I have come across some code that looks like it is forward declaring a struct but I can not find any definition for the struct in the code base. It seems to be used as though the struct was defined. Could someone explain why the below code is valid c++?
What type is Frame? What is the size? I cannot use sizeof() as it will complain it is undefined.
I am trying to convert a similar piece of code to Visual Studio 2015 from 2010. The reinterpret_cast cast is complaining that it cannot be converted due to the fact that
'reinterpret_cast': conversion from 'unsigned long' to 'Frame *' of
greater size
#include <stdio.h>
struct Frame;
int main()
{
unsigned long currentFrame = 5;
Frame* frame = reinterpret_cast<Frame*>(currentFrame);
printf("%p", frame);
}
GCC 4.9.2 was used to compile this example.
I understand the error, but do not understand how the struct is being used. Is it defaulting to int?
The program behaviour is undefined, as a conversion from an unsigned long to Frame* where the former is set to a value not associated with a pointer value that you can set is not in accordance with one of the possibilities mentioned in http://en.cppreference.com/w/cpp/language/reinterpret_cast.
The fact that printf appears to output the address of a pointer is a manifestation of that undefined behaviour.
The fact that Frame is an incomplete type does not matter here. With the exception of nullptr, one past the address of a scalar (i.e. single object or a plain-old-data object), and one past the end of an array, the behaviour on setting a pointer type to memory you don't own is also undefined.
Since you are using Frame just as a pointer the compiler doesn't need to know anything about Frame structure itself. It's like using an opaque pointer to something without caring what's pointed.
The cast fails because unsigned long is not guaranteed to be the same size of a pointer according to operating system and data model (eg LLP64 vs LP64). You should consider using intptr_t from <stdint.h> which is guaranteed to be able to store all the bits of a pointer but I don't see how you could need to reinterpred a literal to a memory address.
Related
I've been running the following code through different compilers:
int main()
{
float **a;
void **b;
b = a;
}
From what I've been able to gather, void ** is not a generic pointer which means that any conversion from another pointer should not compile or at least throw a warning. However, here are my results (all done on Windows):
gcc - Throws a warning, as expected.
g++ - Throws an error, as expected (this is due to the less permissive typing of C++, right?)
MSVC (cl.exe) - Throws no warnings whatsoever, even with /Wall specified.
My question is: Am I missing something about the whole thing and is there any specific reason why MSVC does not produce a warning? MSVC does produce a warning when converting from void ** to float **.
Another thing of note: If I replace a = b with the explicit conversion a = (void **)b, none of the compilers throw a warning. I thought this should be an invalid cast, so why wouldn't there be any warnings?
The reason I am asking this question is because I was starting to learn CUDA and in the official Programming Guide (https://docs.nvidia.com/cuda/cuda-c-programming-guide/index.html#device-memory) the following code can be found:
// Allocate vectors in device memory
float* d_A;
cudaMalloc(&d_A, size);
which should perform an implicit conversion to void ** for &d_A, as the first argument of cudaMalloc is of type void **. Similar code can be found all over the documentation. Is this just sloppy work on NVIDIA's end or am I, again, missing something? Since nvcc uses MSVC, the code compiles without warnings.
Am I missing something about the whole thing and is there any specific reason why MSVC does not produce a warning? MSVC does produce a warning when converting from void ** to float **
This assignment without a cast is a constraint violation, so a standard compliant compiler will print a warning or error. However, MSVC is not fully compliant C implementation.
Another thing of note: If I replace a = b with the explicit conversion a = (void **)b, none of the compilers throw a warning. I thought this should be an invalid cast, so why wouldn't there be any warnings?
Pointer conversions via a cast are allowed in some situations. The C standard says the following in section 6.3.2.3p7:
A pointer to an object type may be converted to a pointer to a different object type. If the
resulting pointer is not correctly aligned for the referenced type, the behavior is
undefined. Otherwise, when converted back again, the result shall compare equal to the
original pointer. When a pointer to an object is converted to a pointer to a character type,
the result points to the lowest addressed byte of the object. Successive increments of the
result, up to the size of the object, yield pointers to the remaining bytes of the object.
So you can convert between pointer types provided there are no alignment issues and you only convert back (unless the target is a char *).
float* d_A;
cudaMalloc(&d_A, size);
...
Is this just sloppy work on NVIDIA's end or am I, again, missing something?
Presumably, this function is dereferencing the given pointer and writing the address of some allocated memory. That would mean it is trying to write to a float * as if it were a void *. This is not the same as the typical conversion to/from a void *. Strictly speaking this looks like undefined behavior, although it "works" because modern x86 processors (when not in real mode) use the same representation for all pointer types.
