I was just experimenting with C++. I was trying to write a small macro so that all the functions that I define are automatically stored in a map so that I can query, at run time, what functions exist and run them too. The code is as follows:
#include <map>
using namespace std;
typedef void (*funcPointer)();
map <char*, funcPointer> funcList;
#define Function(x) void x() { funcList[#x] = x;
#define End }
I was used funcPointer and End only for easy readability and implementation. Now, I can define a function as
Function(helloWorld)
cout << "Hello World";
End
Now, to read the function names as a list and run all the functions, I use the following code:
int main() {
//helloWorld();
for (map<char*, funcPointer>::iterator I = funcList.begin(); I != funcList.end(); I++) {
printf(I->first);
I->second();
}
getchar();
return 0;
}
The problem is, if I keep the first line of main() (helloWorld();) commented, the compiler doesn't compile the function and skips it for optimization, as according to the compiler, it is never used. So, the function list turns up empty. If, instead, I call the function once, every thing works perfectly, except that it prints "Hello World" twice. Also, I wrote the macro specifically so I do not have to do that.
So, is there any way that I can force the compiler to compile a function even if it is not used?
The problem is that the code to register the function is inside the function, so won't happen unless you call the function. You might instead register it by initialising a global variable, which will happen automatically before main begins. This might look something like
struct funcRegistration {
funcRegistration(char * name, funcPointer func) {funcList[name] = func;}
};
#define Function(x) \
void x(); \
funcRegistration x##_registration(#x, x); \
void x() {
The compiler compiles the function, however your map won't be populated unless its called.
Because funcList[#x] = x; comes inside the function block { } after macro expansion.
Related
I am trying to print IT SUCCESS\nET SUCCESS\n using following code but it's failing in compilation with error error: ‘printds’ was not declared in this scope which I know is because it's taking macro input as ds literal. Does anyone know how to do this? The use case is that there are several printXX() functions which should be called based on value passed in macro.
#include <stdio.h>
#define FOO(val) { \
print ## val(); \
}
void printIT() { printf("IT SUCCESS\n"); }
void printET() { printf("ET SUCCESS\n"); }
int main() {
const char* ds = "IT", es = "ET";
FOO(ds); FOO(es);
return 0;
}
You can change
FOO(ds); FOO(es);
to
FOO(IT); FOO(ET);
Because macro substitutions happen before your code is compiled.
But you can define a function called FOO like
#include <stdio.h>
#include <iostream>
using namespace std;
void printIT() { printf("IT SUCCESS\n"); }
void printET() { printf("ET SUCCESS\n"); }
void FOO(const string str)
{
if(str=="IT")
printIT();
else
printET();
}
int main()
{
const char* ds = "IT",*es="ET";
FOO(ds);FOO(es);
return 0;
}
it's taking macro input as ds literal.
Yes, that's to be expected. Preprocessor macros are expanded at compile time. The arguments to function-like macros are the literal source-code tokens that appear between the parentheses in the macro invocation. These have no additional meaning to the preprocessor.
Does anyone know how to do this? The use case is that there are several printXX() functions which should be called based on value passed in macro.
Again, macros are expanded, to source code (approximately), at compile time. The process does not and cannot take into account C++ runtime semantics such as converting variables' identifiers into corresponding values.
If runtime dynamic function dispatch based on variables' values is what you're after then you need an altogether different mechanism. You could use ordinary conditional statements to select between different function calls, for example. If you wanted to be even more dynamic then you could consider preparing a lookup table of function pointers, and using that to select and call appropriate functions.
In comments, you added
I've several methods queryServers, queryNodes, queryTargets which I want to call using above trick.
You may be able to accomplish something similar to what you ask via templates or overloaded functions. These mechanisms, too, operate at compile time, so they have no access to runtime information such as variables' values, but they do know about and rely upon C++ data types.
Alternatively, perhaps you're looking for the Strategy pattern.
The first thing you need to know is that the Macros are preprocessor directives which are a fragment of code with a given name. if you use macro name in your program it will replace that code fragment into that place you use the macro name at compile time first stage called Pre-processing stage.
