <map>
using namespace std;
struct A{
};
struct B{
};
map<const type_info*, string> m;
int main(){
m[&typeid(A)] = "A"; // Registration here
m[&typeid(B)] = "B"; // Registration here
A a;
cout << m[&typeid(a)];
}
To get class name without mangling stuff you can use func macro in constructor:
class MyClass {
const char* name;
MyClass() {
name = __func__;
}
}
Do you want [classname] to be 'one' and [objectname] to be 'A'?
If so, this is not possible. These names are only abstractions for the programmer, and aren't actually used in the binary code that is generated. You could give the class a static variable classname, which you set to 'one' and a normal variable objectname which you would assign either directly, through a method or the constructor. You can then query these methods for the class and object names.
Just write simple template:
template<typename T>
const char* getClassName(T) {
return typeid(T).name();
}
struct A {} a;
void main() {
std::cout << getClassName(a);
}
You could try using "typeid".
This doesn't work for "object" name but YOU know the object name so you'll just have to store it somewhere. The Compiler doesn't care what you namned an object.
Its worth bearing in mind, though, that the output of typeid is a compiler specific thing so even if it produces what you are after on the current platform it may not on another. This may or may not be a problem for you.
The other solution is to create some kind of template wrapper that you store the class name in. Then you need to use partial specialisation to get it to return the correct class name for you. This has the advantage of working compile time but is significantly more complex.
Edit: Being more explicit
template< typename Type > class ClassName
{
public:
static std::string name()
{
return "Unknown";
}
};
Then for each class somethign liek the following:
template<> class ClassName<MyClass>
{
public:
static std::string name()
{
return "MyClass";
}
};
Which could even be macro'd as follows:
#define DefineClassName( className ) \
\
template<> class ClassName<className> \
{ \
public: \
static std::string name() \
{ \
return #className; \
} \
}; \
Allowing you to, simply, do
DefineClassName( MyClass );
Finally to Get the class name you'd do the following:
ClassName< MyClass >::name();
Edit2: Elaborating further you'd then need to put this "DefineClassName" macro in each class you make and define a "classname" function that would call the static template function.
Edit3: And thinking about it ... Its obviously bad posting first thing in the morning as you may as well just define a member function "classname()" as follows:
std::string classname()
{
return "MyClass";
}
which can be macro'd as follows:
DefineClassName( className ) \
std::string classname() \
{ \
return #className; \
}
Then you can simply just drop
DefineClassName( MyClass );
into the class as you define it ...
You can try this:
template<typename T>
inline const char* getTypeName() {
return typeid(T).name();
}
#define DEFINE_TYPE_NAME(type, type_name) \
template<> \
inline const char* getTypeName<type>() { \
return type_name; \
}
DEFINE_TYPE_NAME(int, "int")
DEFINE_TYPE_NAME(float, "float")
DEFINE_TYPE_NAME(double, "double")
DEFINE_TYPE_NAME(std::string, "string")
DEFINE_TYPE_NAME(bool, "bool")
DEFINE_TYPE_NAME(uint32_t, "uint")
DEFINE_TYPE_NAME(uint64_t, "uint")
// add your custom types' definitions
And call it like that:
void main() {
std::cout << getTypeName<int>();
}
An improvement for #Chubsdad answer,
//main.cpp
using namespace std;
int main(){
A a;
a.run();
}
//A.h
class A{
public:
A(){};
void run();
}
//A.cpp
#include <iostream>
#include <typeinfo>
void A::run(){
cout << (string)typeid(this).name();
}
Which will print:
class A*
Here is a trick for getting the name of a class you create:
struct NameTest {
NameTest() : mName {std::source_location::current().function_name()} {
}
void operator()() {
auto src_loc = std::source_location::current();
std::cout << "Class name:\t" << mName //
<< "\nFunc:\t\t" << src_loc.function_name() //
<< "\nLine:\t\t" << src_loc.line() << '\n';
}
const std::string mName;
};
int main() {
NameTest name_test;
name_test();
return 0;
}
output:
Class name: NameTest::NameTest()
Func: void NameTest::operator()()
Line: 81
A little string manipulation will strip the unneeded parts

How to create a macro for a templated function in c++

I want to create a macro for a template function
something on the lines of
#define CHECK_MIN(T)(value, target) checkmin(T value, T target) \
checkmin<T>(T val, T target, __FUNCTION__, __LINE__)
template<typename T> checkmin(T val, T target, const char* functionName, long lineNumber)
{
// Check if the val is less than target
// Construct a std::string using the function name and line name
// throw std:: exception passing the string constructed above.
}
I am not able to get the syntax to achieve this. Any ideas?
You can do it by following way:
#include <iostream>
using namespace std;
template<typename T>
void checkmin(T val, T target, const char* functionName, long lineNumber)
{
// Check if the val is less than target
// Construct a std::string using the function name and line name
// throw std:: exception passing the string constructed above.
}
#define CHECK_MIN(value, target) \
checkmin(value,target, __FUNCTION__, __LINE__);
int main() {
/*Not using new in object definitions so I don't have to delete them afterwards since pointers don't stay in memory*/
CHECK_MIN(5,6);
return 0;
}

