The basic question:
Edit: v-The question-v
class foo {
public:
constexpr foo() { }
constexpr int operator()(const int& i) { return int(i); }
}
Performance is a non-trivial issue. How does the compiler actually compile the above? I know how I want it to be resolved, but how does the specification actually specify it will be resolved?
1) Seeing the type int has a constexpr constructor, create a int object and compile the string of bytes that make the type from memory into the code directly?
2) Replace any calls to the overload with a call to the 'int's constructor that for some unknown reason int doesn't have constexpr constructors? (Inlining the call.)
3) Create a function, call the function, and have that function call 'int's consctructor?
Why I want to know, and how I plan to use the knowledge
edit:v-Background only-v
The real library I'm working with uses template arguments to decide how a given type should be passed between functions. That is, by reference or by value because the exact size of the type is unknown. It will be a user's responsibility to work within the limits I give them, but I want these limits to be as light and user friendly as I can sanely make them.
I expect a simple single byte character to be passed around in which case it should be passed by value. I do not bar 300mega-byte behemoth that does several minuets of recalculation every time a copy constructor is invoked. In which case passing by reference makes more sense. I have only a list of requirements that a type must comply with, not set cap on what a type can or can not do.
Why I want to know the answer to my question is so I can in good faith make a function object that accepts this unknown template, and then makes a decision how, when, or even how much of a object should be copied. Via a virtual member function and a pointer allocated with new is so required. If the compiler resolves constexpr badly I need to know so I can abandon this line of thought and/or find a new one. Again, It will be a user's responsibility to work within the limits I give them, but I want these limits to be as light and user friendly as I can sanely make them.
Edit: Thank you for your answers. The only real question was the second sentence. It has now been answered. Everything else If more background is required, Allow me to restate the above:
I have a template with four argument. The goal of the template is a routing protocol. Be that TCP/IP -unlikely- or node to node within a game -possible. The first two are for data storage. They have no requirement beyond a list of operators for each. The last two define how the data is passed within the template. By default this is by reference. For performance and freedom of use, these can be changed define to pass information by value at a user's request.
Each is expect to be a single byte long. They could in the case of metric for a EIGRP or OSFP like protocol the second template argument could be the compound of a dozen or more different variable. Each taking a non-trival time to copy or recompute.
For ease of use I investigate the use a function object that accepts the third and fourth template to handle special cases and polymorphic classes that would fail to function or copy correctly. The goal to not force a user to rebuild their objects from scratch. This would require planning for virtual function to preform deep copies, or any number of other unknown oddites. The usefulness of the function object depends on how sanely a compiler can be depended on not generate a cascade of function calls.
More helpful I hope?
The C++11 standard doesn't say anything about how constexpr will be compiled down to machine instructions. The standard just says that expressions that are constexpr may be used in contexts where a compile time constant value is required. How any particular compiler chooses to translate that to executable code is an implementation issue.
Now in general, with optimizations turned on you can expect a reasonable compiler to not execute any code at runtime for many uses of constexpr but there aren't really any guarantees. I'm not really clear on what exactly you're asking about in your example so it's hard to give any specifics about your use case.
constexpr expressions are not special. For all intents and purposes, they're basically const unless the context they're used in is constexpr and all variables/functions are also constexpr. It is implementation defined how the compiler chooses to handle this. The Standard never deals with implementation details because it speaks in abstract terms.
Related
In C++ it is possible to declare that a function is const, which means, as far as I understand, that the compiler ensures the function does not modify the object. Is there something analogous in C++ where I can require that a function is pure? If not in C++, is there a language where one can make this requirement?
If this is not possible, why is it possible to require functions to be const but not require them to be pure? What makes these requirements different?
For clarity, by pure I want there to be no side effects and no use of variables other than those passed into the function. As a result there should be no file reading or system calls etc.
Here is a clearer definition of side effects:
No modification to files on the computer that the program is run on and no modification to variables with scope outside the function. No information is used to compute the function other than variables passed into it. Running the function should return the same thing every time it is run.
