Possible duplicates I'll explain at the bottom.
I was wondering if it is possible to do a compile time check to see if a function is called before another function.
My use case looks something like this:
auto f = foo();
if(!f.isOk())
return f.getError();
auto v = f.value();
So in this case I would want to get a compile time error if the user did not call isOk before calling value.
As far as I know and searched it does not seem possible but I wanted to ask here just to be sure I didn't miss any c++ magic.
FauxDupes:
How to check at compile time that a function may be called at compile time?
This is about knowing wether your function is a constexpr function. I want to know if one function has been called before the other has been called.
What you want is not possible directly without changing your design substantially.
What you can do is enforce calling always both by wrapping them in a single call:
??? foo(const F& f) {
return f.isOk() ? f.value() : f.getError();
}
However, this just shifts the problem to choosing the return type. You could return a std::variant or with some changes on the design a std::optional, but whatever you do it will be left to the caller to check what actually has been returned.
Don't assume the most stupid user and don't try to protect them from any possible mistake. Instead assume that they do read documentation.
Having to check whether a returned value is valid is a quite common pattern: functions that return a pointer can return a null-pointer, functions returning an iterator can return the end iterator. Such cases are well documented and a responsible caller will check if the returned value is valid.
To get further inspiration I refer you to std::optional, a quite modern addition to C++, which also heavily relies on the user to know what they are dealing with.
PS: Just as one counter-example, a user might write code like this, which makes it impossible to make the desired check at compile time with your current design:
int n;
std::cin >> n;
auto f = foo();
if(n > 10 && !f.isOk())
return f.getError();
auto v = f.value();
One strategy for this kind of thing is to leverage __attribute__((warn_unused_result)) (for GCC) or _Check_return_ (msvc).
Then, change foo() to return the error condition:
SomeObj obj;
auto result = foo(obj);
This will nudge the caller into handling the error. Of course there are obvious limitations: foo() cannot be a constructor, for example, and the caller cannot use auto for the typename.
One way to ensure order is to transform the temporary dependency into physical dependency:
Move method F::getError() and F::value() into their own structure wrapper (Error, Value).
Change bool F::isOk() to something like:
std::variant<Error, Value> F::isOk()
Then, you cannot use Error::getError or Value::value() before calling isOk, as expected:
auto f = foo();
auto isOk = f.isOk();
if (auto* e = std::get_if<Error>(&isOk)) // Or std::visit
return e->getError();
auto& value = std::get<Value>(&isOk);
auto v = value.value();
Related
I am working on a problem that requires me to return different return-types based on my function parameter values that I provide.
I want to do something like this --
In the code below, doSomething() is an already existing function (used by a lot of clients) which takes mode as a function parameter, and returns std::list<ReturnType> already.
Based on the mode value, I had to create another sub-functionality which returns a shared_future<std::list<ReturnType>>.
How can I change this code so that it can return one of the two return types based on the mode value?
Note: ReturnType is a template typename which we are using for the entire class.
Code:
std::shared_future<std::list<ReturnType> > futureValue() {
return functionReturningSharedFuture();
}
std::list<ReturnType> listValue() {
return functionReturningList();
}
std::list<ReturnType> doSomething(int mode) {
if(mode == 1){
// new functionality that I added
return futureValue(); // This (obviously) errors out as of now
}
else{
// already there previously
return listValue();
}
}
int main() {
doSomething(1);
return 0;
}
How can I change this code so that it can return one of the two return types based on the mode value?
Constraints and Issues:
This issue could've been easily solved by function overloading if we provide an extra function parameter (like a true value), but that extra argument is not useful, since we are already using mode. Also, it isn't considered a good design to add variables which have almost no use.
One of the major constraints is that there are clients who are already using this doSomething() expect a std::list<ReturnType>, and so I cannot return boost::any or std::variant or anything similar.
I tried using std::enable_if, but it wasn't working out since we are getting the mode value at runtime.
We can't use template metaprogramming since that would change the way our function is being called on the client-side. Something that we can't afford to do.
Thank you.
This cannot be done.
You can only have one function with a given signature. If you have calling code that already expects this to return a std::list<ReturnType>, that's it; you're done.
If you could guarantee that all existing calling code looks like
auto l = obj.doSomething(1);
then you could potentially change the return type to something which would look like a std::list to any calling code. But if there's any calling code that looks like
std::list<ReturnType> l = obj.doSomething(1);
then that's off the table.
