I have a problem with below code:
#include <iostream.h>
#include <stdlib.h>
int main()
{
unsigned long long int i,j;
unsigned long long int *Array;
i=j=0;
Array=(unsigned long long int*)malloc(18446744073709551616);
for(i=0ULL;i<18446744073709551616;i++)
*(Array+i)=i;
std::cin>>j;
std::cout<<*(Array+j);
return 0;
}
My compiler (Borland C++ builder 6.0) gives me an Access Violation Error. There are also warnings on the stage of compiling the program. I have never used unsigned long long int, so I have no idea where is the problem in this case.
The issue you're facing is due to the fact that malloc cannot possibly return a valid pointer to a block of memory that's is of the requested size due to memory constraints your system is faced with, thus malloc does what it normally does when it cannot allocate the desired memory -- it returns nullptr. (malloc reference here)
The most relevant portion of the web-page linked is the following:
Return Value:
On success, a pointer to the memory block allocated by the function.
The type of this pointer is always void*, which can be cast to the desired type of data pointer in order to be dereferenceable.
If the function failed to allocate the requested block of memory, a null pointer is returned.
The reason why you are getting an access violation error is due to the fact that you are trying to deference a pointer that is pointing to a null (hence invalid) location in memory.
In the future, I recommend you try to allocate more reasonably sized blocks of memory (For instance, 1kb, 1mb etc). If you wish to use an unsigned long long int you should perhaps look into creating something pertaining to math instead of memory manipulation.
Addendum:
If you want to get the max value for the type you should have done something like the following:
std::numeric_limits<unsigned long long int>::max(); as noted by iBug, the number you input undergoes unsigned integer overflow. Integer overflow for unsigned integers is a defined behavior so the actual value of the magic value 18446744073709551616 is 0; you are malloc'ing 0 bytes.
The behavior for malloc'ing 0 bytes is the following (As per the c standard):
If the size of the space requested is zero, the behavior is implementation defined: either a null pointer is returned, or the behavior is as if the size were some nonzero value, except that the returned pointer shall not be used to access an object.
You still cannot deference the pointer returned or use it as an object.
As an aside: *(Array + i) = i; is equivalent to Array[i] = i; or even i[Array] = i; :)
unsigned long long is non-negative 64-bit integer, its highest possible value is 264-1 = 18446744073709551615. the compiler may not have known what you meant by 18446744073709551616 (value overflowed).
Also, 264 equals to 16 EB, or 16,777,216 TB. I don't know where such a huge storage is available, even if it's not RAM.
All malloc can do is finding that the requested size is too huge to allocate and return you a null pointer. Then when you're trying to access a null pointer, you get an "Access Violation" error.
Related
I have a return value from a library which is a void pointer. I know that it points to a short int; I try to obtain the int value in the following way (replacing the function call with a simple assignment to a void *):
short n = 1;
void* s = &n;
int k = *(int*)s;
I try to cast a void pointer that points to an address in which there is a short and I try to cast the pointer to point to an int and when I do so the output becomes a rubbish value. While I understand why it's behaving like that I don't know if there's a solution to this.
If the problem you are dealing with truly deals with short and int, you can simply avoid the pointer and use:
short n = 1;
int k = n;
If the object types you are dealing with are different, then the solution will depend on what those types are.
Update, in response to OP's comment
In a comment, you said,
I have a function that returns a void pointer and I would need to cast the value accordingly.
If you know that the function returns a void* that truly points to a short object, then, your best bet is:
void* ptr = function_returning_ptr();
short* sptr = reinterpret_cast<short*>(ptr);
int k = *sptr;
The last line work since *sptr evaluates to a short and the conversion of a short to an int is a valid operation. On the other hand,
int k = *(int*)sptr;
does not work since conversion of short* to an int* is not a valid operation.
Your code is subject to undefined behavior, as it violates the so-called strict aliasing rules. Without going into too much detail and simplifying a bit, the rule states that you can not access an object of type X though a pointer to type Z unless types X and Z are related. There is a special exception for char pointer, but it doesn't apply here.
In your example, short and int are not related types, and as such, accessing one through pointer to another is not allowed.