NOTE: I did some more research and encountered pure script
(Thanks for jarod42's comment)
Based on a quick read of the wikipedia article I am under the impression you can require functions be pure in pure script, however I am not completely sure.
Short answer: No. There is no equivalent keyword called pure that constrains a function like const does.
However, if you have a specific global variable you'd like to remain untouched, you do have the option of static type myVar. This will require that only functions in that file will be able to use it, and nothing outside of that file. That means any function outside that file will be constrained to leave it alone.
As to "side effects", I will break each of them down so you know what options you have:
No modification to files on the computer that the program is run on.
You can't constrain a function to do this that I'm aware. C++ just doesn't offer a way to constrain a function like this. You can, however, design a function to not modify any files, if you like.
No modification to variables with scope outside the function.
Globals are the only variables you can modify outside a function's scope that I'm aware of, besides anything passed by pointer or reference as a parameter. Globals have the option of being constant or static, which will keep you from modifying them, but, beyond that, there's really nothing you can do that I'm aware.
No information is used to compute the function other than variables passed into it.
Again, you can't constrain it to do so that I'm aware. However, you can design the function to work like this if you want.
Running the function should return the same thing every time it is run.
I'm not sure I understand why you want to constrain a function like this, but no. Not that I'm aware. Again, you can design it like this if you like, though.
As to why C++ doesn't offer an option like this? I'm guessing reusability. It appears that you have a specific list of things you don't want your function to do. However, the likelihood that a lot of other C++ users as a whole will need this particular set of constraints often is very small. Maybe they need one or two at a time, but not all at once. It doesn't seem like it would be worth the trouble to add it.
The same, however, cannot be said about const. const is used all the time, especially in parameter lists. This is to keep data from getting modified if it's passed by reference, or something. Thus, the compiler needs to know what functions modify the object. It uses const in the function declaration to keep track of this. Otherwise, it would have no way of knowing. However, with using const, it's quite simple. It can just constrain the object to only use functions that guarantee that it remains constant, or uses the const keyword in the declaration if the function.
Thus, const get's a lot of reuse.
Currently, C++ does not have a mechanism to ensure that a function has "no side effects and no use of variables other than those passed into the function." You can only force yourself to write pure functions, as mentioned by Jack Bashford. The compiler can't check this for you.
There is a proposal (N3744 Proposing [[pure]]). Here you can see that GCC and Clang already support __attribute__((pure)). Maybe it will be standardized in some form in the future revisions of C++.
In C++ it is possible to declare that a function is const, which means, as far as I understand, that the compiler ensures the function does not modify the object.
Not quite. The compiler will allow the object to be modified by (potentially ill-advised) use of const_cast. So the compiler only ensures that the function does not accidentally modify the object.
What makes these requirements [constant and pure] different?
They are different because one affects correct functionality while the other does not.
Suppose C is a container and you are iterating over its contents. At some point within the loop, perhaps you need to call a function that takes C as a parameter. If that function were to clear() the container, your loop will likely crash. Sure, you could build a loop that can handle that, but the point is that there are times when a caller needs assurance that the rug will not be pulled out from under it. Hence the ability to mark things const. If you pass C as a constant reference to a function, that function is promising to not modify C. This promise provides the needed assurance (even though, as I mentioned above, the promise can be broken).
I am not aware of a case where use of a non-pure function could similarly cause a program to crash. If there is no use for something, why complicate the language with it? If you can come up with a good use-case, maybe it is something to consider for a future revision of the language.
(Knowing that a function is pure could help a compiler optimize code. As far as I know, it's been left up to each compiler to define how to flag that, as it does not affect functionality.)
I try to create a c++ flex/bison parser. I used this tutorial as a starting point and did not change any bison/flex configurations. I am stuck now to the point of trying to unit test the lexer.
