Prohibit function call within other function at compile time - c++

I want to prohibit that any invocable/function is called within a specific other function:
int foo()
{
return 1;
}
template <typename... Args>
void specific(Args... t){}
specific(1); // OK
specific(foo()); // This should raise a compile error
specific should only get non-invocables, is this possible?
I don't see how to use std::is_invocable for this.
A solution with a macro would also be fine, which C++14 at disposal.
EDIT: Background is that we want to prohibit function calls within logging statements(as macros):
auto a = something();
LOG_MSG("Log something {}", a); //OK
LOG_MSG("Log something {}", something()); // should not be possible

Related

Cleanly handling void as generic return type of labmda in C++

I am trying to write a function like this:
template <typename T>
void testActionBehavesIdentically(Foo& fooA, Foo& fooB, std::function<T(Foo&)> action)
if (std::is_same<T, void>::value)
{
// perform action even if T is void
action(fooA);
action(fooB);
}
else
{
// if T is not void, test also return values
T resultA = action(fooA);
T resultB = action(fooB);
CHECK_EQUAL(resultA, resultB);
}
// tests that the state of both foos is the same
CHECK_EQUAL(fooA, fooB);
}
where T is sometimes void. The compilation (on VS2019) fails with
error C2182: 'resultA': illegal use of type 'void'. Is there a clean way around it? (Hopefully one that will compile in most standard compilers)? Thank you.
Right now what's happening is that the compiler still will compile both if branches (since it doesn't actually know which will be called until runtime). This results in a failure since one of the branches doesn't compile correctly. There are a couple fixes (from the comments):
If your compiler supports it this would be one option:
if constexpr (std::is_same<T, void>::value) {
// perform action even if T is void
action(fooA);
action(fooB);
} else {
...
}
this will only actually compile one branch depending on the type of T.
If your compiler doesn't support it, here's another option:
template <typename T>
void testActionBehavesIdentically(Foo& fooA, Foo& fooB, std::function<T(Foo&)> action) {
// check assuming T != void
}
// overload which will be called if T was void
void testActionBehavesIdentically(Foo& fooA, Foo& fooB, std::function<void(Foo&)> action) {
// check assuming T == void
}
The compiler will match the overload when T is void instead of dispatching to the generic version of the function. Here you could even say "if T is an int do something different" which is kind of neat (but can get messy).

is there a way to store a generic templated function pointer?