The size of a short is only 16 bits the size of a int is 32 bits ( in most cases not always) this means that you are tricking the computer into thinking that your pointer to a short is actually pointing to an integer. This causes it to read more memory that it should and is reading garbage memory. If you cast s to a pointer to a short then deference it it will work.
short n = 1;
void* s = &n;
int k = *(short*)s;
Assuming you have 2 byte shorts and 4 byte ints, There's 3 problems with casting pointers in your method.
First off, the 4 byte int will necessarily pick up some garbage memory when using the short's pointer. If you're lucky the 2 bytes after short n will be 0.
Second, the 4 byte int may not be properly aligned. Basically, the memory address of a 4 byte int has to be a multiple of 4, or else you risk bus errors. Your 2 byte short is not guaranteed to be properly aligned.
Finally, you have a big-endian/little-endian dependency. You can't turn a big-endian short into a little-endian int by just tacking on some 0's at the end.
In the very fortunate circumstance that the bytes following the short are 0, AND the short is integer aligned, AND the system uses little-endian representation, then such a cast will probably work. It would be terrible, but it would (probably) work.
The proper solution is to use the original type and let the compiler cast. Instead of int k = *(int*)s;, you need to use int k = *(short *)s;
char b = 'a';
int *a = (int*)&b;
std::cout << *a;
What could be the content of *a? It is showing garbage value. Can you anyone please explain. Why?
Suppose char takes one byte in memory and int takes two bytes (the exact number of bytes depends of the platform, but usually they are not same for char and int). You set a to point to the memory location same as b. In case of b dereferencing will consider only one byte because it's of type char. In case of a dereferencing will access two bytes and thus will print the integer stored at these locations. That's why you get a garbage: first byte is 'a', the second is random byte - together they give you a random integer value.
Either the first or the last byte should be hex 61 depending on byte order. The other three bytes are garbage. best to change the int to an unsigned int and change the cout to hex.
I don't know why anyone would want to do this.
You initialize a variable with the datatype char ...
a char in c++ should have 1 Byte and an int should contain 2 Byte. Your a points to the address of the b variable... an adress should be defined as any hexadecimal number. Everytime you call this "program" there should be any other hexadecimal number, because the scheduler assigns any other address to your a variable if you start this program new.
Think of it as byte blocks. Char has one byte block (8 bits). If you set a conversion (int*) then you get the next 7 byte blocks from the char's address. Therefore you get 7 random byte blocks which means you'll get a random integer. That's why you get a garbage value.
The code invokes undefined behavior, garbage is a form of undefined behavior, but your program could also cause a system violation and crash with more consequences.
int *a = (int*)&b; initializes a pointer to int with the address of a char. Dereferencing this pointer will attempt to read an int from that address:
If the address is misaligned and the processor does not support misaligned accesses, you may get a system specific signal or exception.
If the address is close enough to the end of a segment that accessing beyond the first byte causes a segment violation, that's what you can get.
If the processor can read the sizeof(int) bytes at the address, only one of those will be a, (0x61 in ASCII) but the others have undetermined values (aka garbage). As a matter of fact, on some architectures, reading from uninitialized memory may cause problems: under valgrind for example, this will cause a warning to be displayed to the user.
All the above are speculations, undefined behavior means anything can happen.
I am an expert C# programmer, but I am very new to C++. I get the basic idea of pointers just fine, but I was playing around. You can get the actual integer value of a pointer by casting it as an int:
int i = 5;
int* iptr = &i;
int ptrValue = (int)iptr;
Which makes sense; it's a memory address. But I can move to the next pointer, and cast it as an int:
int i = 5;
int* iptr = &i;
int ptrValue = (int)iptr;
int* jptr = (int*)((int)iptr + 1);
int j = (int)*iptr;
and I get a seemingly random number (although this is not a good PSRG). What is this number? Is it another number used by the same process? Is it possibly from a different process? Is this bad practice, or disallowed? And if not, is there a use for this? It's kind of cool.
What is this number? Is it another number used by the same process? Is it possibly from a different process?