I have a function in my unit tests that directly calls yylex, and checks the result of it:
private: static void checkIntToken(MyScanner &scanner, Compiler *comp, unsigned long expected, unsigned char size, char isUnsigned, unsigned int line, const std::string &label) {
yy::MyParser::location_type loc;
yy::MyParser::semantic_type semantic; // <---- is seems like the destructor of this variable causes the crash
int type = scanner.yylex(&semantic, &loc, comp);
Assert::equals(yy::MyParser::token::INT, type, label + "__1");
MyIntToken* token = semantic.as<MyIntToken*>();
Assert::equals(expected, token->value, label + "__2");
Assert::equals(size, token->size, label + "__3");
Assert::equals(isUnsigned, token->isUnsigned, label + "__4");
Assert::equals(line, loc.begin.line, label + "__5");
//execution comes to this point, and then, program crashes
}
The error message is:
program: ../src/__autoGenerated__/MyParser.tab.hh:190: yy::variant<32>::~variant() [S = 32]: Assertion `!yytypeid_' failed.
I have tried to follow the logic in the auto-generated bison files, and make some sense out of it. But I did not succeed on that and ultimately gave up. I searched then for any advice on the web about this error message but did not find any.
The location indicated by the error has the following code:
~variant (){
YYASSERT (!yytypeid_);
}
EDIT: The problem disappears only if I remove the
%define parse.assert
option from the bison file. But I am not sure if this is a good idea...
What is the proper way to obtain the value of the token generated by flex, for unit testing purposes?
Note: I've tried to explain bison variant types to the best of my knowledge. I hope it is accurate but I haven't used them aside from some toy experiments. It would be an error to assume that this explanation in any way implies an endorsement of the interface.
The so-called "variant" type provided by bison's C++ interface is not a general-purpose variant type. That was a deliberate decision based on the fact that the parser is always able to figure out the semantic type associated with a semantic value on the parser stack. (This fact also allows a C union to be used safely within the parser.) Recording type information within the "variant" would therefore be redundant. So they don't. In that sense, it is not really a discriminated union, despite what one might expect of a type named "variant".
(The bison variant type is a template with an integer (non-type) template argument. That argument is the size in bytes of the largest type which is allowed in the variant; it does not in any other way specify the possible types. The semantic_type alias serves to ensure that the same template argument is used for every bison variant object in the parser code.)
Because it is not a discriminated union, its destructor cannot destruct the current value; it has no way to know how to do that.
This design decision is actually mentioned in the (lamentably insufficient) documentation for the Bison "variant" type. (When reading this, remember that it was originally written before std::variant existed. These days, it would be std::variant which was being rejected as "redundant", although it is also possible that the existence of std::variant might have had the happy result of revisiting this design decision). In the chapter on C++ Variant Types, we read:
Warning: We do not use Boost.Variant, for two reasons. First, it appeared unacceptable to require Boost on the user’s machine (i.e., the machine on which the generated parser will be compiled, not the machine on which bison was run). Second, for each possible semantic value, Boost.Variant not only stores the value, but also a tag specifying its type. But the parser already “knows” the type of the semantic value, so that would be duplicating the information.
Therefore we developed light-weight variants whose type tag is external (so they are really like unions for C++ actually).
And indeed they are. So any use of a bison "variant" must have a definite type:
You can build a variant with an argument of the type to build. (This is the only case where you don't need a template parameter, because the type is deduced from the argument. You would have to use an explicit template parameter only if the argument were not of the precise type; for example, an integer of lesser rank.)
You can get a reference to the value of known type T with as<T>. (This is undefined behaviour if the value has a different type.)
You can destruct the value of known type T with destroy<T>.
You can copy or move the value from another variant of known type T with copy<T> or move<T>. (move<T> involves constructing and then destructing a T(), so you might not want to do it if T had an expensive default constructor. On the whole, I'm not convinced by the semantics of the move method. And its name conflicts semantically with std::move, but again it came first.)
