porting 32 bit to 64bit code - c++

Was trying to port 32bit to 64bit code I was wondering if there are some standard rules when it comes to porting ?
I have my code compiling in a 64bit environment and now I come across some errors like
cast from pointer to integer of different size [-Werror=pointer-to-int-cast] for
x = (int32_t)y;
And to get of this I use x = (size_t)y; I get rid of the error but is this the correct way. Also in various location I have to cast a variable to (unsigned long long). For example
printf("Total Time : %5qu\n",time->compile_time
This gives an error error: format '%qu' expects argument of type 'long long unsigned int', but argument 2 has type (XYZ).
to get this fixed i do something like
printf("Total Time : %5qu\n",(unsigned long long) time->compile_time
Again is this proper ??

I think it's safe to assume that y is a pointer in this case.
Instead of size_t you should use intptr_t or uintptr_t.
See size_t vs. uintptr_t.
As for your second cast it depends what you mean by proper?
The usual advice is to avoid casting. However, like all things in programming there is a reason that they are available. When working on an implementation of malloc on an embedded system I had to cast pointers to uintptr_t in order to be able to do the necessary arithmetic on them. This code was tested on a 64 bit PC but ran on a 32 bit micro controller. The fact that I used two architectures was the best way to ensure it was somewhat portable code.
Casting though makes your code dependent on how the underlying type is defined! Just like you noticed with your x = (int32_t)y this line made your code dependent on the fact that a pointer was 32 bits wide.
The only advice I can give you is to know your types. If you want to cast, it is ok (so long as you can't make your variable of the correct type to begin with) but it may reduce your portability unless you choose the "correct" type to cast to.
The same goes for the printf cast. If I was you, I would read the definition of %5qu thoroughly first (this may help). Then I would attempt to use an appropriately typed variable (or conversely a different format string) and only if that failed would I resort to a cast.
I have never used %qu but I would interpret it as a 64 bit unsigned int so I would try using uint64_t (because long long is not guaranteed to be 64 bits across all platforms). Although by what I read on Wikipedia the q specifier is platform specific to begin with so it might be wise to change it.
Any more than this and the question becomes too broad (it's good that we stuck to specific examples). If you get stuck, come back with individual types that you want to check and ask questions only about them.

Was it Stroustrup that said they call it a 'cast' because it props up something that's broken ? ;-)
x = (int32_t) y;
In this case you are using an exact width type, so it really depends on what x and y are. The error message suggest that y is a pointer. A pointer is not an int32_t, so the real question is why is y being assigned to x ... it may indicate a potential problem. Casting it away may just cover the problem so that it bites you at run-time rather than compile-time. Figure out what the code thinks it's doing and "re-jigger" the types to fit the code. The reason the error goes away when you use a (size_t) cast is that likely the pointer is 64 bits and size_t is 64 bits, but you could consider that a simple form of random casting luck. The same is true when casting to (unsigned long long). Don't assume that an int is 32 or 64 bits and don't using casting as a porting tool ... it will get you in trouble. It's tough to be more specific based on a single line of code. If you want to post a < 25 line function that has issues; more specific advice may be available.

Related

effect of using sprintf / printf using %ld format string instead of %d with int data type

We have some legacy code that at one point in time long data types were refactored to int data types. During this refactor a number of printf / sprintf format statements were left incorrect as %ld instead of changed to %d. For example:
int iExample = 32;
char buf[200];
sprintf(buf, "Example: %ld", iExample);
This code is compiled on both GCC and VS2012 compilers. We use Coverity for static code analysis and code like in the example was flagged as a 'Printf arg type mismatch' with a Medium level of severity, CWE-686: Function Call With Incorrect Argument Type I can see this would be definitely a problem had the format string been that of an signed (%d) with an unsigned int type or something along these lines.
I am aware that the '_s' versions of sprintf etc are more secure, and that the above code can also be refactored to use std::stringstream etc. It is legacy code however...
I agree that the above code really should be using %d at the very least or refactored to use something like std::stringstream instead.
Out of curiosity is there any situation where the above code will generate incorrect results? As this legacy code has been around for quite some time and appears to be working fine.
UPDATED
Removed the usage of the word STL and just changed it to be std::stringstream.
As far as the standard is concerned, the behavior is undefined, meaning that the standard says exactly nothing about what will happen.
In practice, if int and long have the same size and representation, it will very likely "work", i.e., behave as if the correct format string has been used. (It's common for both int and long to be 32 bits on 32-bit systems).
If long is wider than int, it could still work "correctly". For example, the calling convention might be such that both types are passed in the same registers, or that both are pushed onto the stack as machine "words" of the same size.
Or it could fail in arbitrarily bad ways. If int is 32 bits and long is 64 bits, the code in printf that tries to read a long object might get a 64-bit object consisting of the 32 bits of the actual int that was passed combined with 32 bits of garbage. Or the extra 32 bits might consistently be zero, but with the 32 significant bits at the wrong end of the 64-bit object. It's also conceivable that fetching 64 bits when only 32 were passed could cause problems with other arguments; you might get the correct value for iExample, but following arguments might be fetched from the wrong stack offset.
My advice: The code should be fixed to use the correct format strings (and you have the tools to detect the problematic calls), but also do some testing (on all the C implementations you care about) to see whether it causes any visible symptoms in practice. The results of the testing should be used only to determine the priority of fixing the problems, not to decide whether to fix them or not. If the code visibly fails now, you should fix it now. If it doesn't, you can get away with waiting until later (presumably you have other things to work on).
It's undefined and depends on the implementation. On implementations where int and long have the same size, it will likely work as expected. But just try it on any system with 32-bit int and 64-bit long, especially if your integer is not the last format argument, and you're likely to get problems where printf reads 64 bits where only 32 were provided, the rest quite possibly garbage, and possibly, depending on alignment, the following arguments also cannot get accessed correctly.

