问题
So I'm trying to write a buffering library for the 64th time and I'm starting get into some pretty advanced stuff. Thought I'd ask for some proffesional input on this.
In my first header file I have this:
typedef struct StdBuffer { void* address; } StdBuffer;
extern void StdBufferClear(StdBuffer);
In another header file that #includes the first header file I have this:
typedef struct CharBuffer { char* address; } CharBuffer;
void (*CharBufferClear)(CharBuffer) = (void*) StdBufferClear;
Will declaring this function pointer void interfere with the call? They have matching by value signatures. I have never seen a function pointer declared void before, but its the only way to get it to compile cleanly.
Stackwise it should not make any difference at all from what I learned in assembler coding.
irrelevent OMG! I just said Stackwise on StackOverflow!
Hmm.. Looks like I've assumed too much here. Allow me to reclarify if I may. I don't care what 'type' of data is stored at the address. All that I am concerned with is the size of a 'unit' and how many units are at the address. Take a look at the interface agreement contract for the API if you will:
typedef struct StdBuffer {
size_t width; ///< The number of bytes that complete a data unit.
size_t limit; ///< The maximum number of data units that can be allocated for this buffer.
void * address; ///< The memory address for this buffer.
size_t index; ///< The current unit position indicator.
size_t allocated; ///< The current number of allocated addressable units.
StdBufferFlags flags;///< The API contract for this buffer.
} StdBuffer;
You see, memcpy, memmove and the like don't really care whats at an address all they want is the specifics which I'm clearly keeping track of here.
Have a look now at the first prototype to follow this contract:
typedef struct CharBuffer {
size_t width; ///< The number of bytes that complete a data unit.
size_t limit; ///< The maximum number of data units that can be allocated for this buffer.
char * address; ///< The memory address for this buffer.
size_t index; ///< The current unit position indicator.
size_t allocated; ///< The current number of allocated addressable units.
CharBufferFlags flags;///< The API contract for this buffer.
} CharBuffer;
As you an clearly see the data type is irrelevant in this context. You can say that C handles it differently depending on the case, but at the end of the day, an address is an address, a byte is byte and a long is a long for as long as we are dealing with memory on the same machine.
The purpose of this system when brought together is to remove all of this type based juggling C seems to be so proud of (and rightfully so...) Its just pointless for what I would like to do. Which is create a contract abiding prototype for any standard size of data (1, 2, 4, 8, sizeof(RandomStruct)) located at any address.
Having the ability to perform my own casting with code and manipulate that data with api functions that operate on specific length blocks of memory with specific length memory units. However, the prototype must contain the official data pointer type, because it just doesn't make sense for the end user to have to recast their data every time they would like to do something with that address pointer. It would not make sense to call it a CharBuffer if the pointer was void.
The StdBuffer is a generic type that is never EVER used except within the api itself, to manage all contract abiding data types.
The api that this system will incorporate is from my latest edition of buffering. Which is quite clearly documented here @Google Code I am aware that some things will need to change to bring this all together namely I won't have the ability to manipulate data directly from within the api safely without lots of proper research and opinion gathering.
Which just brought to my attention that I also need a Signed/Unsigned bit flag in the StdBufferFlags Members.
Perhaps the final piece to this puzzle is also in order for your perusal.
/** \def BIT(I)
\brief A macro for setting a single constant bit.
*
* This macro sets the bit indicated by I to enabled.
* \param I the (1-based) index of the desired bit to set.
*/
#define BIT(I) (1UL << (I - 1))
/** \enum StdBufferFlags
\brief Flags that may be applied to all StdBuffer structures.
* These flags determine the contract of operations between the caller
* and the StdBuffer API for working with data. Bits 1-4 are for the
* API control functions. All other bits are undefined/don't care bits.
*
* If your application would like to use the don't care bits, it would
* be smart not to use bits 5-8, as these may become used by the API
* in future revisions of the software.
*/
typedef enum StdBufferFlags {
BUFFER_MALLOCD = BIT(1), ///< The memory address specified by this buffer was allocated by an API
BUFFER_WRITEABLE = BIT(2), ///< Permission to modify buffer contents using the API
BUFFER_READABLE = BIT(3), ///< Permission to retrieve buffer contents using the API
BUFFER_MOVABLE = BIT(4) ///< Permission to resize or otherwise relocate buffer contents using the API
}StdBufferFlags;
回答1:
This code requires a diagnostic:
void (*CharBufferClear)(CharBuffer) = (void*) StdBufferClear;
You're converting a void * pointer to a function pointer without a cast. In C, a void * pointer can convert to pointers to object types without a cast, but not to function pointer types. (In C++, a cast is needed to convert void * to object types also, for added safety.)
What you want here is just to cast between function pointer types, i.e.:
void (*CharBufferClear)(CharBuffer) = (void (*)(CharBuffer)) StdBufferClear;
Then you are still doing the same type punning because the functions are different types. You are trying to call a function which takes a StdBuffer using a pointer to a function which takes a CharBuffer.
This type of code is not well-defined C. Having defeated the type system, you're on your own, relying on testing, examining the object code, or obtaining some assurances from the compiler writers that this sort of thing works with that compiler.
What you learned in assembler coding doesn't apply because assembly languages have only a small number of rudimentary data types such as "machine address" or "32 bit word". The concept that two data structures with an identical layout and low-level representation might be incompatible types does not occur in assembly language.
Even if two types look the same at the low level (another example: unsigned int and unsigned long are sometimes exactly the same) C compilers can optimize programs based on the assumption that the type rules have not been violated. For instance suppose that A and B point to the same memory location. If you assign to an an object A->member, a C compiler can assume that the object B->member is not affected by this, if A->member and B->member have incompatible types, like one being char * and the other void *. The generated code keeps caching the old value of B->member in a register, even though the in-memory copy was overwritten by the assignment to A->member. This is an example of invalid aliasing.
回答2:
The standard does not define the results of casting a function-pointer to void *.
Equally, converting between function pointers and then calling through the wrong one is also undefined behaviour.
回答3:
There are some constructs which any standards-conforming C compiler are required to implement consistently, and there are some constructs which 99% of C compilers do implement consistently, but which standards-conforming compilers would be free to implement differently. Attempting to cast a pointer to a function which takes one type of pointer, into a pointer to a function which takes another type of pointer, falls into the latter category. Although the C standard specifies that a void* and a char* must be the same size, there is nothing that would require that they share the same bit-level storage format, much less the parameter-passing convention. While most machines allow bytes to be accessed in much the same way as words, such ability is not universal. The designer of an application-binary-interface [the document which specifies among other things how parameters are passed to routines] might specify that a char* be passed in a way which maximizes the efficiency of byte access, while a void* should be passed in a way that maximizes the efficiency of word access while retaining the ability to hold an unaligned byte address, perhaps by using a supplemental word to hold a zero or one to indicate LSB/MSB). On such a machine, having a routine that expects a void* called from code that expects to pass a char* could cause the routine to access arbitrary wrong data.
回答4:
No, it doesn't matter what data type is used to store the data. It only matters the type C uses to read and write that data, and that the data is of sufficient size.
来源：https://stackoverflow.com/questions/10305192/what-is-the-effect-of-casting-a-function-pointer-void