For a method with an argument of void* is C++ employing static/reinterpret_cast to make the conversion or is there a different mechanism in play here?
void foo(void* p)
{
// ... use p by casting it back to Base first, using static/reinterpret cast
}
Base* base(new Derived);
foo(base); // at this exact line, is there a static/reinterpret_cast going on under the covers?
I am asking because it seems that on one hand the standard says that for c-style cast, C++ will go and try a C++ cast (static, reinterpret, const) until something that works is found. However I can't find a reasonable explanation as to what goes on when a method with a void* argument is called. On the face of thing there is no cast, so what happens?
The language specification does not express the behavior in terms of static_cast or reinterpret_cast in this case. It just says that the pointer base is implicitly converted to void * type. Conversions that can be performed implicitly are called standard conversions in C++. Conversion of any object pointer type to void * is one of the standard conversions from pointer conversions category.
It is true that this is the same conversion that can be preformed explicitly by static_cast, but static_cast is completely irrelevant in this case. Standard conversions in C++ work "by themselves" without involving any specific cast operators under the cover.
In fact, it is the behavior of static_cast that is defined in terms of standard conversions for such cases, not the other way around.
Related
This can be compiled (despite being UB (right?) because fvp == nullptr)
int f;
void* fvp{};
decltype(f)* fp = static_cast<decltype(f)*>(fvp);
but this cannot
void f() {}
void* fvp{};
decltype(f)* fp = static_cast<decltype(f)*>(fvp);
because (Clang says)
Static_cast from 'void *' to 'decltype(f) ' (aka 'void ()()') is not allowed [bad_cxx_cast_generic]
Why is this the case? And where from the standard do I understand this? I guess [expr.static.cast] is where I should look at; specifically 7, but I'm not sure.
Follow up question on the reason why I wanted to ask this one.
This is not allowed because [expr.static_cast] does not allow it. Indeed, the relevant language is:
No other conversion shall be performed explicitly using a static_cast.
Since there is no conversion listed that would permit conversion from void* to a function pointer type, that conversion is not allowed.
The rationale is that function and data pointers may have different sizes or alignment requirements.
You might consider a reinterpret_cast, which conditionally supports such conversions. But it would be better to use a design that does not require such conversions. If you need to type-erase a function pointer, you will still need reinterpret_cast but can use e.g. void(*)() as the erased type, which is always supported.
where from the standard do I understand this? I guess [expr.static.cast] is where I should look at; specifically 7, but I'm not sure
Basically, there is no guarantee for a full-round trip between void* and a function pointer.
This can be seen from expr.reinterpret.cast#8 which says:
Converting a function pointer to an object pointer type or vice versa is conditionally-supported. The meaning of such a conversion is implementation-defined, except that if an implementation supports conversions in both directions, converting a prvalue of one type to the other type and back, possibly with different cv-qualification, shall yield the original pointer value.
C++17 makes noexcept part of a function's type. It also allows implicit conversions from noexcept function pointers to potentially throwing function pointers.
void (*ptr_to_noexcept)() noexcept = nullptr;
void (*ptr_to_throwing)() = ptr_to_noexcept; // implicit conversion
http://eel.is/c++draft/expr.static.cast#7 says that static_cast can perform the inverse of such a conversion.
void (*noexcept_again)() noexcept = static_cast<void(*)() noexcept>(ptr_to_throwing);
Unfortunately, both GCC and clang tell me otherwise: https://godbolt.org/z/TgrL7q
What is the correct way to do this? Are reinterpret_cast and C style cast my only options?
You might have skipped over the important part:
The inverse of any standard conversion sequence not containing an lvalue-to-rvalue, array-to-pointer, function-to-pointer, null pointer, null member pointer, boolean, or function pointer conversion, can be performed explicitly using static_cast.
Currently, a function pointer conversion includes only the conversion from noexcept to potentially throwing. Because you're doing the inverse of a function pointer conversion, static_cast will not work, just like you can't static_cast a pointer to an array, or any of the other conversions listed there.
So yes, reinterpret_cast would be appropriate and also raises the appropriate alarm bells that should come with discarding noexcept.