You cannot generally cast pointers to integers and back and expect them to be dereferencable. Integers are numbers. Pointers are pointers. They are totally different abstractions and are not compatible.
If integers are not large enough to be able to store the internal representation of pointers (which is likely the case; integers are usually 32 bits long and pointers are usually 64 bits long), or if you modify the integer before casting it back to a pointer, your program exhibits undefined behaviour and as such anything can happen.
See C++: Is it safe to cast pointer to int and later back to pointer again?
Is this bad practice, or disallowed?
Disallowed? Nah.
Bad practice? Terrible practice.
You move beyond i pointer by 4 or 8 bytes and print out the number, which might be another number stored in your program space. The value is unknown and this is Undefined Behavior. Also there is a good chance that you might get an error (that means your program can blow up) [Ever heard of SIGSEGV? The Segmentation violation problem]
You are discovering that random places in memory contain "unknown" data. Not only that, but you may find yourself pointing to memory that your process does not have "rights" to so that even the act of reading the contents of an address can cause a segmentation fault.
In general is you allocate some memory to a pointer (for example with malloc) you may take a look at these locations (which may have random data "from the last time" in them) and modify them. But data that does not belong explicitly to a pointer's block of memory can behave all kings of undefined behavior.
Incidentally if you want to look at the "next" location just to
NextValue = *(iptr + 1);
Don't do any casting - pointer arithmetic knows (in your case) exactly what the above means : " the contents of the next I refer location".
int i = 5;
int* iptr = &i;
int ptrValue = (int)iptr;
int* jptr = (int*)((int)iptr + 1);
int j = (int)*iptr;
You can cast int to pointer and back again, and it will give you same value
Is it possibly from a different process? no it's not, and you can't access memory of other process except using readProcessMemmory and writeProcessMemory under win32 api.
You get other number because you add 1 to the pointer, try to subtract 1 and you will same value.
When you define an integer by
int i = 5;
it means you allocate a space in your thread stack, and initialize it as 5. Then you get a pointer to this memory, which is actually a position in you current thread stack
When you increase your pointer by 1, it means you point to the next location in your thread stack, and you parse it again as an integer,
int* jptr = (int*)((int)iptr + 1);
int j = (int)*jptr;
Then you will get an integer from you thread stack which is close to where you defined your int i.
Of course this is not suggested to do, unless you want to become an hacker and want to exploit stack overflow (here it means what it is, not the site name, ha!)
Using a pointer to point to a random address is very dangerous. You must not point to an address unless you know what you're doing. You could overwrite its content or you may try to modify a constant in read-only memory which leads to an undefined behaviour...
This for example when you want to retrieve the elements of an array. But cannot cast a pointer to integer. You just point to the start of the array and increase your pointer by 1 to get the next element.
int arr[5] = {1, 2, 3, 4, 5};
int *p = arr;
printf("%d", *p); // this will print 1
p++; // pointer arithmetics
printf("%d", *p); // this will print 2
It's not "random". It just means that there are some data on the next address
Reading a 32-bit word from an address A will copy the 4 bytes at [A], [A+1], [A+2], [A+3] into a register. But if you dereference an int at [A+1] then the CPU will load the bytes from [A+1] to [A+4]. Since the value of [A+4] is unknown it may make you think that the number is "random"
Anyway this is EXTREMELY dangerous 💀 since
the pointer is misaligned. You may see the program runs fine because x86 allows for unaligned accesses (with some performance penalty). But most other architectures prohibit unaligned operations and your program will just end in segmentation fault. For more information read Purpose of memory alignment, Data Alignment: Reason for restriction on memory address being multiple of data type size
you may not be allowed to touch the next byte as it may be outside of your address space, is write-only, is used for another variable and you changed its value, or whatever other reasons. You'll also get a segfault in that case
the next byte may not be initialized and reading it will crash your application on some architectures
That's why the C and C++ standard state that reading memory outside an array invokes undefined behavior. See
How dangerous is it to access an array out of bounds?
Access array beyond the limit in C and C++
Is accessing a global array outside its bound undefined behavior?
I know this is a rather noobish question, but no amount of googling or permutations of code seem to work.
