I am little confused with the applicability of reinterpret_cast
vs static_cast
. From what I have read the general rules are to use static cast when the types can be interpreted at compile time hence the word static
. This is the cast the C++ compiler uses internally for implicit casts also.
reinterpret_cast
s are applicable in two scenarios:
convert integer types to pointer types and vice versa
convert one pointer type to another. The general idea I get is this is unportable and should be avoided.
Where I am a little confused is one usage which I need, I am calling C++ from C and the C code needs to hold on to the C++ object so basically it holds a void*
. What cast should be used to convert between the void *
and the Class type?
I have seen usage of both static_cast
and reinterpret_cast
? Though from what I have been reading it appears static
is better as the cast can happen at compile time? Though it says to use reinterpret_cast
to convert from one pointer type to another?
reinterpret_cast
does not happen at run time. They are both compile-time statements. From en.cppreference.com/w/cpp/language/reinterpret_cast: "Unlike static_cast, but like const_cast, the reinterpret_cast expression does not compile to any CPU instructions. It is purely a compiler directive which instructs the compiler to treat the sequence of bits (object representation) of expression as if it had the type new_type."
The C++ standard guarantees the following:
static_cast
ing a pointer to and from void*
preserves the address. That is, in the following, a
, b
and c
all point to the same address:
int* a = new int();
void* b = static_cast<void*>(a);
int* c = static_cast<int*>(b);
reinterpret_cast
only guarantees that if you cast a pointer to a different type, and then reinterpret_cast
it back to the original type, you get the original value. So in the following:
int* a = new int();
void* b = reinterpret_cast<void*>(a);
int* c = reinterpret_cast<int*>(b);
a
and c
contain the same value, but the value of b
is unspecified. (in practice it will typically contain the same address as a
and c
, but that's not specified in the standard, and it may not be true on machines with more complex memory systems.)
For casting to and from void*
, static_cast
should be preferred.
One case when reinterpret_cast
is necessary is when interfacing with opaque data types. This occurs frequently in vendor APIs over which the programmer has no control. Here's a contrived example where a vendor provides an API for storing and retrieving arbitrary global data:
// vendor.hpp
typedef struct _Opaque * VendorGlobalUserData;
void VendorSetUserData(VendorGlobalUserData p);
VendorGlobalUserData VendorGetUserData();
To use this API, the programmer must cast their data to VendorGlobalUserData
and back again. static_cast
won't work, one must use reinterpret_cast
:
// main.cpp
#include "vendor.hpp"
#include <iostream>
using namespace std;
struct MyUserData {
MyUserData() : m(42) {}
int m;
};
int main() {
MyUserData u;
// store global data
VendorGlobalUserData d1;
// d1 = &u; // compile error
// d1 = static_cast<VendorGlobalUserData>(&u); // compile error
d1 = reinterpret_cast<VendorGlobalUserData>(&u); // ok
VendorSetUserData(d1);
// do other stuff...
// retrieve global data
VendorGlobalUserData d2 = VendorGetUserData();
MyUserData * p = 0;
// p = d2; // compile error
// p = static_cast<MyUserData *>(d2); // compile error
p = reinterpret_cast<MyUserData *>(d2); // ok
if (p) { cout << p->m << endl; }
return 0;
}
Below is a contrived implementation of the sample API:
// vendor.cpp
static VendorGlobalUserData g = 0;
void VendorSetUserData(VendorGlobalUserData p) { g = p; }
VendorGlobalUserData VendorGetUserData() { return g; }
void*
for that?
USpoofChecker*
, where USpoofChecker
is an empty struct. However, under the hood, whenever you pass a USpoofChecker*
, it undergoes reinterpret_cast
to an internal C++ type.
struct_a*->void*->struct_a*
is well-defined. On the other hand, struct_a*->void*->struct_b*
and direct atruct_a->struct_b*
isn't.
The short answer: If you don't know what reinterpret_cast
stands for, don't use it. If you will need it in the future, you will know.
Full answer:
Let's consider basic number types.
When you convert for example int(12)
to unsigned float (12.0f)
your processor needs to invoke some calculations as both numbers has different bit representation. This is what static_cast
stands for.
On the other hand, when you call reinterpret_cast
the CPU does not invoke any calculations. It just treats a set of bits in the memory like if it had another type. So when you convert int*
to float*
with this keyword, the new value (after pointer dereferecing) has nothing to do with the old value in mathematical meaning (ignoring the fact that it is undefined behavior to read this value).