The behavior of my code is different in c and c++.
void *(*funcPtr)() = dlsym(some symbol..) ; // (1) works in c but not c++
int (*funcPtr)();
*(void**)(&funcPtr) = dlsym(some symbol..) ; // (2) works in c++
I dont understand how the 2nd casting works in c++ while the 1st casting doesnt work in c++. The error message for (1) show is invalid conversion from void* to void*() in c++.
The problem is that dlsym returns a void* pointer. While in C, any such pointer is implicitly convertible into any other (object!) pointer type (for comparison: casting the result of malloc), this is not the case in C++ (here you need to cast the result of malloc).
For function pointers, though, this cast is not implicit even in C. Apparently, as your code compiles, your compiler added this implicit cast for function pointers, too, as a compiler extension (in consistency with object pointers); however, for being fully standard compliant, it actually should issue a diagnostic, e. g. a compiler warning, mandated by C17 6.5.4.3:
Conversions that involve pointers, other than where permitted by the constraints of 6.5.16.1, shall be specified by means of an explicit cast.
But now instead of casting the target pointer, you rather should cast the result of dlsym into the appropriate function pointer:
int (*funcPtr)() = reinterpret_cast<int(*)()>(dlsym(some symbol..));
or simpler:
auto funcPtr = reinterpret_cast<int(*)()>(dlsym(some symbol..));
or even:
int (*funcPtr)() = reinterpret_cast<decltype(funcPtr)>(dlsym(some symbol..));
(Last one especially interesting if funcPtr has been declared previously.)
Additionally, you should prefer C++ over C style casts, see my variants above. You get more precise control over what type of cast actually occurs.
Side note: Have you noticed that the two function pointers declared in your question differ in return type (void* vs. int)? Additionally, a function not accepting any arguments in C needs to be declared as void f(void); accordingly, function pointers: void*(*f)(void). C++ allows usage of void for compatibility, while skipping void in C has totally different meaning: The function could accept anything, you need to know from elsewhere (documentation) how many arguments of which type actually can be passed to.
Strictly speaking, none of this code is valid in either language.
void *(*funcPtr)() = dlsym(some symbol..) ;
dlsym returns type void* so this is not valid in either language. The C standard only allows for implicit conversions between void* and pointers to object type, see C17 6.3.2.3:
A pointer to void may be converted to or from a pointer to any object type. A pointer to
any object type may be converted to a pointer to void and back again; the result shall
compare equal to the original pointer.
Specifically, your code is a language violation of one of the rules of simple assignment, C17 6.5.16.1, emphasis mine:
the left operand has atomic, qualified, or unqualified pointer type, and (considering
the type the left operand would have after lvalue conversion) one operand is a pointer
to an object type, and the other is a pointer to a qualified or unqualified version of
void, and the type pointed to by the left has all the qualifiers of the type pointed to
by the right;
The reason why it might compile on certain compilers is overly lax compiler settings. If you are for example using gcc, you need to compile with gcc -std=c17 -pedantic-errors to get a language compliant compiler, otherwise it defaults to the non-standard "gnu11" language.
int (*funcPtr)();
*(void**)(&funcPtr) = dlsym(some symbol..) ; // (2) works in c++
Here you explicitly force the function pointer to become type void**. The cast is fine syntax-wise, but this may or may not be a valid pointer conversion on the specific compiler. Again, conversions between object pointers and function pointers are not supported by the C or C++ standard, but you are relying on non-standard extensions. Formally this code invokes undefined behavior in both C and C++.
In practice, lots of compilers have well-defined behavior here, because most systems have the same representation of object pointers and function pointers.
Given that the conversion is valid, you can of course de-reference a void** to get a void* and then assign another void* to it.
The C++ standards mentions that reinterpret_cast is implementation defined, and doesn't give any guarantees except that casting back (using reinterpret_cast) to original type will result in original value passed to first.
C-style casting of at least some types behaves much the same way - casting back and forth results with the same value - Currently I am working with enumerations and ints, but there are some other examples as well.
While C++ standard gives those definitions for both cast-styles, does it also give the same guarantee for mixed casts? If library X returns from function int Y() some enum value, can use any of above casts, without worrying what cast was used to convert initial enum to int in Y's body? I don't have X's source code, so I cannot check (and it can change with next version anyway), and things like that are hardly mentioned in documentation.
I know that under most implementations in such cases both casts behave the same; my question is: what does C++ standard say about such cases - if anything at all.