C++ How do you output an object's type [duplicate]

Is it possible to get the object name too?
#include<cstdio>
class one {
public:
int no_of_students;
one() { no_of_students = 0; }
void new_admission() { no_of_students++; }
};
int main() {
one A;
for(int i = 0; i < 99; i++) {
A.new_admission();
}
cout<<"class"<<[classname]<<" "<<[objectname]<<"has "
<<A.no_of_students<<" students";
}
where I can fetch the names, something like
[classname] = A.classname() = one
[objectname] = A.objectname() = A
Does C++ provide any mechanism to achieve this?
You can display the name of a variable by using the preprocessor. For instance
#include <iostream>
#define quote(x) #x
class one {};
int main(){
one A;
std::cout<<typeid(A).name()<<"\t"<< quote(A) <<"\n";
return 0;
}
outputs
3one A
on my machine. The # changes a token into a string, after preprocessing the line is
std::cout<<typeid(A).name()<<"\t"<< "A" <<"\n";
Of course if you do something like
void foo(one B){
std::cout<<typeid(B).name()<<"\t"<< quote(B) <<"\n";
}
int main(){
one A;
foo(A);
return 0;
}
you will get
3one B
as the compiler doesn't keep track of all of the variable's names.
As it happens in gcc the result of typeid().name() is the mangled class name, to get the demangled version use
#include <iostream>
#include <cxxabi.h>
#define quote(x) #x
template <typename foo,typename bar> class one{ };
int main(){
one<int,one<double, int> > A;
int status;
char * demangled = abi::__cxa_demangle(typeid(A).name(),0,0,&status);
std::cout<<demangled<<"\t"<< quote(A) <<"\n";
free(demangled);
return 0;
}
which gives me
one<int, one<double, int> > A
Other compilers may use different naming schemes.
use typeid(class).name
// illustratory code assuming all includes/namespaces etc
#include <iostream>
#include <typeinfo>
using namespace std;
struct A{};
int main(){
cout << typeid(A).name();
}
It is important to remember that this
gives an implementation defined names.
As far as I know, there is no way to get the name of the object at run time reliably e.g. 'A' in your code.
EDIT 2:
#include <typeinfo>
#include <iostream>
#include <map>
using namespace std;
struct A{
};
struct B{
};
map<const type_info*, string> m;
int main(){
m[&typeid(A)] = "A"; // Registration here
m[&typeid(B)] = "B"; // Registration here
A a;
cout << m[&typeid(a)];
}
To get class name without mangling stuff you can use func macro in constructor:
class MyClass {
const char* name;
MyClass() {
name = __func__;
}
}
Do you want [classname] to be 'one' and [objectname] to be 'A'?
If so, this is not possible. These names are only abstractions for the programmer, and aren't actually used in the binary code that is generated. You could give the class a static variable classname, which you set to 'one' and a normal variable objectname which you would assign either directly, through a method or the constructor. You can then query these methods for the class and object names.
Just write simple template:
template<typename T>
const char* getClassName(T) {
return typeid(T).name();
}
struct A {} a;
void main() {
std::cout << getClassName(a);
}
You could try using "typeid".
This doesn't work for "object" name but YOU know the object name so you'll just have to store it somewhere. The Compiler doesn't care what you namned an object.
Its worth bearing in mind, though, that the output of typeid is a compiler specific thing so even if it produces what you are after on the current platform it may not on another. This may or may not be a problem for you.
The other solution is to create some kind of template wrapper that you store the class name in. Then you need to use partial specialisation to get it to return the correct class name for you. This has the advantage of working compile time but is significantly more complex.
Edit: Being more explicit
template< typename Type > class ClassName
{
public:
static std::string name()
{
return "Unknown";
}
};
Then for each class somethign liek the following:
template<> class ClassName<MyClass>
{
public:
static std::string name()
{
return "MyClass";
}
};
Which could even be macro'd as follows:
#define DefineClassName( className ) \
\
template<> class ClassName<className> \
{ \
public: \
static std::string name() \
{ \
return #className; \
} \
}; \
Allowing you to, simply, do
DefineClassName( MyClass );
Finally to Get the class name you'd do the following:
ClassName< MyClass >::name();
Edit2: Elaborating further you'd then need to put this "DefineClassName" macro in each class you make and define a "classname" function that would call the static template function.
Edit3: And thinking about it ... Its obviously bad posting first thing in the morning as you may as well just define a member function "classname()" as follows:
std::string classname()
{
return "MyClass";
}
which can be macro'd as follows:
DefineClassName( className ) \
std::string classname() \
{ \
return #className; \
}
Then you can simply just drop
DefineClassName( MyClass );
into the class as you define it ...
You can try this:
template<typename T>
inline const char* getTypeName() {
return typeid(T).name();
}
#define DEFINE_TYPE_NAME(type, type_name) \
template<> \
inline const char* getTypeName<type>() { \
return type_name; \
}
DEFINE_TYPE_NAME(int, "int")
DEFINE_TYPE_NAME(float, "float")
DEFINE_TYPE_NAME(double, "double")
DEFINE_TYPE_NAME(std::string, "string")
DEFINE_TYPE_NAME(bool, "bool")
DEFINE_TYPE_NAME(uint32_t, "uint")
DEFINE_TYPE_NAME(uint64_t, "uint")
// add your custom types' definitions
And call it like that:
void main() {
std::cout << getTypeName<int>();
}
An improvement for #Chubsdad answer,
//main.cpp
using namespace std;
int main(){
A a;
a.run();
}
//A.h
class A{
public:
A(){};
void run();
}
//A.cpp
#include <iostream>
#include <typeinfo>
void A::run(){
cout << (string)typeid(this).name();
}
Which will print:
class A*
Here is a trick for getting the name of a class you create:
struct NameTest {
NameTest() : mName {std::source_location::current().function_name()} {
}
void operator()() {
auto src_loc = std::source_location::current();
std::cout << "Class name:\t" << mName //
<< "\nFunc:\t\t" << src_loc.function_name() //
<< "\nLine:\t\t" << src_loc.line() << '\n';
}
const std::string mName;
};
int main() {
NameTest name_test;
name_test();
return 0;
}
output:
Class name: NameTest::NameTest()
Func: void NameTest::operator()()
Line: 81
A little string manipulation will strip the unneeded parts

