Consider the below code :
#include <iostream>
void test_function(const char* caller_name = __FUNCTION__) {
std::cout << caller_name << " ";
}
void func1() {
test_function();
}
int main() {
test_function();
func1();
}
The above code prints nothing.
Is there any way that test_function prints the name of the function from which it is called ?
Eg : For this case, output should be : main func1
I understand that desired output can be achieved if we make the argument to test_function, non default, and pass __FUNCTION__ from the caller function.
This could be achieved by using macro also, but it is not what I am looking for.
Use a macro
#define TEST_FUNCTION() test_function(__FUNCTION__)
void test_function(const char *caller) {
std::cout << caller << '\n';
}
void foo() {
TEST_FUNCTION(); // prints foo
}
int main() {
TEST_FUNCTION(); // prints main
foo();
}
Related
I am attempting to create a wrapper around class functions. The purpose of my wrapper is to test input, output, and enforce order of operations with various calls throughout my program. I am trying to not make any changes to the callee class. Attached is an example of what I am trying to achieve, but unable to figure out.
Main.cpp
#include "func_warpper.h"
#include "func.h"
int main()
{
func_wrapper fw
fun func;
int origValue = 5;
fw.caller([&](int origValue) { func.f(origValue); }, origValue);
int output = func.getResult().number;
std::cout << " value outputed by function 2 : " << output << std::endl;
// output
// note that above line does give me the result I am looking for
// however, I want to be able to get this inside the function of caller
return 0;
}
func.h .... I want this to be unmodified
#ifndef FUN_H
#define FUN_H
class fun
{
public:
struct result
{
int number;
};
fun();
~fun();
void f(int value);
struct result getResult(){return this->testResult;};
private:
struct result testResult;
};
#endif
func.cpp .... I want this to be unmodified
#include "func.h"
fun::fun(){
this->testResult.number = 0;
return;
}
fun::~fun(){
return;
}
void fun::f(int value){
int updateValue = value * 5;
this->testResult.number = updateValue;
}
func_wrapper.h .... I can modify this until the cows come home, please go ham with recommended changes :)
class func_wrapper
{
public:
struct new_result
{
int new_number;
};
func_wrapper();
~func_wrapper();
void caller(std::function<void(int)> clb, int val);
struct new_result getNewResult() { return this->new_testResult; };
private:
struct new_result new_testResult;
};
#endif
func_wrapper.cpp .... same as above, I can modify this until the cows come home, please go ham with recommended changes :)
#include "func_wrapper.h"
func_wrapper::func_wrapper()
{
//ctor
this->new_testResult.new_number = 0;
return;
}
func_wrapper::~func_wrapper()
{
//dtor
}
void func_wrapper::caller(std::function<void(int)> clb, int val)
{
std::cout << " value entered into function: " << val << std::endl;
// clb(val); seems to call the function but does not store locally anything
clb(val);
clb;
// clb; seems to store all the information locally however I seem unable to
// to reach the infromation: clb -> [functor] -> func -> testResult -> number
// would like ...
int output = clb ??? // the result of what gets filled from number struct
// if I attempt to #include func.h
// func func;
// func.getResult().number; locally the answer is zero with or without delay
}
Through several days of searching, I have not found anything that can help with this problem, to include similar enough questions on stack overflow. Any help would be greatly appreciated, thank you.
So, my understanding is that inside func_wrapper::caller you want to be able to access the wrapped class that is inside your callback. Unfortuately, the way you are doing it, is impossible. There is no (legitimate) way to reach inside the function and access its arguments.
However, if you break up the operation into its component parts, you can do what you want. You would want a caller function more like this:
template <typename Type, typename Function>
void caller(Type&& functor, Function function, int val)
{
std::cout << " value entered into function: " << val << std::endl;
std::invoke(function, functor, val);
std::cout << "value inside wrapper: " << functor.getResult().number << "\rn";
}
and then call it like this.
fw.caller(func, &fun::f, origValue);
https://godbolt.org/z/151YfEeoo
#JohnFilleau had mentioned to pass the class object instead of the function from within the class. The following is the solution based on example code that he provided, and I modified to work with the example. I realize the question is confusing but would like to thank both JohnFilleau and Taekahn for the discussion.
In main.cpp
int main()
{
func_wrapper fw;
fun func;
int origValue = 5;
fw.caller2(func, origValue);
return 0:
}
func_wrapper::caller2
void func_wrapper::caller2(fun& fun, int val)
{
std::cout << " value entered into function: " << val << std::endl;
fun.f(val);
int output = fun.getResult().number;
std::cout << " did this work: " << output << std::endl;
}
In the header I had to add
#include "func.h"
with the change to the header as follows
void caller2(fun& fun, int val);
Lets consider the next piece of code:
#include <iostream>
void print()
{
std::cout << "I feel void" << std::endl;
}
void (*func)();
func = print;
This does not compile, since "func does not name a type".
But I already declared about func's type. It's a function pointer that takes no arguments and returns void. Why do I need to name a type again?
Do I always have to initialize a function pointer when declaring it?
No.