Why is “cast from ‘X*’ to ‘Y’ loses precision” a hard error and what is suitable fix for legacy code

1. Why?
Code like this used to work and it's kind of obvious what it is supposed to mean. Is the compiler even allowed (by the specification) to make it an error?
I know that it's loosing precision and I would be happy with a warning. But it still has a well-defined semantics (at least for unsigned downsizing cast is defined) and the user just might want to do it.
2. Workaround
I have legacy code that I don't want to refactor too much because it's rather tricky and already debugged. It is doing two things:
Sometimes stores integers in pointer variables. The code only casts the pointer to integer if it stored an integer in it before. Therefore while the cast is downsizing, the overflow never happens in reality. The code is tested and works.
When integer is stored, it always fits in plain old unsigned, so changing the type is not considered a good idea and the pointer is passed around quite a bit, so changing it's type would be somewhat invasive.
Uses the address as hash value. A rather common thing to do. The hash table is not that large to make any sense to extend the type.
The code uses plain unsigned for hash value, but note that the more usual type of size_t may still generate the error, because there is no guarantee that sizeof(size_t) >= sizeof(void *). On platforms with segmented memory and far pointers, size_t only has to cover the offset part.
So what are the least invasive suitable workarounds? The code is known to work when compiled with compiler that does not produce this error, so I really want to do the operation, not change it.
Notes:
void *x;
int y;
union U { void *p; int i; } u;
*(int*)&x and u.p = x, u.i are not equivalent to (int)x and are not the opposite of (void *)y. On big endian architectures, the first two will return the bytes on lower addresses while the later will work on low order bytes, which may reside on higher addresses.
*(int*)&x and u.p = x, u.i are both strict aliasing violations, (int)x is not.
C++, 5.2.10:
4 - A pointer can be explicitly converted to any integral type large enough to hold it. [...]
C, 6.3.2.3:
6 - Any pointer type may be converted to an integer type. [...] If the result cannot be represented in the integer type, the behavior is undefined. [...]
So (int) p is illegal if int is 32-bit and void * is 64-bit; a C++ compiler is correct to give you an error, while a C compiler may either give an error on translation or emit a program with undefined behaviour.
You should write, adding a single conversion:
(int) (intptr_t) p
or, using C++ syntax,
static_cast<int>(reinterpret_cast<intptr_t>(p))
If you're converting to an unsigned integer type, convert via uintptr_t instead of intptr_t.
This is a tough one to solve "generically", because the "looses precision" indicates that your pointers are larger than the type you are trying to store it in. Which may well be "ok" in your mind, but the compiler is concerned that you will be restoring the int value back into a pointer, which has now lost the upper 32 bits (assuming we're talking 32-bit int and 64-bit pointers - there are other possible combinations).
There is uintptr_t that is size-compatible with whatever the pointer is on the systems, so typically, you can overcome the actual error by:
int x = static_cast<int>(reinterpret_cast<uintptr_t>(some_ptr));
This will first force a large integer from a pointer, and then cast the large integer to a smaller type.
Answer for C
Converting pointers to integers is implementation defined. Your problem is that the code that you are talking about seems never have been correct. And probably only worked on ancient architectures where both int and pointers are 32 bit.
The only types that are supposed to convert without loss are [u]intptr_t, if they exist on the platform (usually they do). Which part of such an uintptr_t is appropriate to use for your hash function is difficult to tell, you shouldn't make any assumptions on that. I would go for something like
uintptr_t n = (uintptr_t)x;
and then
((n >> 32) ^ n) & UINT32_MAX
this can be optimized out on 32 bit archs, and would give you traces of all other bits on 64 bit archs.
For C++ basically the same should apply, just the cast would be reinterpret_cast<std:uintptr_t>(x).