You probably need to rethink your design here.
From the example main, I see doSomething(1);, so maybe at the call site the value of the parameter mode is always known at compile-time. In this case, one option is that you make doSomething a template<int mode> function. I'm thinking about something like this:
#include <iostream>
#include <list>
#include <vector>
// assuming you cannot change this (actually you have changed it in you example, ...)
std::list<int> doSomething(int mode) {
std::cout << "already existing function\n";
return std::list<int>{1,2,3};
}
// then you can put this too
template<int N>
auto doSomething();
template<>
auto doSomething<10>() {
std::cout << "new function\n";
return std::vector<int>{1,2,3};
}
int main() {
auto x = doSomething(3);
auto y = doSomething<10>();
}
Probably another option would be to use a if constexpr intead of if and an auto/decltype(auto) return type in doSomething, but I haven't tried it.
I often use -1 as the invalid value type when returning from a function, where the input yields incorrect output. For instance, writing an indexing function where the index is out of bounds, instead of throwing an exception, -1 can be returned. But when writing a function that has negative values as possible return types, this technique does not work. What is the correct way to return an invalid type value in such instances?
The technique I use mostly is to set the return type to be of type *int, and return a Pointer to NULL. But, that requires all return values to be of a pointer type, which seems like an extra overhead to the function. Is there an accepted standard for returning values in such cases?
In newer C++, I'd suggest using std::optional<>; if you don't yet have it, boost::optional<>.
One option would be to let your function take a bool& as an output parameter used to indicate if the returned value is valid.
int myFunc(bool& valid); // sets 'valid' to true if result is usable, false otherwise
Users can then do
bool valid = false;
Int result = myFunc(valid);
if (!valid) {
// Handle error
}
// Use result
Not the most pretty solution, but it does the job.
Apart from the answer I provided above, there's a very clean, continuation-passing solution (given you're non-virtual):
template<typename Success, typename Failed>
void parse( const std::string& str, Success s, Failed f )
{
auto a = start_parse(str);
if( a.problem() )
return f(); // you _might_ have an error code here
s( finish_parse(str, a) );
}
Then you might customize by:
Success:
[&i] (int i_) { i = i_; }
out(i), where out(int& output_) returns the above lambda for output_
actual code doing something useful
function to continue with
Failed:
[&i]{ i = 0; }, `[&i]{ i = nullopt; }, or any other default value
[] { throw MyFavouriteException(); }
retry logic
std::terminate()
[]{} if you don't care (or if you're 100% sure it'll succeed)
It might look a little verbose, but IMHO:
it's trivial to read
any other schematics can be mimicked, even if there's no default c'tor
easy to change as well
'you don't pay for what you don't use', can surely be optimized away
every schematic is visible and apparent from code:
for default value, caller sets it, not callee or global
std::optional<> and default value are handled the same
for exception, caller knows better what to throw
for no action, you don't have to lookup the implementation to know this
for std::terminate(), well, you know what to expect
if you 'speak' CPS, you might actually continue and save an if / catch / etc.
The only issue I see is constructor initializer lists. Any thoughts on this?
So I ran across this (IMHO) very nice idea of using a composite structure of a return value and an exception - Expected<T>. It overcomes many shortcomings of the traditional methods of error handling (exceptions, error codes).
See the Andrei Alexandrescu's talk (Systematic Error Handling in C++) and its slides.
The exceptions and error codes have basically the same usage scenarios with functions that return something and the ones that don't. Expected<T>, on the other hand, seems to be targeted only at functions that return values.
So, my questions are:
Have any of you tried Expected<T> in practice?
How would you apply this idiom to functions returning nothing (that is, void functions)?
Update:
I guess I should clarify my question. The Expected<void> specialization makes sense, but I'm more interested in how it would be used - the consistent usage idiom. The implementation itself is secondary (and easy).
For example, Alexandrescu gives this example (a bit edited):
string s = readline();
auto x = parseInt(s).get(); // throw on error
auto y = parseInt(s); // won’t throw
if (!y.valid()) {
// ...
}
This code is "clean" in a way that it just flows naturally. We need the value - we get it. However, with expected<void> one would have to capture the returned variable and perform some operation on it (like .throwIfError() or something), which is not as elegant. And obviously, .get() doesn't make sense with void.