The Goal:
decide during runtime which templated function to use and then use it later without needing the type information.
A Partial Solution:
for functions where the parameter itself is not templated we can do:
int (*func_ptr)(void*) = &my_templated_func<type_a,type_b>;
this line of code can be modified for use in an if statement with different types for type_a and type_b thus giving us a templated function whose types are determined during runtime:
int (*func_ptr)(void*) = NULL;
if (/* case 1*/)
func_ptr = &my_templated_func<int, float>;
else
func_ptr = &my_templated_func<float, float>;
The Remaining Problem:
How do I do this when the parameter is a templated pointer?
for example, this is something along the lines of what I would like to do:
int (*func_ptr)(templated_struct<type_a,type_b>*); // This won't work cause I don't know type_a or type_b yet
if (/* case 1 */) {
func_ptr = &my_templated_func<int,float>;
arg = calloc(sizeof(templated_struct<int,float>, 1);
}
else {
func_ptr = &my_templated_func<float,float>;
arg = calloc(sizeof(templated_struct<float,float>, 1);
}
func_ptr(arg);
except I would like type_a, and type_b to be determined during runtime. I see to parts to the problem.
What is the function pointers type?
How do I call this function?
I think I have the answer for (2): simply cast the parameter to void* and the template function should do an implicit cast using the function definition (lease correct me if this won't work as I think it will).
(1) is where I am getting stuck since the function pointer must include the parameter types. This is different from the partial solution because for the function pointer definition we were able to "ignore" the template aspect of the function since all we really need is the address of the function.
Alternatively there might be a much better way to accomplish my goal and if so I am all ears.
Thanks to the answer by #Jeffrey I was able to come up with this short example of what I am trying to accomplish:
template <typename A, typename B>
struct args_st {
A argA;
B argB;
}
template<typename A, typename B>
void f(struct args_st<A,B> *args) {}
template<typename A, typename B>
void g(struct args_st<A,B> *args) {}
int someFunction() {
void *args;
// someType needs to know that an args_st struct is going to be passed
// in but doesn't need to know the type of A or B those are compiled
// into the function and with this code, A and B are guaranteed to match
// between the function and argument.
someType func_ptr;
if (/* some runtime condition */) {
args = calloc(sizeof(struct args_st<int,float>), 1);
f((struct args_st<int,float> *) args); // this works
func_ptr = &g<int,float>; // func_ptr should know that it takes an argument of struct args_st<int,float>
}
else {
args = calloc(sizeof(struct args_st<float,float>), 1);
f((struct args_st<float,float> *) args); // this also works
func_ptr = &g<float,float>; // func_ptr should know that it takes an argument of struct args_st<float,float>
}
/* other code that does stuff with args */
// note that I could do another if statement here to decide which
// version of g to use (like I did for f) I am just trying to figure out
// a way to avoid that because the if statement could have a lot of
// different cases similarly I would like to be able to just write one
// line of code that calls f because that could eliminate many lines of
// (sort of) duplicate code
func_ptr(args);
return 0; // Arbitrary value
}
Can't you use a std::function, and use lambdas to capture everything you need? It doesn't appear that your functions take parameters, so this would work.
ie
std::function<void()> callIt;
if(/*case 1*/)
{
callIt = [](){ myTemplatedFunction<int, int>(); }
}
else
{
callIt = []() {myTemplatedFunction<float, float>(); }
}
callIt();
If I understand correctly, What you want to do boils down to:
template<typename T>
void f(T)
{
}
int somewhere()
{
someType func_ptr;
int arg = 0;
if (/* something known at runtime */)
{
func_ptr = &f<float>;
}
else
{
func_ptr = &f<int>;
}
func_ptr(arg);
}
You cannot do that in C++. C++ is statically typed, the template types are all resolved at compile time. If a construct allowed you to do this, the compiler could not know which templates must be instanciated with which types.
The alternatives are:
inheritance for runtime polymorphism
C-style void* everywhere if you want to deal yourself with the underlying types
Edit:
Reading the edited question:
func_ptr should know that it takes an argument of struct args_st<float,float>
func_ptr should know that it takes an argument of struct args_st<int,float>
Those are incompatible. The way this is done in C++ is by typing func_ptr accordingly to the types it takes. It cannot be both/all/any.
If there existed a type for func_ptr so that it could take arguments of arbitrary types, then you could pass it around between functions and compilation units and your language would suddenly not be statically typed. You'd end up with Python ;-p
Maybe you want something like this:
#include <iostream>
template <typename T>
void foo(const T& t) {
std::cout << "foo";
}
template <typename T>
void bar(const T& t) {
std::cout << "bar";
}
template <typename T>
using f_ptr = void (*)(const T&);
int main() {
f_ptr<int> a = &bar<int>;
f_ptr<double> b = &foo<double>;
a(1);
b(4.2);
}
Functions taking different parameters are of different type, hence you cannot have a f_ptr<int> point to bar<double>. Otherwise, functions you get from instantiating a function template can be stored in function pointers just like other functions, eg you can have a f_ptr<int> holding either &foo<int> or &bar<int>.
Disclaimer: I have already provided an answer that directly addresses the question. In this answer, I would like to side-step the question and render it moot.
As a rule of thumb, the following code structure is an inferior design in most procedural languages (not just C++).
if ( conditionA ) {
// Do task 1A
}
else {
// Do task 1B
}
// Do common tasks
if ( conditionA ) {
// Do task 2A
}
else {
// Do task 2B
}
You seem to have recognized the drawbacks in this design, as you are trying to eliminate the need for a second if-else in someFunction(). However, your solution is not as clean as it could be.
It is usually better (for code readability and maintainability) to move the common tasks to a separate function, rather than trying to do everything in one function. This gives a code structure more like the following, where the common tasks have been moved to the function foo().
if ( conditionA ) {
// Do task 1A
foo( /* arguments might be needed */ );
// Do task 2A
}
else {
// Do task 1B
foo( /* arguments might be needed */ );
// Do task 2B
}
As a demonstration of the utility of this rule of thumb, let's apply it to someFunction(). ... and eliminate the need for dynamic memory allocation ... and a bit of cleanup ... unfortunately, addressing that nasty void* is out-of-scope ... I'll leave it up to the reader to evaluate the end result. The one feature I will point out is that there is no longer a reason to consider storing a "generic templated function pointer", rendering the asked question moot.
// Ideally, the parameter's type would not be `void*`.
// I leave that for a future refinement.
void foo(void * args) {
/* other code that does stuff with args */
}
int someFunction(bool condition) {
if (/* some runtime condition */) {
args_st<int,float> args;
foo(&args);
f(&args); // Next step: pass by reference instead of passing a pointer
}
else {
args_st<float,float> args;
foo(&args);
f(&args); // Next step: pass by reference instead of passing a pointer
}
return 0;
}
Your choice of manual memory management and over-use of the keyword struct suggests you come from a C background and have not yet really converted to C++ programming. As a result, there are many areas for improvement, and you might find that your current approach should be tossed. However, that is a future step. There is a learning process involved, and incremental improvements to your current code is one way to get there.
First, I'd like to get rid of the C-style memory management. Most of the time, using calloc in C++ code is wrong. Let's replace the raw pointer with a smart pointer. A shared_ptr looks like it will help the process along.
// Instead of a raw pointer to void, use a smart pointer to void.
std::shared_ptr<void> args;
// Use C++ memory management, not calloc.
args = std::make_shared<args_st<int,float>>();
// or
args = std::make_shared<args_st<float,float>>();
This is still not great, as it still uses a pointer to void, which is rarely needed in C++ code unless interfacing with a library written in C. It is, though, an improvement. One side effect of using a pointer to void is the need for casts to get back to the original type. This should be avoided. I can address this in your code by defining correctly-typed variables inside the if statement. The args variable will still be used to hold your pointer once the correctly-typed variables go out of scope.
More improvements along this vein can come later.
The key improvement I would make is to use the functional std::function instead of a function pointer. A std::function is a generalization of a function pointer, able to do more albeit with more overhead. The overhead is warranted here in the interest of robust code.
An advantage of std::function is that the parameter to g() does not need to be known by the code that invokes the std::function. The old style of doing this was std::bind, but lambdas provide a more readable approach. Not only do you not have to worry about the type of args when it comes time to call your function, you don't even need to worry about args.
int someFunction() {
// Use a smart pointer so you do not have to worry about releasing the memory.
std::shared_ptr<void> args;
// Use a functional as a more convenient alternative to a function pointer.
// Note the lack of parameters (nothing inside the parentheses).
std::function<void()> func;
if ( /* some runtime condition */ ) {
// Start with a pointer to something other than void.
auto real_args = std::make_shared<args_st<int,float>>();
// An immediate function call:
f(real_args.get());
// Choosing a function to be called later:
// Note that this captures a pointer to the data, not a copy of the data.
// Hence changes to the data will be reflected when this is invoked.
func = [real_args]() { g(real_args.get()); };
// It's only here, as real_args is about to go out of scope, where
// we lose the type information.
args = real_args;
}
else {
// Similar to the above, so I'll reduce the commentary.
auto real_args = std::make_shared<args_st<float,float>>();
func = [real_args]() { g(real_args.get()); };
args = real_args;
}
/* other code that does stuff with args */
/* This code is probably poor C++ style, but that can be addressed later. */
// Invoke the function.
func();
return 0;
}
Your next step probably should be to do some reading on these features so you understand what this code does. Then you should be in a better position to leverage the power of C++.