You can swap the values of two variants which both have the same known type T with swap<T>.
Now, the generated parser understands all these restrictions, and it always knows the real types of the "variants" it has at its disposal. But you might come along and try to do something with one of these objects in a way that violates a constraint. Since the object really doesn't have any way to check the constraint, you'll end up with undefined behaviour which will probably have some disastrous eventual consequence.
So they also implemented an option which allows the "variant" to check the constraints. Unsurprisingly, this consists of adding a discriminator. But since the discriminator is only used to validate and not to modify behaviour, it is not a small integer which chooses between a small number of known alternatives, but rather a pointer to a std::typeid (or NULL if the variant does not yet contain a value.) (To be fair, in most cases alignment constraints mean that using a pointer for this purpose is no more expensive than using a small enum. All the same...)
So that's what you're running into. You enabled assertions with %define parse.assert; that option was provided specifically to prevent you from doing what you are trying to do, which is let the variant object's destructor run before the variant's value is explicitly destructed.
So the "correct" way to avoid the problem is to insert an explicit call at the end of the scope:
// execution comes to this point, and then, without the following
// call, the program will fail on an assertion
semantic.destroy<MyIntType*>();
}
With the parse assertion enabled, the variant object will be able to verify that the types specified as template parameters to semantic.as<T> and semantic.destroy<T> are the same types as the value stored in the object. (Without parse.assert, that too is your responsibility.)
Warning: opinion follows.
In case anyone reading this cares, my preference for using real std::variant types comes from the fact that it is actually quite common for the semantic value of an AST node to require a discriminated union. The usual solution (in C++) is to construct a type hierarchy which is, in some ways, entirely artificial, and it is quite possible that std::variant can better express the semantics.
In practice, I use the C interface and my own discriminated union implementation.
Will it be possible to specialize std::optional for user-defined types? If not, is it too late to propose this to the standard?
My use case for this is an integer-like class that represents a value within a range. For instance, you could have an integer that lies somewhere in the range [0, 10]. Many of my applications are sensitive to even a single byte of overhead, so I would be unable to use a non-specialized std::optional due to the extra bool. However, a specialization for std::optional would be trivial for an integer that has a range smaller than its underlying type. We could simply store the value 11 in my example. This should provide no space or time overhead over a non-optional value.
Am I allowed to create this specialization in namespace std?
The general rule in 17.6.4.2.1 [namespace.std]/1 applies:
A program may add a template specialization for any standard library template to namespace std only if the declaration depends on a user-defined type and the specialization meets the standard library requirements for the original template and is not explicitly
prohibited.
So I would say it's allowed.
N.B. optional will not be part of the C++14 standard, it will be included in a separate Technical Specification on library fundamentals, so there is time to change the rule if my interpretation is wrong.
If you are after a library that efficiently packs the value and the "no-value" flag into one memory location, I recommend looking at compact_optional. It does exactly this.
It does not specialize boost::optional or std::experimental::optional but it can wrap them inside, giving you a uniform interface, with optimizations where possible and a fallback to 'classical' optional where needed.
I've asked about the same thing, regarding specializing optional<bool> and optional<tribool> among other examples, to only use one byte. While the "legality" of doing such things was not under discussion, I do think that one should not, in theory, be allowed to specialize optional<T> in contrast to eg.: hash (which is explicitly allowed).
I don't have the logs with me but part of the rationale is that the interface treats access to the data as access to a pointer or reference, meaning that if you use a different data structure in the internals, some of the invariants of access might change; not to mention providing the interface with access to the data might require something like reinterpret_cast<(some_reference_type)>. Using a uint8_t to store a optional-bool, for example, would impose several extra requirements on the interface of optional<bool> that are different to the ones of optional<T>. What should the return type of operator* be, for example?
Basically, I'm guessing the idea is to avoid the whole vector<bool> fiasco again.