However, you cannot have expression statements such as assignments in namespace scope.
you can do that in main! , you cant do that in namespaces:
void (*func)();
void print()
{
std::cout << "I feel void" << std::endl;
}
int main(int argc, char* argv[])
{
func = print;
return 0;
}
but if you want do that before main execution , you can try below code(I prefer dont use this , but it's helpful sometimes):
#include <iostream>
void (*func)();
void print()
{
std::cout << "I feel void" << std::endl;
}
class Initilizer
{
public:
Initilizer()
{
init();
}
static void init()
{
func = ::print;
}
};
Initilizer obj;
int main(int argc, char* argv[])
{
func(); /*Test func*/
return 0;
}
I am developing a C++ tool that should run transparent to main program. That is: if user simply links the tool to his program the tool will be activated. For that I need to invoke two functions, function a(), before main() gets control and function b() after main() finishes.
I can easily do the first by declaring a global variable in my program and have it initialize by return code of a(). i.e
int v = a() ;
but I cannot find a way to call b() after main() finishes?
Does any one can think of a way to do this?
The tool runs on windows, but I'd rather not use any OS specific calls.
Thank you, George
Use RAII, with a and b called in constructor/destructor.
class MyObj {
MyObj()
{
a();
};
~MyObj()
{
b();
};
};
Then just have an instance of MyObj outside the scope of main()
MyObj obj;
main()
{
...
}
Some things to note.
This is bog-standard C++ and will work on any platform
You can use this without changing ANY existing source code, simply by having your instance of MyObj in a separate compilation unit.
While it will run before and after main(), any other objects constructed outside main will also run at this time. And you have little control of the order
of your object's construction/destruction, among those others.
SOLUTION IN C:
have a look at atexit:
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
void bye(void)
{
printf("That was all, folks\n");
}
int main(void)
{
long a;
int i;
a = sysconf(_SC_ATEXIT_MAX);
printf("ATEXIT_MAX = %ld\n", a);
i = atexit(bye);
if (i != 0) {
fprintf(stderr, "cannot set exit function\n");
exit(EXIT_FAILURE);
}
exit(EXIT_SUCCESS);
}
http://linux.die.net/man/3/atexit
this still implies however that you have access to your main and you can add the atexit call. If you have no access to the main and you cannot add this function call I do not think there is any option.
EDIT:
SOLUTION IN C++:
as sudgested there is a c++ equivalent from std. I simply paste in here an example which i copied from the link available just below the code:
#include <iostream>
#include <cstdlib>
void atexit_handler_1()
{
std::cout << "at exit #1\n";
}
void atexit_handler_2()
{
std::cout << "at exit #2\n";
}
int main()
{
const int result_1 = std::atexit(atexit_handler_1);
const int result_2 = std::atexit(atexit_handler_2);
if ((result_1 != 0) or (result_2 != 0)) {
std::cerr << "Registration failed\n";
return EXIT_FAILURE;
}
std::cout << "returning from main\n";
return EXIT_SUCCESS;
}
http://en.cppreference.com/w/cpp/utility/program/atexit
Isn't any global variable constructed before main and destructed afterward? I made a test struct whose constructor is called before main and the destructor afterward.
#include <iostream>
struct Test
{
Test() { std::cout << "Before main..." << std::endl; }
~Test() { std::cout << "After main..." << std::endl; }
};
Test test;
int main()
{
std::cout << "Returning now..." << std::endl;
return 0;
}
If you're happy to stick with a single compiler and non-standard C/C++, then GCC's __attribute__((constructor)) and __attribute__((destructor)) might be of use:
#include <stdio.h>
void __attribute__((constructor)) ctor()
{
printf("Before main()\n");
}
void __attribute__((destructor)) dtor()
{
printf("After main()\n");
}
int main()
{
printf("main()\n");
return 0;
}
Result:
Before main()
main()
After main()
Alternatively to the destructor, you can use atexit() in a similar manner - in C++, you do not need to have access to main() to register atexit there. You can do that as well it in your a() - for example:
void b(void) {
std::cout << "Exiting.\n";
}
int a(void) {
std::cout << "Starting.\n";
atexit(b);
return 0;
}
// global in your module
int i = a();
That being said, I'd also prefer the global C++ class object, which will call the b() stuff in its destructor.
The program:
#include <iostream>
void foo(void (*bar)()){ bar(); };
void foo(int a = 5)
{
std::cout << a << std::endl;
}
int main()
{
foo(foo); //Error
}
DEMO
I expected that eventually foo(5) will be called. In contrast, the following program works fine:
#include <iostream>
void foo(void (*bar)()){ bar(); };
void foo()
{
std::cout << 5 << std::endl;
}
int main()
{
foo(foo); //OK
}
DEMO
Could you explain that difference?
In the first example although foo has default argument its type is void (bar*)(int). Having the default argument makes it possible to call foo without specifying the argument value explicitly, but there is still an int argument. Simply its value is automatically populated(during compilation).
What's the point of initialising an unnamed C++ parameter? For example:
void foo(int = 0) {}
A declaration has no need of a parameter name. The definition does, however. Also, the default parameter cannot be repeated in the definition. Here's a small program that works (but I don't know why you would want to do something like this, really...):
#include <iostream>
void foo(int = 5);
int main() {
foo();
foo(3);
return 0;
}
void foo(int i) {
std::cout << i << std::endl;
}
The output is
5
3
I can imagine in the context of callback functions the construct might be useful:
#include <iostream>
// Please assume the callback is an external library:
typedef void (*callback_function)(int);
callback_function callback;
void foo(int = 0) {
std::cout << "Hello\n";
}
int main() {
callback = foo;
callback(1);
foo();
}
It will serve as the default parameter value. It belongs in the function declaration.