I came across a line of code written in C++:
long *lbuf = (long*)spiReadBuffer;
And it turns out that "spiReadBuffer" is a byte array with 12 elements. But I am a little confused. I think I am familiar with defining pointers and I can see that "lbuf" is a type "long" pointer. Also I thought for casting we can do something like this:
y = (int) x;
But what if I put a "*" after the "int" just like my first example, where there is one after "long"?
I apologize if this is a really trivial question, but as I went through the type casting and pointers topics I did not come across my case and I did not really understand it.
I would appreciate it if you could guide me or introduce me to any relevant materials or resources.
This is called type punning. It tricks the compiler into reading the memory occupied by an object as if it was of another type.
In your case, the array spiReadBuffer decays to a pointer to its first element, then the pointer is cast and stored. When you dereference this pointer, you will access the beginning of the array as if it were a long.
The problem with this approach is that it triggers undefined behaviour (see strict aliasing). So even though it works in a lot of situations, it can also break without notice.
There are two ways (that I know of) to type-pun safely. The first one is standard-compliant : std::memcpy.
char spiReadBuffer[12];
long rbAsLong;
std::memcpy(&rbAsLong, &spiReadBuffer, sizeof rbAsLong);
// rbAsLong contains the first four bytes of spiReadBuffer, reinterpreted as a long.
The second one involves an extension that is often provided by compilers (but you should check), that extends the behaviour of unions.
union {
char buf[12];
long asLong;
} spiReadBuffer;
The standard states that writing to a member of a union then reading from another member is undefined behaviour. These compiler extensions choose to define it as a safe reinterpretation.
in C/C++ arrays are treated the same way by the compiler:
char spiReadBuffer[12];
char* pBuffer;
the compiler will treat both spiReadBuffer and pBuffer as pointers.
The code snippet
long *lbuf = (long*)spiReadBuffer;
is an example of type casting, only it's for pointer types. A char* is converted to a long*; You could say this is a type of pointer arithmetic because now, you can read sizeof(long) bytes from spiReadBuffer using the long* ( instead of one byte at a time ).
The second snippet you showed : y = (int) x; is also a cast, but not for pointers;
Consider this snippet:
char spiReadBuffer[] = {1,2,3,4,5,6,7,8};
long *lbuf = (long*)spiReadBuffer;
printf ("%08x\n", lbuf[0]);
It will print 04030201 on a little endian architecture or 01020304 on a little endian architecture.
After the long *lbuf = (long*)spiReadBuffer statement lBuf points to the beginning of the spiReadBuffer and lbuf[0] (or *lBuf) allows you to read the first 4 bytes of spiReadBuffer as a long.
Is this statement correct? Can any "TYPE" of pointer can point to any other type?
Because I believe so, still have doubts.
Why are pointers declared for definite types? E.g. int or char?
The one explanation I could get was: if an int type pointer was pointing to a char array, then when the pointer is incremented, the pointer will jump from 0 position to the 2 position, skipping 1 position in between (because int size=2).
And maybe because a pointer just holds the address of a value, not the value itself, i.e. the int or double.
Am I wrong? Was that statement correct?
Pointers may be interchangeable, but are not required to be.
In particular, on some platforms, certain types need to be aligned to certain byte-boundaries.
So while a char may be anywhere in memory, an int may need to be on a 4-byte boundary.
Another important potential difference is with function-pointers.
Pointers to functions may not be interchangeable with pointers to data-types on many platforms.
It bears repeating: This is platform-specific.
I believe Intel x86 architectures treat all pointers the same.
But you may well encounter other platforms where this is not true.
Every pointer is of some specific type. There's a special generic pointer type void* that can point to any object type, but you have to convert a void* to some specific pointer type before you can dereference it. (I'm ignoring function pointer types.)
You can convert a pointer value from one pointer type to another. In most cases, converting a pointer from foo* to bar* and back to foo* will yield the original value -- but that's not actually guaranteed in all cases.
You can cause a pointer of type foo* to point to an object of type bar, but (a) it's usually a bad idea, and (b) in some cases, it may not work (say, if the target types foo and bar have different sizes or alignment requirements).
You can get away with things like:
int n = 42;
char *p = (char*)&n;
which causes p to point to n -- but then *p doesn't give you the value of n, it gives you the value of the first byte of n as a char.
The differing behavior of pointer arithmetic is only part of the reason for having different pointer types. It's mostly about type safety. If you have a pointer of type int*, you can be reasonably sure (unless you've done something unsafe) that it actually points to an int object. And if you try to treat it as an object of a different type, the compiler will likely complain about it.