So, what would your code look like if you had another function, say toUpper(s), which modifies the string in-place and has no return value?
Have any of you tried Expected; in practice?
It's quite natural, I used it even before I saw this talk.
How would you apply this idiom to functions returning nothing (that is, void functions)?
The form presented in the slides has some subtle implications:
The exception is bound to the value.
It's ok to handle the exception as you wish.
If the value ignored for some reasons, the exception is suppressed.
This does not hold if you have expected<void>, because since nobody is interested in the void value the exception is always ignored. I would force this as I would force reading from expected<T> in Alexandrescus class, with assertions and an explicit suppress member function. Rethrowing the exception from the destructor is not allowed for good reasons, so it has to be done with assertions.
template <typename T> struct expected;
#ifdef NDEBUG // no asserts
template <> class expected<void> {
std::exception_ptr spam;
public:
template <typename E>
expected(E const& e) : spam(std::make_exception_ptr(e)) {}
expected(expected&& o) : spam(std::move(o.spam)) {}
expected() : spam() {}
bool valid() const { return !spam; }
void get() const { if (!valid()) std::rethrow_exception(spam); }
void suppress() {}
};
#else // with asserts, check if return value is checked
// if all assertions do succeed, the other code is also correct
// note: do NOT write "assert(expected.valid());"
template <> class expected<void> {
std::exception_ptr spam;
mutable std::atomic_bool read; // threadsafe
public:
template <typename E>
expected(E const& e) : spam(std::make_exception_ptr(e)), read(false) {}
expected(expected&& o) : spam(std::move(o.spam)), read(o.read.load()) {}
expected() : spam(), read(false) {}
bool valid() const { read=true; return !spam; }
void get() const { if (!valid()) std::rethrow_exception(spam); }
void suppress() { read=true; }
~expected() { assert(read); }
};
#endif
expected<void> calculate(int i)
{
if (!i) return std::invalid_argument("i must be non-null");
return {};
}
int main()
{
calculate(0).suppress(); // suppressing must be explicit
if (!calculate(1).valid())
return 1;
calculate(5); // assert fails
}
Even though it might appear new for someone focused solely on C-ish languages, to those of us who had a taste of languages supporting sum-types, it's not.
For example, in Haskell you have:
data Maybe a = Nothing | Just a
data Either a b = Left a | Right b
Where the | reads or and the first element (Nothing, Just, Left, Right) is just a "tag". Essentially sum-types are just discriminating unions.
Here, you would have Expected<T> be something like: Either T Exception with a specialization for Expected<void> which is akin to Maybe Exception.
Like Matthieu M. said, this is something relatively new to C++, but nothing new for many functional languages.
I would like to add my 2 cents here: part of the difficulties and differences are can be found, in my opinion, in the "procedural vs. functional" approach. And I would like to use Scala (because I am familiar both with Scala and C++, and I feel it has a facility (Option) which is closer to Expected<T>) to illustrate this distinction.
In Scala you have Option[T], which is either Some(t) or None.
In particular, it is also possible to have Option[Unit], which is morally equivalent to Expected<void>.
In Scala, the usage pattern is very similar and built around 2 functions: isDefined() and get(). But it also have a "map()" function.
I like to think of "map" as the functional equivalent of "isDefined + get":
if (opt.isDefined)
opt.get.doSomething
becomes
val res = opt.map(t => t.doSomething)
"propagating" the option to the result
I think that here, in this functional style of using and composing options, lies the answer to your question:
So, what would your code look like if you had another function, say toUpper(s), which modifies the string in-place and has no return value?
Personally, I would NOT modify the string in place, or at least I will not return nothing. I see Expected<T> as a "functional" concept, that need a functional pattern to work well: toUpper(s) would need to either return a new string, or return itself after modification:
auto s = toUpper(s);
s.get(); ...
or, with a Scala-like map
val finalS = toUpper(s).map(upperS => upperS.someOtherManipulation)
if you don't want to follow a functional route, you can just use isDefined/valid and write your code in a more procedural way:
auto s = toUpper(s);
if (s.valid())
....
If you follow this route (maybe because you need to), there is a "void vs. unit" point to make: historically, void was not considered a type, but "no type" (void foo() was considered alike a Pascal procedure). Unit (as used in functional languages) is more seen as a type meaning "a computation". So returning a Option[Unit] does make more sense, being see as "a computation that optionally did something". And in Expected<void>, void assumes a similar meaning: a computation that, when it does work as intended (where there are no exceptional cases), just ends (returning nothing). At least, IMO!
