I have a special class to be used as return type of methods, containing the wanted value or in case of failure an error message which is even cascading from earlier errors. It works as expected.
As the returned type is differently complex I like to use the keyword auto. But when using a lot of methods I have to create new return variables.
A typical part of code looks like this:
auto ret1 = methodA();
if(ret1.isValid()...
auto ret2 = methodB();
if(ret2.isValid()...
I dont like to always create a new variable. But I like the elegant way of error handling. Using a more dump return type like an error code in integer would solve the problem but then I have no benefit from the error handling return type.
Is there any trick to reuse the first return variable ret1?
You would have to create new scopes to reuse the variable name for a different variable, like:
{
auto ret = methodA();
if (ret.isValid()) ...
}
{
auto ret = methodB();
if (ret.isValid()) ...
}
You can also take advantage of the scope created by if, placing the init-statement inside:
if (auto ret = methodA(); ret.isValid()) ...
if (auto ret = methodB(); ret.isValid()) ...
auto is not a type.
It is a keyword, that says "put the type here for me, by deducing it from the initial value". That occurs during compilation, once.
You cannot reuse ret1 to store an object of a different type, whether you use auto or not.
This shouldn't really be a problem. If you're concerned about "running out of names", or "having many similar names", your names are not descriptive enough and/or your scopes aren't tight enough.
auto is not a type. In auto foo = bar(); the compiler simply figures out what type bar() actually returns and substitutes that in. So if bar() returns int then that's the type of foo, if it returns bool then that is the type of foo. And once the type that auto should be replaced with (the first time) has been determined, then it can never change. auto doesn't mean "variable type" it just means "hey compiler, I'm too lazy to figure out the type to put here, please do it for me", but there is no difference what-so-ever compared to you just writing the final type yourself.
So, you can reuse the variable if what you assign to it the second time is of the same type as the first time - otherwise not.
I dont like to always create a new variable.
Much better is to create a const variable:
const auto ret1 = methodA();
if(ret1.isValid()...
const auto ret2 = methodB();
if(ret2.isValid()...
In this case you need to make const all the methods like isValid, but that is even better: "is" shouldn't have side effects and modify the state.
Next step is to remove the temp variable at all:
if(methodA().isValid()) {
...
}
if(methodB().isValid()) {
...
}
The alternative is to wrap each function call into a block:
{
const auto ret = methodA();
if(ret.isValid()...
}
{
const auto ret = methodB();
if(ret.isValid()...
}
This allows you to reuse the const variable name.
Each block becomes a candidate for extraction into a separate function (see Uncle Bob in "Clean Code").
Related
In the project I am currently working on I find myself writing a lot of code that looks like the following, where, get_optional_foo is returning an std::optional:
//...
auto maybe_foo = get_optional_foo(quux, ...)
if (!maybe_foo.has_value())
return {};
auto foo = maybe_foo.value()
//...
// continue on, doing things with foo...
I want to bail out of the function if I get a null option; otherwise, I want to assign a non-optional variable to the value. I've started using the convention of naming the optional with a maybe_ prefix but am wondering if there is some way of doing this such that I don't need to use a temporary for the optional at all? This variable is only ever going to be used to check for a null option and dereference if there is a value.
You don't need an intermediate object. std::optional supports a pointer interface to access it so you can just use it like:
//...
auto foo = get_optional_foo(quux, ...)
if (!foo)
return {};
//...
foo->some_member;
(*foo).some_member;
Slightly different than what you are asking, but consider:
if (auto foo = get_optional_foo(1)) {
// ...
return foo->x;
} else {
return {};
}
This places the main body of the function in an if() block, which may be more readable.
Shortest I can think of:
auto maybe_foo = get_optional_foo(quux, ...)
if (!maybe_foo) return {};
auto &foo = *maybe_foo; // alternatively, use `*maybe_foo` below
If you have multiple optionals in the function and it's very unlikely they'll be empty you can wrap the whole thing with a try - catch.
try {
auto &foo = get_optional_foo(quux, ...).value();
auto &bar = get_optional_bar(...).value();
...
} catch (std::bad_optional_access &e) {
return {};
}
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++.
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();
I have a routine that does some moderately expensive operations, and the client could consume the result as either a string, integer, or a number of other data types. I have a public data type that is a wrapper around an internal data type. My public class looks something like this:
class Result {
public:
static Result compute(/* args */) {
Result result;
result.fData = new ExpensiveInternalObject(/* args */);
return result;
}
// ... constructors, destructor, assignment operators ...
std::string toString() const { return fData->toString(); }
int32_t toInteger() const { return fData->toInteger(); }
double toDouble() const { return fData->toDouble(); }
private:
ExpensiveInternalObject* fData;
}
If you want the string, you can use it like this:
// Example A
std::string resultString = Result::compute(/*...*/).toString();
If you want more than one of the return types, you do it like this:
// Example B
Result result = Result::compute(/*...*/);
std::string resultString = result.toString();
int32_t resultInteger = result.toInteger();
Everything works.