Basically, we have distinct pointer types for the same reasons we have other distinct types: so we can keep track of what kind of value is stored in each object, with help from the compiler.
(There have been languages that only have untyped generic pointers. In such a language, it's more difficult to avoid type errors, such as storing a value of one type and accidentally accessing it as if it were of another type.)
Any pointer can refer to any location in memory, so technically the statement is correct. With that said, you need to be careful when reinterpreting pointer types.
A pointer basically has two pieces of information: a memory location, and the type it expects to find there. The memory location could be anything. It could be the location where an object or value is stored; it could be in the middle of a string of text; or it could just be an arbitrary block of uninitialised memory.
The type information in a pointer is important though. The array and pointer arithmetic explanation in your question is correct -- if you try to iterate over data in memory using a pointer, then the type needs to be correct, otherwise you may not iterate correctly. This is because different types have different sizes, and may be aligned differently.
The type is also important in terms of how data is handled in your program. For example, if you have an int stored in memory, but you access it by dereferencing a float* pointer, then you'll probably get useless results (unless you've programmed it that way for a specific reason). This is because an int is stored in memory differently from the way a float is stored.
Can any "TYPE" of pointer can point to any other type?
Generally no. The types have to be related.
It is possible to use reinterpret_cast to cast a pointer from one type to another, but unless those pointers can be converted legally using a static_cast, the reinterpret_cast is invalid. Hence you can't do Foo* foo = ...; Bar* bar = (Bar*)foo; unless Foo and Bar are actually related.
You can also use reinterpret_cast to cast from an object pointer to a void* and vice versa, and in that sense a void* can point to anything -- but that's not what you seem to be asking about.
Further you can reinterpret_cast from object pointer to integral value and vice versa, but again, not what you appear to be asking.
Finally, a special exception is made for char*. You can initialize a char* variable with the address of any other type, and perform pointer math on the resulting pointer. You still can't dereference thru the pointer if the thing being pointed to isn't actually a char, but it can then be casted back to the actual type and used that way.
Also keep in mind that every time you use reinterpret_cast in any context, you are dancing on the precipice of a cliff. Dereferencing a pointer to a Foo when the thing it actually points to is a Bar yields Undefined Behavior when the types are not related. You would do well to avoid these types of casts at all costs.
Some pointers are more equal than others...
First of all, not all pointers are necessarily the same thing. Function pointers can be something very different from data pointers, for instance.
Aside: Function pointers on PPC
On the PPC platform, this was quite obvious: A function pointer was actually two pointers under the hood, so there was simply no way to meaningfully cast a function pointer to a data pointer or back. I.e. the following would hold:
int* dataP;
int (*functionP)(int);
assert(sizeof(dataP) == 4);
assert(sizeof(functionP) == 8);
assert(sizeof(dataP) != sizeof(functionP));
//impossible:
//dataP = (int*)functionP; //would loose information
//functionP = (int (*)(int))dataP; //part of the resulting pointer would be garbage
Alignment
Furthermore, there is problems with alignment: Depending on the platform some data types may need to be aligned in memory. This is especially common with vector data types, but could apply to any type larger than a byte. For instance, if an int must be 4 byte aligned, the following code might crash:
char a[4];
int* alias = (int*)a;
//int foo = *alias; //may crash because alias is not aligned properly
This is not an issue if the pointer comes from a malloc() call, as that is guaranteed to return sufficiently aligned pointers for all types:
char* a = malloc(sizeof(int));
int* alias = (int*)a;
*alias = 0; //perfectly legal, the pointer is aligned
Strict aliasing and type punning
Finally, there are strict aliasing rules: You must not access an object of one type through a pointer to another type. Type punning is forbidden:
assert(sizeof(float) == sizeof(uint32_t));
float foo = 42;
//uint32_t bits = *(uint32_t*)&foo; //type punning is illegal
If you absolutely must reinterpret a bit pattern as another type, you must use memcpy():
assert(sizeof(float) == sizeof(uint32_t));
float foo = 42;
uint32_t bits;
memcpy(&bits, &foo, sizeof(bits)); //bit pattern reinterpretation is legal when copying the data
To allow memcpy() and friends to actually be implementable, the C/C++ language standards provide for an exception for char types: You can cast any pointer to a char*, copy the char data over to another buffer, and then access that other buffer as some other type. The results are implementation defined, but the standards allow it. Use cases are mostly general data manipulation routines like I/O, etc.
TL;DR:
Pointers are much less interchangeable than you think. Don't reinterpret pointers in any other way than to/from char* (check alignment in the "from" case). And even that does not work for function pointers.