C++ preprocesor macro for accumulating comma-separated strings

I need to do the following:
const char* my_var = "Something";
REGISTER(my_var);
const char* my_var2 = "Selse";
REGISTER(my_var2);
...
concst char* all[] = { OUTPUT_REGISTERED }; // inserts: "my_var1, my_var2, ..."
REGISTER and OUTPUT_REGISTERED are preprocesor macros. This would be great for large number of strings, like ~100. Is it possible to accomplish this?
PS. The code belongs to level-0 "block" – i.e. it is not inside any function. AFAIK, I cannot call regular functions there.
#include <iostream>
#include <vector>
using namespace std;
vector<const char*>& all()
{
static vector<const char*> v;
return v;
}
struct string_register
{
string_register(const char* s)
{
all().push_back(s);
}
};
#define REGISTER3(x,y,sr) string_register sr ## y(x)
#define REGISTER2(x,y) REGISTER3(x,y,sr)
#define REGISTER(x) REGISTER2(x, __COUNTER__)
REGISTER("foo");
REGISTER("bar");
int main()
{
}

How can I get the class name from a C++ object?

Is it possible to get the object name too?
#include<cstdio>
class one {
public:
int no_of_students;
one() { no_of_students = 0; }
void new_admission() { no_of_students++; }
};
int main() {
one A;
for(int i = 0; i < 99; i++) {
A.new_admission();
}
cout<<"class"<<[classname]<<" "<<[objectname]<<"has "
<<A.no_of_students<<" students";
}
where I can fetch the names, something like
[classname] = A.classname() = one
[objectname] = A.objectname() = A
Does C++ provide any mechanism to achieve this?
You can display the name of a variable by using the preprocessor. For instance
#include <iostream>
#define quote(x) #x
class one {};
int main(){
one A;
std::cout<<typeid(A).name()<<"\t"<< quote(A) <<"\n";
return 0;
}
outputs
3one A
on my machine. The # changes a token into a string, after preprocessing the line is
std::cout<<typeid(A).name()<<"\t"<< "A" <<"\n";
Of course if you do something like
void foo(one B){
std::cout<<typeid(B).name()<<"\t"<< quote(B) <<"\n";
}
int main(){
one A;
foo(A);
return 0;
}
you will get
3one B
as the compiler doesn't keep track of all of the variable's names.
As it happens in gcc the result of typeid().name() is the mangled class name, to get the demangled version use
#include <iostream>
#include <cxxabi.h>
#define quote(x) #x
template <typename foo,typename bar> class one{ };
int main(){
one<int,one<double, int> > A;
int status;
char * demangled = abi::__cxa_demangle(typeid(A).name(),0,0,&status);
std::cout<<demangled<<"\t"<< quote(A) <<"\n";
free(demangled);
return 0;
}
which gives me
one<int, one<double, int> > A
Other compilers may use different naming schemes.
use typeid(class).name
// illustratory code assuming all includes/namespaces etc
#include <iostream>
#include <typeinfo>
using namespace std;
struct A{};
int main(){
cout << typeid(A).name();
}
It is important to remember that this
gives an implementation defined names.
As far as I know, there is no way to get the name of the object at run time reliably e.g. 'A' in your code.
EDIT 2:
#include <typeinfo>
#include <iostream>
#include <map>
using namespace std;
struct A{
};
struct B{
};
map<const type_info*, string> m;
int main(){
m[&typeid(A)] = "A"; // Registration here
m[&typeid(B)] = "B"; // Registration here
A a;
cout << m[&typeid(a)];
}
To get class name without mangling stuff you can use func macro in constructor:
class MyClass {
const char* name;
MyClass() {
name = __func__;
}
}
Do you want [classname] to be 'one' and [objectname] to be 'A'?
If so, this is not possible. These names are only abstractions for the programmer, and aren't actually used in the binary code that is generated. You could give the class a static variable classname, which you set to 'one' and a normal variable objectname which you would assign either directly, through a method or the constructor. You can then query these methods for the class and object names.