In your example, it might not be too bad, as the access type is still your_integer_type& (or pointer). But in that case, simply designing your integer type to allow for a "zombie" or "undetermined" value instead of relying on optional<> to do the job for you, with its extra overhead and requirements, might be the safest choice.
Make it easy to opt-in to space savings
I have decided that this is a useful thing to do, but a full specialization is a little more work than necessary (for instance, getting operator= correct).
I have posted on the Boost mailing list a way to simplify the task of specializing, especially when you only want to specialize some instantiations of a class template.
http://boost.2283326.n4.nabble.com/optional-Specializing-optional-to-save-space-td4680362.html
My current interface involves a special tag type used to 'unlock' access to particular functions. I have creatively named this type optional_tag. Only optional can construct an optional_tag. For a type to opt-in to a space-efficient representation, it needs the following member functions:
T(optional_tag) constructs an uninitialized value
initialize(optional_tag, Args && ...) constructs an object when there may be one in existence already
uninitialize(optional_tag) destroys the contained object
is_initialized(optional_tag) checks whether the object is currently in an initialized state
By always requiring the optional_tag parameter, we do not limit any function signatures. This is why, for instance, we cannot use operator bool() as the test, because the type may want that operator for other reasons.
An advantage of this over some other possible methods of implementing it is that you can make it work with any type that can naturally support such a state. It does not add any requirements such as having a move constructor.
You can see a full code implementation of the idea at
https://bitbucket.org/davidstone/bounded_integer/src/8c5e7567f0d8b3a04cc98142060a020b58b2a00f/bounded_integer/detail/optional/optional.hpp?at=default&fileviewer=file-view-default
and for a class using the specialization:
https://bitbucket.org/davidstone/bounded_integer/src/8c5e7567f0d8b3a04cc98142060a020b58b2a00f/bounded_integer/detail/class.hpp?at=default&fileviewer=file-view-default
(lines 220 through 242)
An alternative approach
This is in contrast to my previous implementation, which required users to specialize a class template. You can see the old version here:
https://bitbucket.org/davidstone/bounded_integer/src/2defec41add2079ba023c2c6d118ed8a274423c8/bounded_integer/detail/optional/optional.hpp
and
https://bitbucket.org/davidstone/bounded_integer/src/2defec41add2079ba023c2c6d118ed8a274423c8/bounded_integer/detail/optional/specialization.hpp
The problem with this approach is that it is simply more work for the user. Rather than adding four member functions, the user must go into a new namespace and specialize a template.
In practice, all specializations would have an in_place_t constructor that forwards all arguments to the underlying type. The optional_tag approach, on the other hand, can just use the underlying type's constructors directly.
In the specialize optional_storage approach, the user also has the responsibility of adding proper reference-qualified overloads of a value function. In the optional_tag approach, we already have the value so we do not have to pull it out.
optional_storage also required standardizing as part of the interface of optional two helper classes, only one of which the user is supposed to specialize (and sometimes delegate their specialization to the other).
The difference between this and compact_optional
compact_optional is a way of saying "Treat this special sentinel value as the type being not present, almost like a NaN". It requires the user to know that the type they are working with has some special sentinel. An easily specializable optional is a way of saying "My type does not need extra space to store the not present state, but that state is not a normal value." It does not require anyone to know about the optimization to take advantage of it; everyone who uses the type gets it for free.
The future
My goal is to get this first into boost::optional, and then part of the std::optional proposal. Until then, you can always use bounded::optional, although it has a few other (intentional) interface differences.
I don't see how allowing or not allowing some particular bit pattern to represent the unengaged state falls under anything the standard covers.
If you were trying to convince a library vendor to do this, it would require an implementation, exhaustive tests to show you haven't inadvertently blown any of the requirements of optional (or accidentally invoked undefined behavior) and extensive benchmarking to show this makes a notable difference in real world (and not just contrived) situations.
Of course, you can do whatever you want to your own code.
These should probably be in different questions, but they're related so...