Overloaded template function doesn't select the correct version for a specific type

Situation is: I have a generic function but would like an altered version for a specific type.
I would instinctively write a widget const& w parameter to be able to accept any widget by reference, but that doesn't compile in that case, because the compiler ignored the specific overload (2) and used generic (1) instead. It does compile if I remove the const in the params of (2), why is that?
godbolt
struct widget {
int widget_size;
};
// (1)
// compiler uses this instead of (2)
template <typename Arg>
int get_size(Arg && arg) {
return arg.size;
}
// (2)
int get_size(widget const& arg) {
return arg.widget_size;
}
int main() {
widget w;
get_size(w);
}
because the compiler ignored the specific overload .
Please note that Version 2 is a better match as argument passed lacks const keyword and version 1 also lacks const keyword.
that doesn't compile in that case.
Widget class doesn't have size member variable in it. It's got widget_size , which is causing compilation error.
int get_size(Arg && arg) {
return arg.widget_size; // See this
}
It does compile if I remove the const in the params of (2), why is that?
This is because get_size(w); started matching version 2, thereby omitting any need for compiler to check for widget_size in version 1.

How to compile-time detect functions that overloaded important template?

Suppose we have a template:
template <class T>
void VeryImportantFunction(T t) {
// something
}
Somewhere it is called with something like:
// ..
int a = 12345;
VeryImportantFunction(a);
// ..
It is very big project with tons of source code, and occasionally somewhere in deep of the code appears a new header with overloaded function:
void VeryImportantFunction(int t) {
// totally another behavior
}
And code fragment above will call overloaded function, because it have more priority.
Can we somehow disable or in another way compile-time detect functions that can overload our important template?
Your question is unclear, but here's my take on it.
If you want to hit the template overload, you can simply invoke the function by explicitly specifying the template parameters:
int a = 12345;
VeryImportantFunction<int>(a);
If you want this from happening again in the future, then make VeryImportantFunction either a lambda or a struct - those cannot be overloaded "externally":
inline const auto VeryImportantFunction = [](auto x){ /* ... */ };
// no one can overload this!
If you want to know all the overloads of VeryImportantFunction without external tooling, then call it in a completely wrong way - the compiler error will likely show all considered overloads:
VeryImportantFunction(5, 5, 5, 5);
// error... will likely show all candidates
Write
inline void VeryImportantFunction(int t)
{
VeryImportantFunction<int>(t); // call the template function
}
immediately after your template definition.
Then if someone has written their own version of void VeryImportantFunction(int t), you'll get a compiler error.