So recently I've been dealing with WinAPI's WriteProcessMemory and ReadProcessMemory which both take LPVOID as their second argument which is the actual address in the memory of another process we want to read from.
Considering that, would this integer to pointer conversion be correct C++ code (without undefined behavior, etc.):
#include <windows.h>
int main()
{
WriteProcessMemory(/*...*/,
//want to read from memory address 0x1234 in the target process
reinterpret_cast<void*>(static_cast<uintptr_t>(0x1234)),
/*...*/);
return 0;
}
AFAIK, the standard says that uintptr_t is an integer type which is able to hold a pointer without the value of the pointer being changed. Since it is an integer, I should be able to store 0x1234 in it. The standard also says that reinterpret_casting an uintptr_t to a void* will also leave the value unchanged.
On Windows platforms the pointer is stored exactly in the same way as the same bit-count integer type. The difference is how the compiler interpretes the value of a pointer and an int. So no conversion is neccessary.
Casting it to void */LPVOID is just to make the compiler happy and accept the integer value in a place for pointer and no conversion is made. Going through static_cast+reinterpret_cast is overkill in this case.
As long as you are sure that the correct address is 0x1234, conversion is not needed. Just a plain (LPVOID)0x1234 or (void *)0x1234 should do.
[edit] As noted in the comments the constant may get truncated. Then no conversion can fix the pointer value. A way to handle this is to explicitly specify the type used in the constant:
E.g. (void *)0x1234567812345678ULL will be OK because the compiler is instructed to use 64bit unsigned type.
Is it safe to cast pointer to int and later back to pointer again?
How about if we know if the pointer is 32 bit long and int is 32 bit long?
long* juggle(long* p) {
static_assert(sizeof(long*) == sizeof(int));
int v = reinterpret_cast<int>(p); // or if sizeof(*)==8 choose long here
do_some_math(v); // prevent compiler from optimizing
return reinterpret_cast<long*>(v);
}
int main() {
long* stuff = new long(42);
long* ffuts = juggle(stuff);
std::cout << "Is this always 42? " << *ffuts << std::endl;
}
Is this covered by the Standard?
No.
For instance, on x86-64, a pointer is 64-bit long, but int is only 32-bit long. Casting a pointer to int and back again makes the upper 32-bit of the pointer value lost.
You may use the intptr_t type in <cstdint> if you want an integer type which is guaranteed to be as long as the pointer. You could safely reinterpret_cast from a pointer to an intptr_t and back.
Yes, if... (or "Yes, but...") and no otherwise.
The standard specifies (3.7.4.3) the following:
A pointer value is a safely-derived pointer [...] if it is the result of a well-defined pointer conversion or reinterpret_cast of a safely-derived pointer value [or] the result of a reinterpret_cast of an integer representation of a safely-derived pointer value
An integer value is an integer representation of a safely-derived pointer [...] if its type is at least as large as std::intptr_t and [...] the result of a reinterpret_cast of a safely-derived pointer value [or]
the result of a valid conversion of an integer representation of a safely-derived pointer value [or] the result of an additive or bitwise operation, one of whose operands is an integer representation of a
safely-derived pointer value
A traceable pointer object is [...] an object of an integral type that is at least as large as std::intptr_t
The standard further states that implementations may be relaxed or may be strict about enforcing safely-derived pointers. Which means it is unspecified whether using or dereferencing a not-safely-derived pointer invokes undefined behavior (that's a funny thing to say!)
Which alltogether means no more and no less than "something different might work anyway, but the only safe thing is as specified above".
Therefore, if you either use std::intptr_t in the first place (the preferrable thing to do!) or if you know that the storage size of whatever integer type you use (say, long) is at least the size of std::intptr_t, then it is allowable and well-defined (i.e. "safe") to cast to your integer type and back. The standard guarantees that.
If that's not the case, the conversion from pointer to integer representation will probably (or at least possibly) lose some information, and the conversion back will not give a valid pointer. Or, it might by accident, but this is not guaranteed.
An interesting anecdote is that the C++ standard does not directly define std::intptr_t at all; it merely says "the same as 7.18 in the C standard".
The C standard, on the other hand, states "designates a signed integer type with the property that any valid
pointer to void can be converted to this type, then converted back to pointer to void, and the result will compare equal to the original pointer".
Which means, without the rather complicated definitions above (in particular the last bit of the first bullet point), it wouldn't be allowable to convert to/from anything but void*.
Yes and no.