Why do we need to write constexpr at all? Given a set of restrictions couldn't a compiler evaluate code to see if it satisfies the constexpr requirements, and treat it as constexpr if it does? As a purely documentation keyword I'm not sure it holds up because I can't think of a case where I (the user of someone else's constexpr function) should really care if it's run time or not.
Here's my logic: If it's an expensive function I think as a matter of good practice I should treat it as such regardless of whether I give it compile-time constant input or not. That might mean calling it during load time and saving off the result, instead of calling it during a critical point in the execution. The reason is because constexpr doesn't actually guarantee to me that it will not be executed in run time in the first place — so perhaps a new/different mechanism should do that.
The constexpr restrictions seem to exclude many, if not most, functions from being compile-time evaluated which logically could be. I've read this is at least in part (or perhaps wholly?) to prevent infinite looping and hanging the compiler. But, if this is the reason, is it legitimate?
Shouldn't a compiler be able to compute if, for any given constexpr function with the given inputs used, it loops infinitely? This is not solving the halting problem for any input. The input to a constexpr function is compile time constant and finite, so the compiler only has to check for infinite looping for a finite set of input: the input actually used. It should be a regular compilation error if you write a compile-time infinite loop.
I asked a very similar question, Why do we need to mark functions as constexpr?
When I pressed Richard Smith, a Clang author, he explained:
The constexpr keyword does have utility.
It affects when a function template specialization is instantiated (constexpr function template specializations may need to be instantiated if they're called in unevaluated contexts; the same is not true for non-constexpr functions since a call to one can never be part of a constant expression). If we removed the meaning of the keyword, we'd have to instantiate a bunch more specializations early, just in case the call happens to be a constant expression.
It reduces compilation time, by limiting the set of function calls that implementations are required to try evaluating during translation. (This matters for contexts where implementations are required to try constant expression evaluation, but it's not an error if such evaluation fails -- in particular, the initializers of objects of static storage duration.)
This all didn't seem convincing at first, but if you work through the details, things do unravel without constexpr. A function need not be instantiated until it is ODR-used, which essentially means used at runtime. What is special about constexpr functions is that they can violate this rule and require instantiation anyway.
Function instantiation is a recursive procedure. Instantiating a function results in instantiation of the functions and classes it uses, regardless of the arguments to any particular call.
If something went wrong while instantiating this dependency tree (potentially at significant expense), it would be difficult to swallow the error. Furthermore, class template instantiation can have runtime side-effects.
Given an argument-dependent compile-time function call in a function signature, overload resolution may incur instantiation of function definitions merely auxiliary to the ones in the overload set, including the functions that don't even get called. Such instantiations may have side effects including ill-formedness and runtime behavior.
It's a corner case to be sure, but bad things can happen if you don't require people to opt-in to constexpr functions.
As for constexpr objects, certain types can produce core constant expressions which are usable in constant expression contexts without having been declared constexpr. But you don't really want the compiler to try evaluating every single expression at compile time. That's what constant propagation is for. On the other hand it seems pretty essential to document when something needs to happen at compile time.
[Note, I totally changed my answer]
To answer your second question, there are two cases for the compiler here:
The compiler has to be able to handle any arbitrary constexpr function(s). In this case you still have the halting problem because the set of inputs is all combinations of constexpr functions and calls to them.
The compiler can handle a finite set of constexpr function(s). In this case the compiler can in fact determine whether some programs will result in infinite loops, while other programs will be uncompilable (since they aren't in the set of valid inputs).
So presumably the restrictions are in place so that it satisfies case 2 for a reasonable amount of compiler effort.
There are both technical and ideological reasons behind this decision.
Not always do we want constexpr ourselves by default - it can take
too much compiling time. That's first. Just imagine you implemented
isPrime function and you have 100 calls with big constexpr
values passed in. I think you don't (in most cases) want to make
compiler compiling this for a couple of minutes longer because it
decided that you need those values in compile-time by itself. But if
it's exactly the case - specify constexpr modifier manually. And this adds the next point:
backward compatibility - it's unwise to assume that every possible C++98 program author who converted this program to C++11 wantsconstexpr.
