I am new to C++ and also in MPI programming. I have confused about this code block in C++
int count;
count=4;
local_array=(int*)malloc(count*sizeof(int));
Why are we using sizeof(int) here in MPI programming?
I could see that you're trying to allocate 4 ints here.
If you look at malloc's signature, it takes the number of bytes for its first parameter. As stated here, int data type takes 4 bytes.
Therefore, if you want 4 ints, you could have typed local_array=(int*)malloc(count*4);. But not everyone remembers that int actualy takes 4 bytes. That's why you use sizeof to find out the actual size of the object or type.
Related
I am working on a project where data is read from memory. Some of this data are integers, and there was a problem accessing them at unaligned addresses. My idea would be to use memcpy for that, i.e.
uint32_t readU32(const void* ptr)
{
uint32_t n;
memcpy(&n, ptr, sizeof(n));
return n;
}
The solution from the project source I found is similar to this code:
uint32_t readU32(const uint32_t* ptr)
{
union {
uint32_t n;
char data[4];
} tmp;
const char* cp=(const char*)ptr;
tmp.data[0] = *cp++;
tmp.data[1] = *cp++;
tmp.data[2] = *cp++;
tmp.data[3] = *cp;
return tmp.n;
}
So my questions:
Isn't the second version undefined behaviour? The C standard says in 6.2.3.2 Pointers, at 7:
A pointer to an object or incomplete type may be converted to a pointer to a different
object or incomplete type. If the resulting pointer is not correctly aligned 57) for the
pointed-to type, the behavior is undefined.
As the calling code has, at some point, used a char* to handle the memory, there must be some conversion from char* to uint32_t*. Isn't the result of that undefined behaviour, then, if the uint32_t* is not corrently aligned? And if it is, there is no point for the function as you could write *(uint32_t*) to fetch the memory. Additionally, I think I read somewhere that the compiler may expect an int* to be aligned correctly and any unaligned int* would mean undefined behaviour as well, so the generated code for this function might make some shortcuts because it may expect the function argument to be aligned properly.
The original code has volatile on the argument and all variables because the memory contents could change (it's a data buffer (no registers) inside a driver). Maybe that's why it does not use memcpy since it won't work on volatile data. But, in which world would that make sense? If the underlying data can change at any time, all bets are off. The data could even change between those byte copy operations. So you would have to have some kind of mutex to synchronize access to this data. But if you have such a synchronization, why would you need volatile?
Is there a canonical/accepted/better solution to this memory access problem? After some searching I come to the conclusion that you need a mutex and do not need volatile and can use memcpy.
P.S.:
# cat /proc/cpuinfo
processor : 0
model name : ARMv7 Processor rev 10 (v7l)
BogoMIPS : 1581.05
Features : swp half thumb fastmult vfp edsp neon vfpv3 tls
CPU implementer : 0x41
CPU architecture: 7
CPU variant : 0x2
CPU part : 0xc09
CPU revision : 10
This code
uint32_t readU32(const uint32_t* ptr)
{
union {
uint32_t n;
char data[4];
} tmp;
const char* cp=(const char*)ptr;
tmp.data[0] = *cp++;
tmp.data[1] = *cp++;
tmp.data[2] = *cp++;
tmp.data[3] = *cp;
return tmp.n;
}
passes the pointer as a uint32_t *. If it's not actually a uint32_t, that's UB. The argument should probably be a const void *.
The use of a const char * in the conversion itself is not undefined behavior. Per 6.3.2.3 Pointers, paragraph 7 of the C Standard (emphasis mine):
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.
The use of volatile with respect to the correct way to access memory/registers directly on your particular hardware would have no canonical/accepted/best solution. Any solution for that would be specific to your system and beyond the scope of standard C.
Implementations are allowed to define behaviors in cases where the Standard does not, and some implementations may specify that all pointer types have the same representation and may be freely cast among each other regardless of alignment, provided that pointers which are actually used to access things are suitably aligned.
Unfortunately, because some obtuse compilers compel the use of "memcpy" as an
escape valve for aliasing issues even when pointers are known to be aligned,
the only way compilers can efficiently process code which needs to make
type-agnostic accesses to aligned storage is to assume that any pointer of a type requiring alignment will always be aligned suitably for such type. As a result, your instinct that approach using uint32_t* is dangerous is spot on. It may be desirable to have compile-time checking to ensure that a function is either passed a void* or a uint32_t*, and not something like a uint16_t* or a double*, but there's no way to declare a function that way without allowing a compiler to "optimize" the function by consolidating the byte accesses into a 32-bit load that will fail if the pointer isn't aligned.
I read the book "Linux Kernel Development", and find some functions that make me confused, listed as bellow:
struct page *alloc_pages(gfp_t gfp_mask, unsigned int order)
void __free_pages(struct page *page, unsigned int order)
unsigned long __get_free_pages(gfp_t gfp_mask, unsigned int order)
void free_pages(unsigned long addr, unsigned int order)
The problem is the use of the two underline in the function name, and how the function pairs.
1. when will the linux kernel uses two underline in its function name?