#include <stdio.h>
#define FOO(val) { \
print ## val(); \
}
void printIT() { printf("IT SUCCESS\n"); }
void printET() { printf("ET SUCCESS\n"); }
int main() {
const char* ds = "IT", es = "ET";
FOO(ds); FOO(es);
return 0;
}
In your code, you try to input ds and es variables into FOO function like macro. But ds and es variables declared in program stack only when you run the program. At the compile-time, it just treats them as only texts. Therefore macro function input it takes as text ds and es and replaced with val. That's why you got the compile time error. The following code fragment I have changed is working as you expected.
#include <stdio.h>
#define FOO(val) { \
print ## val(); \
}
void printIT() { printf("IT SUCCESS\n"); }
void printET() { printf("ET SUCCESS\n"); }
int main() {
const char* ds = "IT", *es = "ET";
FOO(IT); FOO(ET);
return 0;
}
If you are interested you can find more about Macros in the following resources.
GCC online documentation and Article about Macros . Also, you can view the preprocessed code using g++ -E (your cpp file name). Thanks.
Possible solution without MACRO:
void printIT() { printf("IT SUCCESS\n"); }
void printET() { printf("ET SUCCESS\n"); }
void foo(std::string_view s)
{
static const std::map<std::string_view, void(*)()> funcs{
{"IT", &printIT},
{"ET", &printET}
};
auto it = funcs.find(s);
if (it != funcs.end()) {
(*it->second)();
}
}
int main() {
const char* ds = "IT";
const char* es = "ET";
foo(ds); foo(es);
}
Demo
I want to define a macro like
#define AUTO_FUC(x) void x(){}
to define a function named x.
I think the function name will be replaced by x passed by arg x.
AUTO_FUC("test")//it should equal to void test(){} I think
But it didn't.
So how to do this in C++?
I use MinGW and use the macro twice with different args.
AUTO_FUC("A")
AUTO_FUC("B")
It didn't give me function A() and B().Compiler told me void someClass::x() cannot be overloaded.So I know the macro didn't work.
Thanks ,I realize the void "A"() is not a correct function. I forgot I pass a macro as arg and the marco is a string.
And further question. What I really want is create a function with name according to a variable value.And variable value can be write like string a = a;? It should have " am I right? So If I use a string like "A", how can I use the A to create the function void A() when pass the value to marco?
#include <string>
#include <iostream>
#define AUTO_FUC(x) \
void x() \
{\
std::cout<<"hi";\
}\
int main()
{
std::string A = "A";
AUTO_FUC(A); //it's equal to AUTO_FUC("A") it wrong
A(); //So it's undefined;
}
The problem is that you're not passing in A and B (identifiers). You're passing in "A" and "B" (string literals). Macros are textual replacement. So they expand to:
void "A"(){}
void "B"(){}
Surely you'll agree that's not correct C++ code. Change it to this:
AUTO_FUC(A)
AUTO_FUC(B)
[ UPDATE: you've edited your question and now show code where the macro parameters are quoted. Next time, try to post your actual code when asking the question! ;-P ]
Your code works just fine... see it running at ideone.com here
#include <iostream>
#define AUTO_FUC(x) void x(){ std::cout << "hi!\n"; }
AUTO_FUC(test)
int main()
{
test();
}
stdout
hi!
There are a few legal ways which can we declare a function in C++.
Some of the legal ways are:
void function ();
void function (void);
dataType function (dataType);
and so on...
Recently, I came across a function declaration as such:
void (function) (); //Take note of the braces around the function name
I have never seen somehting like this before, when I tested it in a C++ compiler, it runs without any warning or compilation errors.
My question is: Why is void (function) (); a legal way to decalre a function prototype? Is there any special meaning to declare a function in this way? Or does it just work normally like any other function declaration?
One difference is that enclosing it in parenthesis prevents the function-like macros expansion by the preprocessor. As mentioned in the other answers it makes no difference though to the actual compiler.
For instance:
// somewhere buried deep in a header
#define function(a, b) a + b
// your code
void function() { // this expands the macro and gives compilation error
}
void (function)() { // this does not expand and works as expected
}
This comes in handy for instance when the bright minds behind the Microsoft Visual Studio library decided to provide function-like macros for things like min and max. (There are other ways like #undef to go around this).
Note that object-like macros (e.g. #define function 3 + 4) are still expanded.
The preprocessor is just a dumb text replacement tool (as opposed to the compiler which is just a (smart) text replacement tool). It takes the macro definition and replaces it everywhere. He is not aware of the semantics of what he replaces.
