Are char*, int*, long* or even long long* of same size (on a given platform)?
- 2My curiosity begs the question: Why do you care? :-)Pete– Pete2009-08-06 20:29:41 +00:00Commented Aug 6, 2009 at 20:29
- char c[100]; int *i; i = &c[0]; i = 0x5555u; question: c[0] == 0x55u? c[1] == 0x55u?c-stranger– c-stranger2009-08-06 21:25:38 +00:00Commented Aug 6, 2009 at 21:25
- Assigning the address of a char to a pointer to int is just not a good idea....KTC– KTC2009-08-06 23:45:07 +00:00Commented Aug 6, 2009 at 23:45
- 1@c-stranger: most certainly not guaranteeed. If the machine is big-endian, then the first two bytes of c would be 0. Also, there is no guarantee that c is aligned as strictly as an int would be aligned - though it probably is.Jonathan Leffler– Jonathan Leffler2009-08-07 03:48:46 +00:00Commented Aug 7, 2009 at 3:48
- 5This is important question if you are writing portable c, have a struct with pointers, and you want the memory to be aligned. brewx.qualcomm.com/bws/content/gi/common/appseng/en/…Tero Niemi– Tero Niemi2012-10-05 21:48:04 +00:00Commented Oct 5, 2012 at 21:48
8 Answers
They're not guaranteed to be the same size, although on the platforms I have experience with they usually are.
C 2011 online draft:
6.2.5 Types
...
28 A pointer tovoidshall have the same representation and alignment requirements as a pointer to a character type.48) Similarly, pointers to qualified or unqualified versions of compatible types shall have the same representation and alignment requirements. All pointers to structure types shall have the same representation and alignment requirements as each other. All pointers to union types shall have the same representation and alignment requirements as each other. Pointers to other types need not have the same representation or alignment requirements.
48) The same representation and alignment requirements are meant to imply interchangeability as arguments to functions, return values from functions, and members of unions.
1 Comment
Not necessarily. The standard does not guarantee sizeof(double*) == sizeof(int*). For instance, assume a processor with two address buses with different widths (like on some Harvard architecture processors) it could have pointers with different sizes.
14 Comments
void* cannot be dereferenced directly - you have to cast it back to the original type. Presumably, even if representation of pointers actually differs between types, the original representation will be restored when you cast back.int and double (it could be any other two types). The point is that sizes of pointers themselves can differ.sizeof on one of them in order to figure out how many bytes to allocate when allocating an array of pointers, since malloc expects a number of bytes.sizeof (except perhaps when applied to a variable-length array type) has exactly zero overhead; it resolves to a constant. sizeof (some_type*) is better than 8 simply because it's more explicit -- and it means the code might be portable to other systems.Note what the C standard says - as quoted by John Bode. Note, too, that the C standard says nothing about the sizes of pointers to functions at all.
The POSIX standard lays down some extra requirements:
2.12.3 Pointer Types
All function pointer types shall have the same representation as the type pointer to void. Conversion of a function pointer to void * shall not alter the representation. A void * value resulting from such a conversion can be converted back to the original function pointer type, using an explicit cast, without loss of information.
Note: The ISO C standard does not require this, but it is required for POSIX conformance.
1 Comment
dlsym() rather than a more general statement. Sometime (maybe this decade), I need to write up the change.On 16-bit embedded processors which have banked (paged) RAM and/or flash, the use of pages may lead to pointers being different sizes - though this is independent of the size of the data they point to.
For example, on Freescale's HCS12 processor which has banked flash, data pointers are all 16 bits.
However, function pointers are 16 bits for near pointers (for code in the same page as the calling function, or in unbanked flash), or 24 bits for far pointers (for code in a different page), incorporating the page number in the address.
Things are complicated if you want to store constant data in paged flash, as due to the limitation on data pointer size, the function using the data has to be in the same page as the data being accessed.
It is conceivable that a 16-bit processor with banked RAM would likewise have different sizes for near and far data pointers.
1 Comment
far or not. Function pointers is a bit more advanced since not only the size of the pointer differs from 16 and 24, but you must also use the correct instructions on the assembler level: JSR versus CALL and RTS versus RTC.There is no such guarantee in either C or C++ ISO standards, but in practice, I've yet to see a platform where this doesn't hold.
Note that regardless of this, reinterpret_cast'ing one pointer to another will more often than not lead to U.B., with a few exceptions (void*, and unsigned char* for PODs). So would any union tricks. So the obvious question is: why would you care?
Comments
Generally yes, All pointers to anything, whether they point to a int or a long or a string or an array of strings or a function, point to a single memory address, which is the same size on a machine. That is because the Processer on a machine has a an address register that these pointers are loaded into, and the size of that address register controls the size of the pointers.
The only exception might be in cases like old Intel 8088 16-bit machines where there was a two step process to determine the memory address, involving a 16 bit segment pointer, (which identified a 64K block of memory within the 1MByte address space), and then a second 16 bit memory address to identify the specific memory address within that segment. These two 16 bit addresses were then combined to get the complete 20 bit memory address. In that scenario, then, I imagine it might be possible to distinquiah between the individual 16 bit addresses, and the combined 20-bit address.
5 Comments
char* (let's assume CHAR_BIT == 8) might need to hold both a word pointer and a byte offset within the word. I've worked on Cray vector systems that did something similar, except that the byte offset was stored in the otherwise unused high-order 3 bits of the 64-bit word.