However, I want to modify this class such that there is no need to allocate memory on the heap if the user needs only one of the result types. For example, I want Example A to essentially do the equivalent of,
auto result = ExpensiveInternalObject(/* args */);
std::string resultString = result.toString();
I've thought about structuring the code such that the args are saved into the instance of Result, make the ExpensiveInternalObject not be calculated until the terminal functions (toString/toInteger/toDouble), and overload the terminal functions with rvalue reference qualifiers, like this:
class Result {
// ...
std::string toString() const & {
if (fData == nullptr) {
const_cast<Result*>(this)->fData = new ExpensiveInternalObject(/*...*/);
}
return fData->toString();
}
std::string toString() && {
auto result = ExpensiveInternalObject(/*...*/);
return result.toString();
}
// ...
}
Although this avoids the heap allocation for the Example A call site, the problem with this approach is that you have to start thinking about thread safety issues. You'd probably want to make fData an std::atomic, which adds overhead to the Example B call site.
Another option would be to make two versions of compute() under different names, one for the Example A use case and one for the Example B use case, but this isn't very friendly to the user of the API, because now they have to study which version of the method to use, and they will get poor performance if they choose the wrong one.
I can't make ExpensiveInternalObject a value field inside Result (as opposed to a pointer) because doing so would require exposing too many internals in the public header file.
Is there a way to make the first function, compute(), know whether its return value is going to become an rvalue reference or whether it is going to become an lvalue, and have different behavior for each case?
You can achieve the syntax you asked for using a kind of proxy object.
Instead of a Result, Result::compute could return an object that represents a promise of a Result. This Promise object could have a conversion operator that implicitly converts to a Result so that "Example B" still works as before. But the promise could also have its own toString(), toInteger(), ... member functions for "Example A":
class Result {
public:
class Promise {
private:
// args
public:
std::string toString() const {
auto result = ExpensiveInternalObject(/* args */);
return result.toString();
}
operator Result() {
Result result;
result.fData = new ExpensiveInternalObject(/* args */);
return result;
}
};
// ...
};
Live demo.
This approach has its downsides though. For example, what if, instead you wrote:
auto result = Result::compute(/*...*/);
std::string resultString = result.toString();
int32_t resultInteger = result.toInteger();
result is now not of Result type but actually a Result::Promise and you end up computing ExpensiveInternalObject twice! You can at least make this to fail to compile by adding an rvalue reference qualifier to the toString(), toInteger(), ... member functions on Result::Promise but it is not ideal.
Considering you can't overload a function by its return type, and you wanted to avoid making two different versions of compute(), the only thing I can think of is setting a flag in the copy constructor of Result. This could work with your particular example, but not in general. For example, it won't work if you're taking a reference, which you can't disallow.
I have the following code:
Foo a;
if (some_fairly_long_condition) {
a = complicated_expression_to_make_foo_1();
} else {
a = complicated_expression_to_make_foo_2();
}
I have two issues with this:
a is a const and should be declared so
the "empty" constructor, Foo() is called for no reason (maybe this is optimised away?)
One way to fix it is by using the ternary operator:
const Foo a = some_fairly_long_condition?
complicated_expression_to_make_foo_1():
complicated_expression_to_make_foo_2();
Is this good practice? How do you go about it?
To answer the second part of your question:
I usually put the initialization code into a lambda:
const Foo a = [&]()->Foo{
if (some_fairly_long_condition) {
return complicated_expression_to_make_foo_1();
} else {
return complicated_expression_to_make_foo_2();
}
}();
In most cases you should even be able to omit the trailing return type, so you can write
const Foo a = [&](){ ...
As far as the first part is concerned:
I'd say that greatly depends on how complex your initialization code is. If all three parts are really complicated expressions (and not just a function call each) then the solution with the ternary operator becomes an unreadable mess, while the lambda method (or a separate named function for that matter) allows you to break up those parts into the respective sub expressions.
If the problem is to avoid ternaty operator and your goal is to define the constant a, this code is an option:
Foo aux;
if (some_fairly_long_condition) {
aux = complicated_expression_to_make_foo_1();
} else {
aux = complicated_expression_to_make_foo_2();
}
const Foo a(aux);
It is a good solution, without any new feature ---as lambdas--- and including the code inline, as you want.