For instance:
// somewhere buried deep in a header
#define function 3 + 2
// your code
void function() {
}
The preprocessor sees the word function and textually replaces it with the string 3 + 2. He is unaware that function is a id-name part of a function declaration and definition. After the preprocess phase there come the actual compile phases. So the compiler actually sees:
// your code
void 3 + 2() {
}
which does not make any sense to him and gives an error.
For function-like macros
// somewhere buried deep in a header
#define function(a, b) a + b
The preprocessor does the same except that it expects two ‘tokens’ enclosed in parenthesis separated by comma (the parameters) and does the replacement. (again no semantics aware):
int d = function(2, 3);
//will be replaced by the preprocessor to:
int d = 2 + 3; // passes compilation phase
void function();
// the preprocessor doesn’t find the arguments for function so it gives an error.
However if it encounters (function) it will not try to expand it (it ignores it). It is just a rule.
it's the same as
void function();
you can declare it as
void ((function)) ();
if you want :)
be careful not to mix this up with the function pointer declaration syntax.
There is nothing special about it, it means exactly the same as the version without parentheses. It is just an artifact of how the syntax is declared. Usually you see the use of parentheses around the function name when a function pointer is declared, e.g.
void (*function_pointer)() = nullptr;
// a function pointer to a function taking and returning void
in contrast to
void *function();
// a function declaration of a function taking void and returning void*
I think it works the same as a normal function because function pointers are declared like: void (*function)() so if you leave out the * then it should be just a function.
It corresponds to the C++ grammar. If to simplify then one of the rules for defining of the declarator looks as
declarator:
(declarator)
So you can write for example
void (function) ();
or
void ( (function) () );
or even the following way
struct A
{
void ( ( function )() const );
};
I think you may find that was:
void (*function) ();
since there is no benefit to using void (function)(); or void (((((function)))))(); for that matter, they're equivalent. If I'm mistaken and it's not a typo, the answer is that you can put as many parentheses around the function name as you like, subject to compiler limitations, as per the code for output6() below.
If I'm not mistaken, that one with the * actually declares a function pointer which can be used to hold a pointer to a function. It does not declare a function at all, just a pointer that can be used to reference a function.
Like an int pointer (for example), the function pointer can point to an arbitrary function, parameters notwhithstanding.
So for example:
#include <iostream>
void (((((output6)))))() { std::cout << 6; }
void output7() { std::cout << 7; }
void output8() { std::cout << 8; }
void (*fn)();
int main() {
fn = &output6; fn();
fn = &output7; fn();
fn = &output8; fn();
std::cout << '\n';
}
would output 678.
I'm working on a project that delivers statistics to the user. I created a class called Dog,
And it has several functions. Speak, woof, run, fetch, etc.
I want to have a function that spits out how many times each function has been called. I'm also interested in the constructor calls and destructor calls as well.
I have a header file which defines all the functions, then a separate .cc file that implements them. My question is, is there a way to keep track of how many times each function is called?
I have a function called print that will fetch the "statistics" and then output them to standard output. I was considering using static integers as part of the class itself, declaring several integers to keep track of those things. I know the compiler will create a copy of the integer and initialize it to a minimum value, and then I'll increment the integers in the .cc functions.
I also thought about having static integers as a global variable in the .cc. Which way is easier? Or is there a better way to do this?
Any help is greatly appreciated!
Using static member variables is the way to go. However, the compiler will not "create a copy of the integer and initialize it to a minimum value"; you'll have to provide a definition for each one in the .cc file and initialize it to 0 there. (Things are a bit different if you're using C++11, but the basic idea is the same.)
There's no reason to use static global variables instead of static members.