Is it possible to ignore [[nodiscard]] in a special case?

C++17 has a new attribute, [[nodiscard]].
Suppose, that I have a Result struct, which has this attribute:
struct [[nodiscard]] Result {
};
Now, if I call a function which returns Result, I got a warning if I don't check the returned Result:
Result someFunction();
int main() {
someFunction(); // warning here, as I don't check someFunction's return value
}
This program generates:
warning: ignoring return value of function declared with 'nodiscard'
attribute [-Wunused-result]
So far, so good. Now suppose, that I have a special function, for which I still want to return Result, but I don't want this warning generated, if the check is omitted:
Result someNonCriticalFunction();
int main() {
someNonCriticalFunction(); // I don't want to generate a warning here
}
It is because, someNonCriticalFunction() does something non-critical (for example, something like printf - I bet that no-one checks printf's return value all the time); most cases, I don't care if it fails. But I still want it to return Result, as in some rare cases, I do need its Result.
Is it possible to do this somehow?
Possible solutions which I don't like:
I would not like calling it as (void)someNonCriticalFunction(), because this function is called a lot of times, it is awkward
creating a wrapper around someNonCriticalFunction(), which calls (void)someNonCriticalFunction(): I don't want to have a differently named function just because of this
removing [[nodiscard]] from Result, and add it to every function which returns Result
Why not make use of std::ignore from the <tuple> header—that would make the discard explicit:
[[nodiscard]] int MyFunction() { return 42; }
int main()
{
std::ignore = MyFunction();
return 0;
}
Compiler explorer of this code snippet: https://godbolt.org/z/eGPsjajz8
CPP Reference for std::ignore: https://en.cppreference.com/w/cpp/utility/tuple/ignore
I recommend the option you ruled out:
"removing [[nodiscard]] from Result, and add it to every function which returns Result."
But since you don't seem happy with it, here's another solution, using bog-standard inheritance:
struct [[nodiscard]] Result {
};
struct DiscardableResult: public Result {
};
For the functions where you can discard the result, use DiscardableResult as return type:
Result func1();
DiscardableResult func2();
func1(); // will warn
func2(); // will not warn
They say that every problem in computer science can be solved by adding another layer of indirection:
template <bool nodiscard=true>
struct Result;
template <>
struct Result<false> {
// the actual implementation
};
template <>
struct [[nodiscard]] Result<true>
: Result<false>
{
using Result<false>::Result;
};
This is effectively making Result conditionally [[nodiscard]], which allows:
Result<true> someFunction();
Result<false> someNonCriticalFunction();
int main() {
someFunction(); // warning here
someNonCriticalFunction(); // no warning here
}
Although really, this is identical to:
removing [[nodiscard]] from Result, and add it to every function which returns Result
which gets my vote to begin with.
You can suppress the warning with another C++17 attribute, namely [[maybe_unused]]
[[nodiscard]] int MyFunction() { return 42; }
int main()
{
[[maybe_unused]] auto v = MyFunction();
return 0;
}
This way you also avoid the confusing dependency to std::tuple which comes with std::ignore, even CppCoreGuidelines is openly recommending to use std::ignore for ignoring [[nodiscard]] values:
Never cast to (void) to ignore a [[nodiscard]]return value. If you
deliberately want to discard such a result, first think hard about
whether that is really a good idea (there is usually a good reason the
author of the function or of the return type used [[nodiscard]] in the
first place). If you still think it's appropriate and your code
reviewer agrees, use std::ignore = to turn off the warning which is
simple, portable, and easy to grep.
Looking at C++ reference, officially std::ignore is only specified to be used in std::tie when unpacking tuples.
While the behavior of std::ignore outside of std::tie is not formally
specified, some code guides recommend using std::ignore to avoid
warnings from unused return values of [[nodiscard]] functions.
cast the result to a (void *).
int main()
{
(void *)someFunction(); //Warning will be gone.
}
This way you "used" your result as far as the compiler is concerned. Great for when you are using a library where nodiscard has been used and you really don't care to know the result.