Just write simple template:
template<typename T>
const char* getClassName(T) {
return typeid(T).name();
}
struct A {} a;
void main() {
std::cout << getClassName(a);
}
You could try using "typeid".
This doesn't work for "object" name but YOU know the object name so you'll just have to store it somewhere. The Compiler doesn't care what you namned an object.
Its worth bearing in mind, though, that the output of typeid is a compiler specific thing so even if it produces what you are after on the current platform it may not on another. This may or may not be a problem for you.
The other solution is to create some kind of template wrapper that you store the class name in. Then you need to use partial specialisation to get it to return the correct class name for you. This has the advantage of working compile time but is significantly more complex.
Edit: Being more explicit
template< typename Type > class ClassName
{
public:
static std::string name()
{
return "Unknown";
}
};
Then for each class somethign liek the following:
template<> class ClassName<MyClass>
{
public:
static std::string name()
{
return "MyClass";
}
};
Which could even be macro'd as follows:
#define DefineClassName( className ) \
\
template<> class ClassName<className> \
{ \
public: \
static std::string name() \
{ \
return #className; \
} \
}; \
Allowing you to, simply, do
DefineClassName( MyClass );
Finally to Get the class name you'd do the following:
ClassName< MyClass >::name();
Edit2: Elaborating further you'd then need to put this "DefineClassName" macro in each class you make and define a "classname" function that would call the static template function.
Edit3: And thinking about it ... Its obviously bad posting first thing in the morning as you may as well just define a member function "classname()" as follows:
std::string classname()
{
return "MyClass";
}
which can be macro'd as follows:
DefineClassName( className ) \
std::string classname() \
{ \
return #className; \
}
Then you can simply just drop
DefineClassName( MyClass );
into the class as you define it ...
You can try this:
template<typename T>
inline const char* getTypeName() {
return typeid(T).name();
}
#define DEFINE_TYPE_NAME(type, type_name) \
template<> \
inline const char* getTypeName<type>() { \
return type_name; \
}
DEFINE_TYPE_NAME(int, "int")
DEFINE_TYPE_NAME(float, "float")
DEFINE_TYPE_NAME(double, "double")
DEFINE_TYPE_NAME(std::string, "string")
DEFINE_TYPE_NAME(bool, "bool")
DEFINE_TYPE_NAME(uint32_t, "uint")
DEFINE_TYPE_NAME(uint64_t, "uint")
// add your custom types' definitions
And call it like that:
void main() {
std::cout << getTypeName<int>();
}
An improvement for #Chubsdad answer,
//main.cpp
using namespace std;
int main(){
A a;
a.run();
}
//A.h
class A{
public:
A(){};
void run();
}
//A.cpp
#include <iostream>
#include <typeinfo>
void A::run(){
cout << (string)typeid(this).name();
}
Which will print:
class A*
Here is a trick for getting the name of a class you create:
struct NameTest {
NameTest() : mName {std::source_location::current().function_name()} {
}
void operator()() {
auto src_loc = std::source_location::current();
std::cout << "Class name:\t" << mName //
<< "\nFunc:\t\t" << src_loc.function_name() //
<< "\nLine:\t\t" << src_loc.line() << '\n';
}
const std::string mName;
};
int main() {
NameTest name_test;
name_test();
return 0;
}
output:
Class name: NameTest::NameTest()
Func: void NameTest::operator()()
Line: 81
A little string manipulation will strip the unneeded parts