Be aware that reading or modifying values after reinterprt_cast'ing are very often Undefined Behavior. In most cases, you should use pointer or reference to std::byte
(starting from C++17) if you want to achieve the bit representation of some data, it is almost always a legal operation. Other "safe" types are char
and unsigned char
, but I would say it shouldn't be used for that purpose in modern C++ as std::byte
has better semantics.
Example: It is true that reinterpret_cast
is not portable because of one reason - byte order (endianness). But this is often surprisingly the best reason to use it. Let's imagine the example: you have to read binary 32bit number from file, and you know it is big endian. Your code has to be generic and works properly on big endian (e.g. some ARM) and little endian (e.g. x86) systems. So you have to check the byte order. It is well-known on compile time so you can write You can write a function to achieve this:
constexpr
function:
/*constexpr*/ bool is_little_endian() {
std::uint16_t x=0x0001;
auto p = reinterpret_cast<std::uint8_t*>(&x);
return *p != 0;
}
Explanation: the binary representation of x
in memory could be 0000'0000'0000'0001
(big) or 0000'0001'0000'0000
(little endian). After reinterpret-casting the byte under p
pointer could be respectively 0000'0000
or 0000'0001
. If you use static-casting, it will always be 0000'0001
, no matter what endianness is being used.
EDIT:
In the first version I made example function is_little_endian
to be constexpr
. It compiles fine on the newest gcc (8.3.0) but the standard says it is illegal. The clang compiler refuses to compile it (which is correct).
short
takes 16 bits in memory. Corrected.
The meaning of reinterpret_cast
is not defined by the C++ standard. Hence, in theory a reinterpret_cast
could crash your program. In practice compilers try to do what you expect, which is to interpret the bits of what you are passing in as if they were the type you are casting to. If you know what the compilers you are going to use do with reinterpret_cast
you can use it, but to say that it is portable would be lying.
For the case you describe, and pretty much any case where you might consider reinterpret_cast
, you can use static_cast
or some other alternative instead. Among other things the standard has this to say about what you can expect of static_cast
(§5.2.9):
An rvalue of type “pointer to cv void” can be explicitly converted to a pointer to object type. A value of type pointer to object converted to “pointer to cv void” and back to the original pointer type will have its original value.
So for your use case, it seems fairly clear that the standardization committee intended for you to use static_cast
.
reinterpret_crash
. No way a compiler bug will stop me from crashing my reinterpreting program. I'll report a bug ASAP!</irony>
template<class T, U> T reinterpret_crash(U a) { return *(T*)nullptr; }
One use of reinterpret_cast is if you want to apply bitwise operations to (IEEE 754) floats. One example of this was the Fast Inverse Square-Root trick:
https://en.wikipedia.org/wiki/Fast_inverse_square_root#Overview_of_the_code
It treats the binary representation of the float as an integer, shifts it right and subtracts it from a constant, thereby halving and negating the exponent. After converting back to a float, it's subjected to a Newton-Raphson iteration to make this approximation more exact:
float Q_rsqrt( float number )
{
long i;
float x2, y;
const float threehalfs = 1.5F;
x2 = number * 0.5F;
y = number;
i = * ( long * ) &y; // evil floating point bit level hacking
i = 0x5f3759df - ( i >> 1 ); // what the deuce?
y = * ( float * ) &i;
y = y * ( threehalfs - ( x2 * y * y ) ); // 1st iteration
// y = y * ( threehalfs - ( x2 * y * y ) ); // 2nd iteration, this can be removed
return y;
}
This was originally written in C, so uses C casts, but the analogous C++ cast is the reinterpret_cast.
error: invalid cast of an rvalue expression of type 'int64_t {aka long long int}' to type 'double&' reinterpret_cast<double&>((reinterpret_cast<int64_t&>(d) >> 1) + (1L << 61))
- ideone.com/6S4ijc
reinterpret_cast
with memcpy
, is it still UB?
memcpy
would definitely make it legal.