INT_PTR in 64 bit conversion

I am converting a large project of multiple c++ applications from 32 bit MFC to 64 bit. These applications will be required to be compiled in both 32 and 64 bit for the forseeable future.
I've run across these types INT_PTR and UINT_PTR. I have two questions.
Is it considered best practice to use these types as a "default" type for general integer purposes, such as loop counters, etc?
I understand that the size of these types are related to the pointer size of the environment you are compiling for, but it seems confusing to use them for general purpose integer. For example for (INT_PTR i = 0; i<10; i++) ... ; i isn't a pointer or pointer related, so the name of the type is confusing to me. Are there better predefines to be used in this situation or should I make my own?
My compiler is VS2010.
Thanks
INT_PTR and similar have a very specific use-case: They model a data type that is large enough and with appropriate alignment requirements to hold an integer or a pointer type. It is used in situations where a method has parameters or a return type that is either an integral data type or a pointer (like GetWindowLongPtr).
For loops there is no generic advice other than this: Use the most appropriate type. If your loop index is used to index into a container, make it a size_t. If your loop index is an integer that runs across all values in between x and y, use the type of x and y, or one that is compatible with both. An INT_PTR is not really an appropriate loop index data type.
You might want to take a look at this: http://msdn.microsoft.com/en-us/library/windows/desktop/aa383751%28v=vs.85%29.aspx/ "Windows Data Types"
You might be a bit confused about the purpose for this type. It's there to ensure that if you cast a pointer to an int (which you probably shouldn't do anyway) you have an int type that fits. Windows is a bit odd in that (if memory serves) int is still 32 bits when compiling for 64 bits. If you need an int of a specific size I'd use the exact width types (in stdint.h)
If your loop is not related to pointers (that is, it is not used as an index of an array or a vector or anything like that), use int whenever a generic integer type is required. It's a good thing to make sure your code doesn't depend on size of int (although for MS VS int is 32 bit for both platforms).
Use size_t, vector<...>::size_type or other appropriate type for index-related loops.
You may use the datatype "intptr_t". It's a platform specific datatype. It holds 4 byte values in 32 bit platform while 8 byte values in 64 bit platform.

What does this C++ construct do?

Somewhere in lines of code, I came across this construct...
//void* v = void* value from an iterator
int i = (int)(long(v))
What possible purpose can this contruct serve?
Why not simply use int(v) instead? Why the cast to long first?
It most possibly silences warnings.
Assuming a 32bit architecture with sizeof(int) < sizeof(long) and sizeof(long) == sizeof(void *) you possibly get a warning if you cast a void * to an int and no warning if you cast a void * to a long as you're not truncating. You then get a warning assigning a long to an int (possible truncation) which is removed by then explicitly casting from a long to an int.
Without knowing the compiler it's hard to say, but I've certainly seen multi-step casts required to prevent warnings. Why not try converting the construct to what you think it should be and see what the compiler says (of course that only helps you to work out what was in the mind of the original programmer if you're using the same compiler and same warning level as they were).
It does eeevil.
On most architectures, a pointer can be considered to be just another kind of number. On most architectures, long is as many bits as a pointer, so there is a 1-to-1 map between long values and pointers. But violations, especially of the second rule, are not uncommon!
long(v) is an alias for reinterpret_cast<long>(v), which carries no guarantees. Not really fit for any purpose, unless your ABI spec says otherwise.
However, for whatever reason, whoever wrote that code prefers int to long. So they again cross their fingers and hope that no essential information is thrown out in the bits that may possibly be lost in the int to long cast.
Two uses of this are creating a unique object identifier, or trying to somehow package the pointer for some kind of arithmetic otherwise unsupported by pointers.
An opaque identifier can be a void*, so casting to integral type is unnecessary.
"Extracting" an integer from a pointer (for e.g. a division operation) can always be done by subtracting a base pointer to obtain a difference of type ptrdiff_t, which is usually long.

How should I handle "cast from ‘void*’ to ‘int’ loses precision" when compiling 32-bit code on 64-bit machine?