The second point is that deciding if the function can be constexpr
would take compiling time by itself. And if it was trying to do that for every possible function it would take some additional time overhead. Even more, often compiler
couldn't decide if the given function can be constexpr at all, so
your first assumption is not correct.
I'm experimenting with variable arguments in C++, using va_args. The idea is useful, and is indeed something I've used a lot in C# via the params functionality. One thing that frustrates me is the following excerpt regarding va_args, above:
Notice also that va_arg does not determine either whether the retrieved argument is the last argument passed to the function (or even if it is an element past the end of that list).
I find it hard to believe that there is no way to programmatically determine the number of variable arguments passed to the function from within that function itself. I would like to perform something like the following:
void fcn(int arg1 ...)
{
va_list argList;
va_start(argList, arg1);
int numRemainingParams = //function that returns number of remaining parameters
for (int i=0; i<numRemainingParams; ++i)
{
//do stuff with params
}
va_end(argList);
}
To reiterate, the documentation above suggests that va_arg doesn't determine whether the retrieved arg is the last in the list. But I feel this information must be accessible in some manner.
Is there a standard way of achieving this?
I find it hard to believe that there is no way to programmatically determine the number of variable arguments passed to the function from within that function itself.
Nonetheless, it is true. C/C++ do not put markers on the end of the argument list, so the called function really does not know how many arguments it is receiving. If you need to mark the end of the arguments, you must do so yourself by putting some kind of marker at the end of the list.
The called function also has no idea of the types or sizes of the arguments provided. That's why printf and friends force you to specify the precise datatype of the value to interpolate into the format string, and also why you can crash a program by calling printf with a bad format string.
Note that parameter passing is specified by the ABI for a particular platform, not by the C++/C standards. However, the ABI must allow the C++/C standards to be implementable. For example, an ABI might want to pass parameters in registers for efficiency, but it might not be possible to implement va_args easily in that case. So it's possible that arguments are also shadowed on the stack. In almost no case is the stack marked to show the end of the argument list, though, since the C++/C standards don't require this information to be made available, and it would therefore be unnecessary overhead.
The way variable arguments work in C and C++ is relatively simple: the arguments are just pushed on the stack and it is the callee's responsibility to somewhat figure out what arguments there are. There is nothing in the standard which provides a way to determine the number of arguments. As a result, the number of arguments are determined by some context information, e.g., the number of elements referenced in a format string.
Individual compilers may know how many elements there are but there is no standard interface to obtain this value.
What you could do instead, however, is to use variadic templates: you can determine very detailed information on the arguments being passed to the function. The interface looks different and it may be necessary to channel the arguments into some sort of data structure but on the upside it would also work with types you cannot pass using variable arguments.
No, there isn't. That's why variable arguments are not safe. They're a part of C, which lacks the expressiveness to achieve type safety for "convenient" variadic functions. You have to live with the fact that C contains constructions whose very correctness depends on values and not just on types. That's why it is an "unsafe language".
Don't use variable arguments in C++. It is a much stronger language that allows you to write equally convenient code that is safe.
No, there's no such way. If you have such a need, it's probably best to pack those function parameters in a std::vector or a similar collection which can be iterated.
The variable argument list is a very old concept inherited from the C history of C++. It dates back to the time where C programmers usually had the generated assembler code in mind.
At that time the compiler did not check at all if the data you passed to a function when calling it matched the data types the function expected to receive. It was the programmer's responsibility to do that right. If, for example, the caller called the function with a char and the function expected an int the program crashed, although the compiler didn't complain.
Today's type checking prevents these errors, but with a variable argument list you go back to those old concepts including all risks. So, don't use it if you can avoid it somehow.
The fact that this concept is several decades old is probably the reason that it feels wrong compared to modern concepts of safe code.