C++ defines the semantic of the C cast syntax in terms of static_cast, const_cast and reinterpret_cast. So you get the same guaranteed for the same operation whatever syntax you use to achieve it.
reinterpret_cast can only be used for specific conversions:
Pointer to (sufficiently large) integer, and the reverse
Function pointer to function pointer
Object pointer to object pointer
Pointer-to-member to pointer-to-member
lvalue expression to reference
plus (conditionally) function pointer to object pointer and the reverse. In most cases, the converted value is unspecified, but there is a guarantee that a conversion followed by its reverse will yield the original value.
In particular, you can't use reinterpret_cast to convert between integer an enumeration types; the conversion must be done using static_cast (or implicitly, when converting an unscoped enumeration to an integer type), which is well defined for sufficiently large integer types. The only possible problem is if the library did something completely insane such as return reinterpret_cast<int&>(some_enum);
A C-style cast will perform either a static_cast or a reinterpret_cast, followed by a const_cast, as necessary; so any conversion that's well-defined by static_cast is also well-defined by a C-style cast.
No, reinterpret_cast is not equivalent to a C style cast. C style casts allow casting away const-volatile (so it includes the functionality of const_cast) not allowed in reinterpret_cast. If static_cast is allowed between the source and destination types, it will perform a static_cast which has different semantics than reinterpret_cast. It the conversion is not allowed, it will fallback to reinterpret_cast. Finally there is a corner case where the C cast cannot be represented in terms of any of the other casts: it ignores access specifiers.
Some examples that illustrate differences:
class b0 { int a; };
class b1 { int b; };
class b2 { int c; };
class d : public b0, public b1, b2 {};
int main() {
d x;
assert( static_cast<b1*>(&x) == (b1*)&x );
assert( reinterpret_cast<b1*>(&x) != (b1*)&x ); // Different value
assert( reinterpret_cast<b2*>(&x) != (b2*)&x ); // Different value,
// cannot be done with static_cast
const d *p = &x;
// reinterpret_cast<b0*>(p); // Error cannot cast const away
(b0*)p; // C style can
}
int i = 1000;
void *p = &i;
int *x = static_cast<int*>(p);
int *y = reinterpret_cast<int*>(p);
which cast should be used to convert from void* to int* and why?
static_cast provided that you know (by design of your program) that the thing pointed to really is an int.
static_cast is designed to reverse any implicit conversion. You converted to void* implicitly, therefore you can (and should) convert back with static_cast if you know that you really are just reversing an earlier conversion.
With that assumption, nothing is being reinterpreted - void is an incomplete type, meaning that it has no values, so at no point are you interpreting either a stored int value "as void" or a stored "void value" as int. void* is just an ugly way of saying, "I don't know the type, but I'm going to pass the pointer on to someone else who does".
reinterpret_cast if you've omitted details that mean you might actually be reading memory using a type other than the type is was written with, and be aware that your code will have limited portability.
By the way, there are not very many good reasons for using a void* pointer in this way in C++. C-style callback interfaces can often be replaced with either a template function (for anything that resembles the standard function qsort) or a virtual interface (for anything that resembles a registered listener). If your C++ code is using some C API then of course you don't have much choice.
In current C++, you can't use reinterpret_cast like in that code. For a conversion of void* to int* you can only use static_cast (or the equivalent C-style cast).
For a conversion between different function type pointers or between different object type pointers you need to use reinterpret_cast.
In C++0x, reinterpret_cast<int*>(p) will be equivalent to static_cast<int*>(p). It's probably incorporated in one of the next WPs.
It's a misconception that reinterpret_cast<T*>(p) would interpret the bits of p as if they were representing a T*. In that case it will read the value of p using p's type, and that value is then converted to a T*. An actual type-pun that directly reads the bits of p using the representation of type T* only happens when you cast to a reference type, as in reinterpret_cast<T*&>(p).
As far as I know, all current compilers allow to reinterpret_cast from void* and behave equivalent to the corresponding static_cast, even though it is not allowed in current C++03. The amount of code broken when it's rejected will be no fun, so there is no motivation for them to forbid it.
When should static_cast, dynamic_cast, const_cast and reinterpret_cast be used? gives some good details.
From the semantics of your problem, I'd go with reinterpret, because that's what you actually do.