I have a package that compiles and works fine on a 32-bit machine. I am now trying to get it to compile on a 64-bit machine and find the following error-
error: cast from ‘void*’ to ‘int’ loses precision
Is there a compiler flag to suppress these errors? or do I have to manually edit these files to avoid these casts?
The issue is that, in 32bits, an int (which is a 32bit integer) will hold a pointer value.
When you move to 64bit, you can no longer store a pointer in an int - it isn't large enough to hold a 64bit pointer. The intptr_t type is designed for this.
Your code is broken. It won't become any less broken by ignoring the warnings the compiler gives you.
What do you think will happen when you try to store a 64-bit wide pointer into a 32-bit integer? Half your data will get thrown away. I can't imagine many cases where that is the correct thing to do, or where it won't cause errors.
Fix your code. Or stay on the 32-bit platform that the code currently works on.
If your compiler defines intptr_t or uintptr_t, use those, as they are integer types guaranteed to be large enough to store a pointer.
If those types are not available, size_t or ptrdiff_t are also large enough to hold a pointer on most (not all) platforms. Or use long (is typically 64-bit on 64-bit platforms on the GCC compiler) or long long (a C99 types which most, but not all compilers, support in C++), or some other implementation-defined integral type that is at least 64 bits wide on a 64-bit platform.
My guess is OP's situation is a void* is being used as general storage for an int, where the void* is larger than the int. So eg:
int i = 123;
void *v = (void*)i; // 64bit void* being (ab)used to store 32bit value
[..]
int i2 = (int)v; // we want our 32bits of the 64bit void* back
Compiler doesn't like that last line.
I'm not going to weigh in on whether it's right or wrong to abuse a void* this way. If you really want to fool the compiler, the following technique seems to work, even with -Wall:
int i2 = *((int*)&v);
Here it takes the address of v, converts the address to a pointer of the datatype you want, then follows the pointer.
It's an error for a reason: int is only half as big as void* on your machine, so you can't just store a void* in an int. You would loose half of the pointer and when the program later tries to get the pointer out of that int again, it won't get anything useful.
Even if the compiler wouldn't give an error the code most likely wouldn't work. The code needs to be changed and reviewed for 64bit compatibility.
Casting a pointer to an int is horrible from a portability perspective. The size of int is defined by the mix of compiler and architecture. This is why the stdint.h header was created, to allow you to explicitly state the size of the type you're using across many different platforms with many different word sizes.
You'd be better off casting to a uintptr_t or intptr_t (from stdint.h, and choose the one that best matches the signedness you need).
You can try to use intptr_t for best portability instead of int where pointer casts are required, such as callbacks.
You do not want to suppress these errors because most likely, they are indicating a problem with the code logic.
If you suppresses the errors, this could even work for a while. While the pointer points to an address in the first 4 GB, the upper 32 bits will be 0 and you won't lose any data. But once you get an address > 4GB, your code will start 'mysteriously' not working.
What you should do is modify any int that can hold a pointer to intptr_t.
You have to manually edit those files in order to replace them with code that isn't likely to be buggy and nonportable.
Suppressing the warnings are a bad idea, but there may be a compiler flag to use 64-bit ints, depending on your compiler and architecture, and this is a safe way to fix the problem (assuming of course that the code didn't also assume ints are 32-bit). For gcc, the flag is -m64.
The best answer is still to fix the code, I suppose, but if it's legacy third-party code and these warnings are rampant, I can't see this refactoring as being a very efficient use of your time. Definitely don't cast pointers to ints in any of your new code, though.
As defined by the current C++ standard, there is no integer type which is guaranteed to hold a pointer. Some platforms will have an intptr_t, but this is not a standard feature of C++. Fundamentally, treating the bits of a pointer as if they were an integer is not a portable thing to do (although it can be made to work on many platforms).
If the reason for the cast is to make the pointer opaque, then void* already achieves this, so the code could use void* instead of int. A typedef might make this a little nicer in the code
typedef void * handle_t;
If the reason for the cast is to do pointer arithmetic with byte granularity, then the best way is probably to cast to a (char const *) and do the math with that.
If the reason for the cast is to achieve compatibility with some existing library (perhaps an older callback interface) which cannot be modified, then I think you need to review the documentation for that library. If the library is capable of supporting the functionality that you require (even on a 64-bit platform), then its documentation may address the intended solution.
I faced similar problem. I solved it in the following way:
#ifdef 64BIT
typedef uint64_t tulong;
#else
typedef uint32_t tulong;
#endif
void * ptr = NULL; //Whatever you want to keep it.
int i;
i = (int)(tulong)ptr;
I think, the problem is of typecasting a pointer to a shorter data type. But for a larger type to int, it works fine.
I converted this problem from typecasting a pointer to long to typecasting a 64-bit integer to 32-bit integer and it worked fine. I am still in the search of a compiler option in GCC/Clang.
Sometimes it is sensible to want to split up a 64-bit item into 2 32-bit items. This is how you would do it:
Header file:
//You only need this if you haven't got a definition of UInt32 from somewhere else
typedef unsigned int UInt32;
//x, when cast, points to the lower 32 bits
#define LO_32(x) (*( (UInt32 *) &x))
//address like an array to get to the higher bits (which are in position 1)
#define HI_32(x) (*( ( (UInt32 *) &x) + 1))
Source file:
//Wherever your pointer points to
void *ptr = PTR_LOCATION
//32-bit UInt containing the upper bits
UInt32 upper_half = HI_32(ptr);
//32-bit UInt containing the lower bits
UInt32 lower_half = LO_32(ptr);