So, using Expected or Option[Unit] could be seen as computations that maybe produced a result, or maybe not. Chaining them will prove it difficult:
auto c1 = doSomething(s); //do something on s, either succeed or fail
if (c1.valid()) {
auto c2 = doSomethingElse(s); //do something on s, either succeed or fail
if (c2.valid()) {
...
Not very clean.
Map in Scala makes it a little bit cleaner
doSomething(s) //do something on s, either succeed or fail
.map(_ => doSomethingElse(s) //do something on s, either succeed or fail
.map(_ => ...)
Which is better, but still far from ideal. Here, the Maybe monad clearly wins... but that's another story..
I've been pondering the same question since I've watched this video. And so far I didn't find any convincing argument for having Expected, for me it looks ridiculous and against clarity&cleanness. I have come up with the following so far:
Expected is good since it has either value or exceptions, we not forced to use try{}catch() for every function which is throwable. So use it for every throwing function which has return value
Every function that doesn't throw should be marked with noexcept. Every.
Every function that returns nothing and not marked as noexcept should be wrapped by try{}catch{}
If those statements hold then we have self-documented easy to use interfaces with only one drawback: we don't know what exceptions could be thrown without peeking into implementation details.
Expected impose some overheads to the code since if you have some exception in the guts of your class implementation(e.g. deep inside private methods) then you should catch it in your interface method and return Expected. While I think it is quite tolerable for the methods which have a notion for returning something I believe it brings mess and clutter to the methods which by design have no return value. Besides for me it is quite unnatural to return thing from something that is not supposed to return anything.
It should be handled with compiler diagnostics. Many compilers already emit warning diagnostics based on expected usages of certain standard library constructs. They should issue a warning for ignoring an expected<void>.
Okay, this is just a minor caveat. I am currently working with the lovely ArcSDK from ESRI. Now to get a value from any of their functions, you basically have to pass the variable, you want to assign the value to.
E.g.:
long output_width;
IRasterProps->get_Width(&output_width);
Its such a minor thing, but when you have to pick out around 30 different pieces of data from their miscellaneous functions, it really starts to get annoying.
So what i was wondering is it possible to somehow by the magic of STL or C++ change this into:
long output_width = IRasterProps->get_Width(<something magical>);
All of the functions return void, otherwise the off chance some of them might return a HRESULT, which i can safely ignore. Any ideas?
***EDIT****
Heres the final result i got which works :)!!!!!
A magic(P p, R (__stdcall T::*f)(A *)) {
A a;
((*p).*f)(&a);
return a;
}
I know I've already answered, but here's another way. It's better in that it's faster (no boost::function overhead) and avoids the binders (since people seem to have an aversion to them), but is worse in that it's much less general (since it only works for one-argument member functions).
template <typename P, typename T, typename A>
A magic(P p, void (T::*f)(A &)) {
A a;
((*p).*f)(a);
return a;
}
Which you'd call like this:
long output_width = magic(raster_props_object, &IRasterProps::get_Width);
Or, if you happen to be using GCC, we can use some more tricks:
#define MORE_MAGIC(p,f) ({ \
typedef __typeof(*(p)) big_ugly_identifier; \
magic((p),(&big_ugly_identifier::f)); \
})
Which will let us do this:
long output_width = MORE_MAGIC(raster_props_object, get_Width);
(Bonus points if the naming conventions made you think of a PDP-10.)
EDIT: Updated to take any pointer-like type, so it will now work with shared_ptr, iterators, and hopefully _com_ptr.
EDIT: Oops, they're pointers, not references. Here's a version (or overload) that deals with that, and allows -- by ignoring -- arbitrarily-typed return values.
template <typename P, typename T, typename A, typename R>
A magic(P p, R (T::*f)(A *)) {
A a;
((*p).*f)(&a);
return a;
}
This is not quite what you specified because you need to wrap get() around the method, but it works:
template<class T, class S>
T get(S fun(T&)) {
T result;
fun(result);
return result;
}
void foo(int& x) {
x = 5;
}
bool bar(char& x) {
x = 'c';
return false;
}
int main() {
int x = get(foo);
char y = get(bar);
return 0;
}
Can you derive from IRasterProps? Being that the case you can construct your own interface to it.