Here is a variant of Avi Ginsburg's program which clearly illustrates the property of reinterpret_cast
mentioned by Chris Luengo, flodin, and cmdLP: that the compiler treats the pointed-to memory location as if it were an object of the new type:
#include <iostream>
#include <string>
#include <iomanip>
using namespace std;
class A
{
public:
int i;
};
class B : public A
{
public:
virtual void f() {}
};
int main()
{
string s;
B b;
b.i = 0;
A* as = static_cast<A*>(&b);
A* ar = reinterpret_cast<A*>(&b);
B* c = reinterpret_cast<B*>(ar);
cout << "as->i = " << hex << setfill('0') << as->i << "\n";
cout << "ar->i = " << ar->i << "\n";
cout << "b.i = " << b.i << "\n";
cout << "c->i = " << c->i << "\n";
cout << "\n";
cout << "&(as->i) = " << &(as->i) << "\n";
cout << "&(ar->i) = " << &(ar->i) << "\n";
cout << "&(b.i) = " << &(b.i) << "\n";
cout << "&(c->i) = " << &(c->i) << "\n";
cout << "\n";
cout << "&b = " << &b << "\n";
cout << "as = " << as << "\n";
cout << "ar = " << ar << "\n";
cout << "c = " << c << "\n";
cout << "Press ENTER to exit.\n";
getline(cin,s);
}
Which results in output like this:
as->i = 0
ar->i = 50ee64
b.i = 0
c->i = 0
&(as->i) = 00EFF978
&(ar->i) = 00EFF974
&(b.i) = 00EFF978
&(c->i) = 00EFF978
&b = 00EFF974
as = 00EFF978
ar = 00EFF974
c = 00EFF974
Press ENTER to exit.
It can be seen that the B object is built in memory as B-specific data first, followed by the embedded A object. The static_cast
correctly returns the address of the embedded A object, and the pointer created by static_cast
correctly gives the value of the data field. The pointer generated by reinterpret_cast
treats b
's memory location as if it were a plain A object, and so when the pointer tries to get the data field it returns some B-specific data as if it were the contents of this field.
One use of reinterpret_cast
is to convert a pointer to an unsigned integer (when pointers and unsigned integers are the same size):
int i;
unsigned int u = reinterpret_cast<unsigned int>(&i);
You could use reinterprete_cast to check inheritance at compile time.
Look here: Using reinterpret_cast to check inheritance at compile time
template <class outType, class inType>
outType safe_cast(inType pointer)
{
void* temp = static_cast<void*>(pointer);
return static_cast<outType>(temp);
}
I tried to conclude and wrote a simple safe cast using templates. Note that this solution doesn't guarantee to cast pointers on a functions.
reinterpret_cast
already does in this situation: "An object pointer can be explicitly converted to an object pointer of a different type.[72] When a prvalue v
of object pointer type is converted to the object pointer type “pointer to cv T
”, the result is static_cast<cv T*>(static_cast<cv void*>(v))
." -- N3797.
c++2003
standard I can NOT find that reinterpret_cast
does static_cast<cv T*>(static_cast<cv void*>(v))
C++03
it was C++98
. Tons of projects used old C++ instead of portable C. Sometimes you have to care about portability. For example you have to support the same code on Solaris, AIX, HPUX, Windows. Where it comes to compiler dependency and portability it is tricky. So a good example of introducing a portability hell is to use a reinterpret_cast
in your code
First you have some data in a specific type like int here:
int x = 0x7fffffff://==nan in binary representation
Then you want to access the same variable as an other type like float: You can decide between
float y = reinterpret_cast<float&>(x);
//this could only be used in cpp, looks like a function with template-parameters
or
float y = *(float*)&(x);
//this could be used in c and cpp
BRIEF: it means that the same memory is used as a different type. So you could convert binary representations of floats as int type like above to floats. 0x80000000 is -0 for example (the mantissa and exponent are null but the sign, the msb, is one. This also works for doubles and long doubles.
OPTIMIZE: I think reinterpret_cast would be optimized in many compilers, while the c-casting is made by pointerarithmetic (the value must be copied to the memory, cause pointers couldn't point to cpu- registers).
NOTE: In both cases you should save the casted value in a variable before cast! This macro could help:
#define asvar(x) ({decltype(x) __tmp__ = (x); __tmp__; })
reinterpret_cast
form int
to float&
is undefined behaviour.
memcpy
to purely register operations when possible; the casts are easy (but are also UB — if the value is used — as pointed out all over this page).
Quick answer: use static_cast
if it compiles, otherwise resort to reinterpret_cast
.
Read the FAQ! Holding C++ data in C can be risky.
In C++, a pointer to an object can be converted to void *
without any casts. But it's not true the other way round. You'd need a static_cast
to get the original pointer back.
Success story sharing
b
is not unspecified anymore in C++11 when usingreinterpret_cast
. And in C++03 a cast ofint*
tovoid*
was forbidden to be done withreinterpret_cast
(although compilers did not implement that and it was impractical, hence was changed for C++11).