The language specification explicitly states that it is safe (meaning that in the end you will get the original pointer value) as long as the size of the integral type is sufficient to store the [implementation-dependent] integral representation of the pointer.
So, in general case it is not "safe", since in general case int can easily turn out to be too small. In your specific case it though it might be safe, since your int might be sufficiently large to store your pointer.
Normally, when you need to do something like that, you should use the intptr_t/uintptr_t types, which are specifically introduced for that purpose. Unfortunately, intptr_t/uintptr_t are not the part of the current C++ standard (they are standard C99 types), but many implementations provide them nevertheless. You can always define these types yourself, of course.
In general, no; pointers may be larger than int, in which case there's no way to reconstruct the value.
If an integer type is known to be large enough, then you can; according to the Standard (5.2.10/5):
A pointer converted to an integer of sufficient size ... and back to the same pointer type will have its original value
However, in C++03, there's no standard way to tell which integer types are large enough. C++11 and C99 (and hence in practice most C++03 implementations), and also Boost.Integer, define intptr_t and uintptr_t for this purpose. Or you could define your own type and assert (preferably at compile time) that it's large enough; or, if you don't have some special reason for it to be an integer type, use void*.
Is it safe? Not really.
In most circumstances, will it work? Yes
Certainly if an int is too small to hold the full pointer value and truncates, you won't get your original pointer back (hopefully your compiler will warn you about this case, with GCC truncating conversions from pointer to integers are hard errors). A long, or uintptr_t if your library supports it, may be better choices.
Even if your integer type and pointer types are the same size, it will not necessarily work depending on your application runtime. In particular, if you're using a garbage collector in your program it might easily decide that the pointer is no longer outstanding, and when you later cast your integer back to a pointer and try to dereference it, you'll find out the object was already reaped.
Absolutely not. Doing some makes a bad assumption that the size of an int and a pointer are the same. This is almost always no the case on 64 bit platforms. If they are not the same a precision loss will occur and the final pointer value will be incorrect.
MyType* pValue = ...
int stored = (int)pValue; // Just lost the upper 4 bytes on a 64 bit platform
pValue = (MyType*)stored; // pValue is now invalid
pValue->SomeOp(); // Kaboom
No, it is not (always) safe (thus not safe in general). And it is covered by the standard.
ISO C++ 2003, 5.2.10:
A pointer can be explicitly converted to any integral type large enough to hold it. The mapping function is implementation-defined.
A value of integral type or enumeration type can be explicitly converted to a pointer. A pointer converted to an integer of sufficient size (if any such exists on the implementation) and back to the same pointer type will have its original value; mappings between pointers and integers are otherwise implementation-defined.
(The above emphases are mine.)
Therefore, if you know that the sizes are compatible, then the conversion is safe.
#include <iostream>
// C++03 static_assert.
#define ASSURE(cond) typedef int ASSURE[(cond) ? 1 : -1]
// Assure that the sizes are compatible.
ASSURE(sizeof (int) >= sizeof (char*));
int main() {
char c = 'A';
char *p = &c;
// If this program compiles, it is well formed.
int i = reinterpret_cast<int>(p);
p = reinterpret_cast<char*>(i);
std::cout << *p << std::endl;
}
Use uintptr_t from "stdint.h" or from "boost/stdint.h". It is guaranteed to have enough storage for a pointer.
No it is not. Even if we rule out the architecture issue, size of a pointer and an integer have differences. A pointer can be of three types in C++ : near, far, and huge. They have different sizes. And if we talk about an integer its normally of 16 or 32 bit. So casting integer into pointers and vice-verse is not safe. Utmost care has to be taken, as there very much chances of precision loss. In most of the cases an integer will be short of space to store a pointer, resulting in loss of value.
If your going to be doing any system portable casting, you need to use something like Microsofts INT_PTR/UINT_PTR, the safety after that relies on the target platforms and what you intend doing to the INT_PTR. generally for most arithmatic char* or uint_8* works better while being typesafe(ish)
To an int ? not always if you are on a 64 bit machine then int is only 4 bytes, however pointers are 8 bytes long and thus you would end up with a different pointer when you cast it back from int.
There are however ways to get around this. You can simply use an 8 byte long data type ,which would work whether or not you are on 32/64 bit system, such as unsigned long long unsigned because you don't want sign extension on 32-bit systems.
It is important to note that on Linux unsigned long will always be pointer size* so if you are targeting Linux systems you could just use that.
*According to cppreference and also tested it myself but not on all Linux and Linux like systems
If the issue is that you want to do normal math on it, probably the safest thing to do would be to cast it to a pointer to char (or better yet, * uint8_t), do your math, and then cast it back.