EDIT: Following on the concept you can probably also apply the Adapter design pattern (or even a Facade if you wish to apply a common interface to several like-minded classes of the SDK).
Looks like a COM object to me.
Visual C++ supports an #import directive to import the type library, and create high-legel wrappers. So you either end up with
width = ptr->GetWidth();
or - even better -
width = ptr->Width;
If a function fails, the HRESULT returned will be transformed into an _com_error exception.
I've used that successfully on many OS and 3rd party COM objects, makes them much easier to use.
Note that you control the wrapper generation through options, the first thing I do is usually adding a rename_namespace or no_namespace, because otherwise the symbold end up in a namespace depending on the typelib name, which is usually ugly.
also, unless you use named_guids option, you might needto change CLSID_xxx and IID_xxx constants to __uuidof(xxx).
EDIT: In retrospect, I'm not sure this one will actually work, since I don't think the template arguments will deduce. Buyer Beware.
Sure! What you need is something to which you can pass a function that will call it and return you the outputted value.
Here's the easy, if less efficient way:
template <typename T>
T magic(boost::function<void(T&)> f) {
T x;
f(x);
return x;
}
Which you'd then call like this using boost::lambda:
long output_width = magic(raster_props_object->*&IRasterProps::get_Width);
Or like this, using boost::bind:
long output_width = magic(bind(&IRasterProps::get_Width, raster_props_object, _1));
You can get rid of boost::function, but that's uglier. Probably worth it, though.
Don't think this is possible. Assigning void to a long should be an error in any case.
Remember, it's probably more performant to pass-by-reference than to return a large object. (won't really make a difference with long's though)
Compiling this:
void foo(long &a) {
}
int main(void) {
long a=0;
a = foo(a);
return 0;
}
gives this error:
g++ x.cc
x.cc: In function ‘int main()’:
x.cc:9: error: void value not ignored as it ought to be
I'm not aware of something insane you could do, precisely like you're asking, and if there was some insane hackery that did work on some peculiar platform I'm pretty sure in a code-review I'd hate it.
It may may more sense to either...
define some trivial inline function wrappers around the APIs you care about
make a specialized class descend from IRasterProps (or whatever) that provides the appropriate accessor methods.
Either of those will impact maintenance time of the code but would safely and cleanly give you the call syntax you are looking for.
I have to return to the previous level of the recursion. is the syntax like below right?
void f()
{
// some code here
//
return;
}
Yes, you can return from a void function.
Interestingly, you can also return void from a void function. For example:
void foo()
{
return void();
}
As expected, this is the same as a plain return;. It may seem esoteric, but the reason is for template consistency:
template<class T>
T default_value()
{
return T();
}
Here, default_value returns a default-constructed object of type T, and because of the ability to return void, it works even when T = void.
Sure. You just shouldn't be returning an actual value.
Yes, you can use that code to return from the function. (I have to be very verbose here to make Stack Overflow not say that my answer is too short)
Yes, that will return from the function to the previous level of recursion. This is going to be very basic explanation, but when you call a function you are creating a new call stack. In a recursive function you are simply adding to that call stack. By returning from a function, whether you return a value or not, you are moving the stack pointer back to the previous function on the stack. It's sort of like a stack of plates. You keep putting plates on it, but than returning moves the top plate.
You could also verify this by using a debugger. Just put a few break points in your code and step through it. You can verify yourself that it works.
The simple answer to this is YES! C++ recognise void method as a function with no return. It basically tells the compiler that whatever happens, once you see the return; break and leave the method....
Yes, sometimes you may wish to return void() instead of just nothing.
Consider a void function that wants to call some pass-through void functions without a bunch of if-else.
return
InputEvent == E_Pressed ? Controller->Grip() :
InputEvent == E_Released ? Controller->Release() :
InputEvent == E_Touched ? Controller->Touch() : void();
You shouldn't have to have the return there, the program will return to the previous function by itself, go into debug mode and step through and you can see it yourself.
On the other hand i don't think having a return there will harm the program at all.
As everyone else said, yes you can. In this example, return is not necessary and questionably serves a purpose. I think what you are referring to is an early return in the middle of a function. You can do that too however it is bad programming practice because it leads to complicated control flow (not single-entry single-exit), along with statements like break. Instead, just skip over the remainder of the function using conditionals like if/else().