foo.h:
class Foo {
static int countCtor_;
static int countDtor_;
static int countprint_:
Foo();
~Foo();
static void print();
};
foo.cc:
#include <iostream>
#include "foo.h"
int Foo::countCtor_ = 0;
int Foo::countDtor_ = 0;
int Foo::countprint_ = 0;
Foo::Foo() {
++countCtor_;
// Something here
}
Foo::~Foo() {
++countDtor_;
// Something here
}
void Foo::print() {
++countprint_;
std::cout << "Ctor: " << countCtor_ << "\n"
<< "Dtor: " << countDtor_ << "\n"
<< "print: " << countprint_ << "\n";
}
But if you've got a lot of functions, the repetition involved is a bit annoying—it's very easy to accidentally do ++countBar_ when you meant ++countBaz_ (especially if you copy and paste the boilerplate), so you may want something a bit fancier, such as a static map and a macro that increments counts[__FUNC__], so you can just use the exact same line in each function. Like this:
foo.h:
#include <map>
class Foo {
static std::map<const char*, int> counts_;
Foo();
~Foo();
void print();
};
foo.cc:
#include <iostream>
#include "foo.h"
std::map<const char *, int> Foo::counts_;
#define INC_COUNT_() do { ++counts_[__FUNC__]; } while (0)
Foo::Foo() {
INC_COUNT_();
// Something here
}
Foo::~Foo() {
INC_COUNT_();
// Something here
}
void Foo::print() {
INC_COUNT_();
for (std::map<const char *, int>::const_iterator it = counts_.begin();
it != counts_.end(); ++it) {
std::cout << it->first << ": " << it->second << "\n";
}
}
In the example code above, __FUNC__ is a placeholder. Unfortunately, there is no standard-compliant value you can use in its place. Most compilers have some subset of __func__, __FUNC__, __FUNCTION__, __FUNCSIG__, and __PRETTY_FUNCTION__. However, none of those are standard in C++03. C++11 does standardize __func__, but only as an "implementation-defined string", which isn't guaranteed to be useful, or even unique. On top of that, the values will be different on different compilers. Also, some of them may be macros rather than identifiers, to make things more fun.
If you want truly portable code, in C++11, you can use something like string(__func__) + ":" + STRINGIZE(__LINE__)—this will be somewhat ugly, but at least each function will have a unique name. And in C++03, there is no equivalent. If you just need "portable enough", consult the documentation for every compiler you use, or rely on something like autoconf.
Is there any reason you can't use standard profiling tools that will count these calls for you? Something like gprof?
Otherwise static integers would be the way to go.
Assuming you want these statistics tracked all the time in your program, you could use an unordered_map of your function names:
std::unordered_map<const char *, unsigned> stats;
void foo () {
// use __FUNCDNAME__ for MSVC
++stats[__PRETTY_FUNCTION__];
//...
}
The use of compiler specific function name specifiers is purposefully there to get the decorated function names. This is so that overloaded function names get counted as separate functions.
This technique allows you to add new functions easily without thinking about anything else, but there is a small additional cost if there are hash collisions (which can be remedied somewhat by sizing the stats map to be larger). There is no hash computed on the string, since the key is a pointer type, it just uses the pointer value itself as the hash.
If this is just one-off code for profiling, then you should first try to use the code profiling tools available on your platform.
You can put static locals inside the methods themselves, that seems cleaner since these variables aren't logically connected to the class so there's no reason to make them members.
Additionaly, you could have a macro to simplify the work. I normally don't recommend using macros, but this seems like an appropriate use:
#define DEFINE_COUNTER \
static int noCalls = 0; \
noCalls++;
void foo()
{
DEFINE_COUNTER
}
Use a library that implements the Observer Pattern or Method Call Interception. You can choose one from this list, or use something like Vitamin.
Certain situations in my code, I end up invoking the function only if that function is defined, or else I should not. How can I achieve this?
like:
if (function 'sum' exists ) then invoke sum ()
Maybe the other way around to ask this question is how to determine if function is defined at runtime and if so, then invoke?
When you declare 'sum' you could declare it like:
#define SUM_EXISTS
int sum(std::vector<int>& addMeUp) {
...
}
Then when you come to use it you could go:
#ifdef SUM_EXISTS
int result = sum(x);
...
#endif
I'm guessing you're coming from a scripting language where things are all done at runtime. The main thing to remember with C++ is the two phases:
Compile time
Preprocessor runs
template code is turned into real source code
source code is turned in machine code
runtime
the machine code is run
So all the #define and things like that happen at compile time.
....
If you really wanted to do it all at runtime .. you might be interested in using some of the component architecture products out there.
Or maybe a plugin kind of architecture is what you're after.