2. why alloc_pages is paired with __free_pages, but not free_pages?
As you can notice:
alloc_pages() / __free_pages() takes "page *" (page descriptor) as argument.
They are ususally used internally by some infrastrcture kernel code, like page fault handler, which wish to manipulate page descriptor instead of memory block content.
__get_free_pages() / free_pages() takes "unsigned long" (virtual address of memory block) as argument
They could be used by code which wish to use the memory block itself, after allocation, you can read / write to this memory block.
As for their name and double underscore "__", you don't need to bother too much. Sometimes kernel functions were named casually without too much consideration when they were first written. And when people think of that the names are not proper, but later those functions are already used wildly in kernel, and kernel guys are simply lazy to change them.
Ok I asked this the other day. But the answers I recieved when asked made me realize I was not fit to question it yet until I did some hard research.
So here I am yet again to retry this....
In examples of malloc I seen something as such...
#include <stdio.h>
int main()
{
int ptr_doe;
ptr_doe = (int *) malloc(sizeof(int));
}
I read that it was not neccesary for the:
(this *) malloc(sizeof(int));
And that only
(int *) malloc(sizeof(\\this));
is neccesary. Is the casting before calling the malloc function ever neccesary?
And how do we know how much memory we need to allocate and what the hell is this?
malloc(10 * sizeof(int));
is it multiplying 4 bytes by 10? and when is it neccesary to use malloc? How does it work internally? Thanks for any help guys
1.
Is the casting before calling the malloc function ever neccesary?
If you use a currect compiler, No. See answers to Do I cast the result of malloc?
2.
how do we know how much memory we need to allocate
It depends on your needs.
3.
malloc(10 * sizeof(int));
means to allocate memory big enough to store 10 int value.
4.
is it multiplying 4 bytes by 10?
If sizeof(int) equals 4, yes.
5.
when is it neccesary to use malloc?
If only at runtime you can know how much memory your program needs.
6.
How does it work internally?
Briefly, malloc() will ask operating system for enough memory, record some information it needs, and return you a pointer to the start of that memory segment.
I am new in vc++... I have one doubt in vc++. what is the size of GetTickCount() function.The return type of GetTickCount() is DWORD. Please anyone Answer for my question.
Thanks in Advance
The size of a function means the number of bytes occupied by the code that belongs to the function. You can find this out using a debugger like Windbg. But this is not useful information in most cases. To get the size of a data type, you can use the sizeof operator. Since the return type of GetTickCount is DWORD (4 bytes), you can either do sizeof(DWORD) or sizeof(GetTickCount()) to get its size. There is also a function by the name GetTickCount64 which returns ULONGLONG which is a 64-bit unsigned value (8 bytes).
GetTickCount() returns a DWORD which is 4 bytes. The function itself can be represented using its start address (function pointer) which will have size equal to size of void* which is 4 bytes on 32-bit systems and 8 bytes on 64-bit systems. Finding the size of code that the function occupies can be problematic and is rarely needed.
I have recently been reading about a family of automatic memory management techniques that rely on storing information in the pointer returned by the allocator, i.e. few bits of header e.g. to differentiate between pointers or to store thread-related information (note that I'm not talking about limited-field reference counting here, only immutable information).
I'd like to toy with these techniques. Now, to implement them, I need to be able to return pointers with a specific shape from my allocator. I suppose I could play with the least weight bits but this would require padding that looks extremely memory consuming, so I believe that I should play with the heaviest bits. However, I have no good idea on how to do this. Is there a way for me to, call malloc or malloc_create_zone or some related function and request a pointer that always starts with the given bits?
Thanks everyone!
The amount of information you can actually store in a pointer is pretty limited (typically one or two bits per pointer). And every attempt to dereference the pointer has to first mask out the magic information. The technique is often called tagging, BTW.
#define TAG_MASK 0x3
#define CONS_TAG 0x1
#define STRING_TAG 0x2
#define NUMBER_TAG 0x3
typedef uintptr_t value_t;
typedef struct cons {
value_t car;
value_t cdr;
} cons_t;
value_t
create_cons(value_t t1, value_t t2)
{
cons_t* pair = malloc(sizeof(cons_t));
value_t addr = (value_t)pair;
pair->car = t1;
pair->cdr = t2;
return addr | CONS_TAG;
}
value_t
car_of_cons(value_t v)
{
if ((v % TAG_MASK) != CONS_TAG) error("wrong type of argument");
return ((cons_t*) (v & ~TAG_MASK))->car;
}
One advantage of this technique is, that you can directly infer the type of the object from the pointer itself. You don't need to dereference it (say, in order to read a special type field or similar). Many language implementations using this scheme also have a special tag combination for "immediate" numbers and other small values, which can be represented direcly using the "pointer".
The disadvatage is, that the amount of information, which can be stored, is pretty limited. Also, as the example code shows, you have to be aware of the tagging in every access to the object, and need to "untag" the pointer before actually using it.
The use of the least significant bits for tagging stemms from the observation, that on most platforms, all pointer to malloced memory is actually aligned on a non-byte boundary (usually 8 bytes), so the least significant bits are always zero.