Using GCC you can:
void func(int argc, char *argv[]) __attribute__((weak)); // weak declaration must always be present
// optional definition:
/*void func(int argc, char *argv[]) {
printf("FOUND THE FUNCTION\n");
for(int aa = 0; aa < argc; aa++){
printf("arg %d = %s \n", aa, argv[aa]);
}
}*/
int main(int argc, char *argv[]) {
if (func){
func(argc, argv);
} else {
printf("did not find the function\n");
}
}
If you uncomment func it will run it otherwise it will print "did not find the function\n".
While other replies are helpful advices (dlsym, function pointers, ...), you cannot compile C++ code referring to a function which does not exist. At minimum, the function has to be declared; if it is not, your code won't compile. If nothing (a compilation unit, some object file, some library) defines the function, the linker would complain (unless it is weak, see below).
But you should really explain why you are asking that. I can't guess, and there is some way to achieve your unstated goal.
Notice that dlsym often requires functions without name mangling, i.e. declared as extern "C".
If coding on Linux with GCC, you might also use the weak function attribute in declarations. The linker would then set undefined weak symbols to null.
addenda
If you are getting the function name from some input, you should be aware that only a subset of functions should be callable that way (if you call an arbitrary function without care, it will crash!) and you'll better explicitly construct that subset. You could then use a std::map, or dlsym (with each function in the subset declared extern "C"). Notice that dlopen with a NULL path gives a handle to the main program, which you should link with -rdynamic to have it work correctly.
You really want to call by their name only a suitably defined subset of functions. For instance, you probably don't want to call this way abort, exit, or fork.
NB. If you know dynamically the signature of the called function, you might want to use libffi to call it.
I suspect that the poster was actually looking for something more along the lines of SFINAE checking/dispatch. With C++ templates, can define to template functions, one which calls the desired function (if it exists) and one that does nothing (if the function does not exist). You can then make the first template depend on the desired function, such that the template is ill-formed when the function does not exist. This is valid because in C++ template substitution failure is not an error (SFINAE), so the compiler will just fall back to the second case (which for instance could do nothing).
See here for an excellent example: Is it possible to write a template to check for a function's existence?
use pointers to functions.
//initialize
typedef void (*PF)();
std::map<std::string, PF> defined_functions;
defined_functions["foo"]=&foo;
defined_functions["bar"]=&bar;
//if defined, invoke it
if(defined_functions.find("foo") != defined_functions.end())
{
defined_functions["foo"]();
}
If you know what library the function you'd like to call is in, then you can use dlsym() and dlerror() to find out whether or not it's there, and what the pointer to the function is.
Edit: I probably wouldn't actually use this approach - instead I would recommend Matiu's solution, as I think it's much better practice. However, dlsym() isn't very well known, so I thought I'd point it out.
You can use #pragma weak for the compilers that support it (see the weak symbol wikipedia entry).
This example and comment is from The Inside Story on Shared Libraries and Dynamic Loading:
#pragma weak debug
extern void debug(void);
void (*debugfunc)(void) = debug;
int main() {
printf(“Hello World\n”);
if (debugfunc) (*debugfunc)();
}
you can use the weak pragma to force the linker to ignore unresolved
symbols [..] the program compiles and links whether or not debug()
is actually defined in any object file. When the symbol remains
undefined, the linker usually replaces its value with 0. So, this
technique can be a useful way for a program to invoke optional code
that does not require recompiling the entire application.
So another way, if you're using c++11 would be to use functors:
You'll need to put this at the start of your file:
#include <functional>
The type of a functor is declared in this format:
std::function< return_type (param1_type, param2_type) >
You could add a variable that holds a functor for sum like this:
std::function<int(const std::vector<int>&)> sum;
To make things easy, let shorten the param type:
using Numbers = const std::vectorn<int>&;
Then you could fill in the functor var with any one of:
A lambda:
sum = [](Numbers x) { return std::accumulate(x.cbegin(), x.cend(), 0); } // std::accumulate comes from #include <numeric>
A function pointer:
int myFunc(Numbers nums) {
int result = 0;
for (int i : nums)
result += i;
return result;
}
sum = &myFunc;
Something that 'bind' has created:
struct Adder {
int startNumber = 6;
int doAdding(Numbers nums) {
int result = 0;
for (int i : nums)
result += i;
return result;
}
};
...
Adder myAdder{2}; // Make an adder that starts at two
sum = std::bind(&Adder::doAdding, myAdder);
Then finally to use it, it's a simple if statement:
if (sum)
return sum(x);
In summary, functors are the new pointer to a function, however they're more versatile. May actually be inlined if the compiler is sure enough, but generally are the same as a function pointer.
When combined with std::bind and lambda's they're quite superior to old style C function pointers.
But remember they work in c++11 and above environments. (Not in C or C++03).
In C++, a modified version of the trick for checking if a member exists should give you what you're looking for, at compile time instead of runtime:
#include <iostream>
#include <type_traits>
namespace
{
template <class T, template <class...> class Test>
struct exists
{
template<class U>
static std::true_type check(Test<U>*);
template<class U>
static std::false_type check(...);
static constexpr bool value = decltype(check<T>(0))::value;
};
template<class U, class = decltype(sum(std::declval<U>(), std::declval<U>()))>
struct sum_test{};
template <class T>
void validate_sum()
{
if constexpr (exists<T, sum_test>::value)
{
std::cout << "sum exists for type " << typeid(T).name() << '\n';
}
else
{
std::cout << "sum does not exist for type " << typeid(T).name() << '\n';
}
}
class A {};
class B {};
void sum(const A& l, const A& r); // we only need to declare the function, not define it
}
int main(int, const char**)
{
validate_sum<A>();
validate_sum<B>();
}
Here's the output using clang:
sum exists for type N12_GLOBAL__N_11AE
sum does not exist for type N12_GLOBAL__N_11BE
I should point out that weird things happened when I used an int instead of A (sum() has to be declared before sum_test for the exists to work, so maybe exists isn't the right name for this). Some kind of template expansion that didn't seem to cause problems when I used A. Gonna guess it's ADL-related.
This answer is for global functions, as a complement to the other answers on testing methods. This answer only applies to global functions.
First, provide a fallback dummy function in a separate namespace. Then determine the return type of the function-call, inside a template parameter. According to the return-type, determine if this is the fallback function or the wanted function.
If you are forbidden to add anything in the namespace of the function, such as the case for std::, then you should use ADL to find the right function in the test.
For example, std::reduce() is part of c++17, but early gcc compilers, which should support c++17, don't define std::reduce(). The following code can detect at compile-time whether or not std::reduce is declared. See it work correctly in both cases, in compile explorer.
#include <numeric>
namespace fallback
{
// fallback
std::false_type reduce(...) { return {}; }
// Depending on
// std::recuce(Iter from, Iter to) -> decltype(*from)
// we know that a call to std::reduce(T*, T*) returns T
template <typename T, typename Ret = decltype(reduce(std::declval<T*>(), std::declval<T*>()))>
using return_of_reduce = Ret;
// Note that due to ADL, std::reduce is called although we don't explicitly call std::reduce().
// This is critical, since we are not allowed to define any of the above inside std::
}
using has_reduce = fallback::return_of_reduce<std::true_type>;
// using has_sum = std::conditional_t<std::is_same_v<fallback::return_of_sum<std::true_type>,
// std::false_type>,
// std::false_type,
// std::true_type>;
#include <iterator>
int main()
{
if constexpr (has_reduce::value)
{
// must have those, so that the compile will find the fallback
// function if the correct one is undefined (even if it never
// generates this code).
using namespace std;
using namespace fallback;
int values[] = {1,2,3};
return reduce(std::begin(values), std::end(values));
}
return -1;
}
In cases, unlike the above example, when you can't control the return-type, you can use other methods, such as std::is_same and std::contitional.
For example, assume you want to test if function int sum(int, int) is declared in the current compilation unit. Create, in a similar fashion, test_sum_ns::return_of_sum. If the function exists, it will be int and std::false_type otherwise (or any other special type you like).
using has_sum = std::conditional_t<std::is_same_v<test_sum_ns::return_of_sum,
std::false_type>,
std::false_type,
std::true_type>;
Then you can use that type:
if constexpr (has_sum::value)
{
int result;
{
using namespace fallback; // limit this only to the call, if possible.
result = sum(1,2);
}
std::cout << "sum(1,2) = " << result << '\n';
}
NOTE: You must have to have using namespace, otherwise the compiler will not find the fallback function inside the if constexpr and will complain. In general, you should avoid using namespace since future changes in the symbols inside the namespace may break your code. In this case there is no other way around it, so at least limit it to the smallest scope possible, as in the above example