I have a structure which is defined like this.
typedef struct
{
int rate;
int duration;
} DummyStructure;
Now, i try to use code similar to this.
//
DummyStructure* structure;
DummyStructure* structure2;
long int point;
//
structure = (DummyStructure*)malloc(sizeof(DummyStructure));
structure->rate = 19;
structure->duration = 92;
point = (long int)&structure;
// now i'd like to set structure2 to the same memory location as 1.
// point is the 8-byte int location (i assume?) of structure.
// so naturally i'd assume that by casting as a DummyStructure pointer
// that long int would act as a pointer to that 1.
// It doesn't.
structure2 = (DummyStructure*)point;
I stress that i've tried every permutation of ** and * that is possible. I just don't get it. Either it doesn't compile, or it does, and when it does i end up with seemingly random numbers for the fields contained in structure2. I assume that somehow i'm winding up with an incorrect memory location, but how else can you get it except from using the &?
I have the memory location, right? How do i set the structure to that location?
EDIT; I forgot to mention (and subsequent answers have asked) but i'm intending to use this to wrap libvorbis for jni. Using jni means that i can't pass-back any of the structs that libvorbis does, but it requires them for its core functions. Therefore my wrapper is going to use vorbis directly to make the structs, and i pass back to java the pointer to them so that when i need to fill the buffer with more sound, i can simply re-reference the struct objects from the integer value of the pointer.
Why are you trying to cast pointers to integers and back? Is it just to learn, to figure something out, to work around some (untold) restriction, or what? It's a rather strange thing to be doing in a "plain" program such as this, as there is no point.
One likely cause of your problems is that there's no guarantee that a pointer will even fit in a long int. You can check by adding this code:
printf("a regular pointer is %u bytes, long int is %u",
(unsigned int) sizeof structure, (unsigned int) sizeof point);
If the numbers printed are different, that's probably the largest cause of your problems.
If you're using C99, you should #include <stdint.h> and then use the intptr_t type instead of unsigned long to hold a pointer, in any case.
structure is already a pointer, so you don't have to take the address there:
long int point = reinterpret_cast<long int>(structure);
DummyStructure* structure2 = reinterpret_cast<DummyStructure*>(point);
structure is already a pointer. You just want to do point = (long int) structure; (although, realistically, why a long int is involved at all, I don't know. It's a lot easier to just do structure2=structure; which works fine since structure and structure2 are both pointers.)
When you do &structure you're getting the memory location where the pointer itself is stored, which is why it isn't the correct value. You really probably don't want to ever use &structure unless it's being passed into a function which is going to change which DummyStructure structure points to.
Others have answered your question, but I'd like to make a more general comment. You mention JNI; in this case, you don't want long int, but jlong (which will be a typedef to either long int or long long int, depending on the machine. The problem is that long will have a different size, depending on the machine, and will map to a different Java type. Of course, you're counting on the fact that jlong will be big enough to hold a pointer, but since jlong is 64 bits, this seems like a safe bet for the immediate future (and beyond—I don't see a time coming where 64 bits of addressing doesn't suffice).
Also, I would suggest you borrow an idea from Swig, and avoid the subtleties of pointer to integral conversions, by using something like the following:
jlong forJNI = 0;
*reinterpret_cast<DummyStructure*>( &forJNI ) = structure;
// ...
structure2 = *reinterpret_cast<DummyStructure*>( &forJNI );
This is ugly, but it is guaranteed to work (with one caveat) for all
systems where sizeof(DummyStructure*) <= 64.
Just be sure to compile with strict aliasing turned off. (You have to
do this anytime you cast between pointers and ints. IMHO, you shouldn't
have to in cases where the casts are visible to the compiler, but some
compiler writers prefer breaking code intentionally, even when the
intent is clear.)
Long ints aren't the same as pointers. Why don't you just do:
DummyStructure** point;
structure = malloc(sizeof(DummyStructure));
structure->rate = 19;
structure->duration = 92;
point = &structure;
structure2 = *point;
The problem is probably a combination of the fact that 1) you don't dereference point. structure2 is a pointer to structure which is itself a pointer. You'd have to do:
structure2 = *((DummyStructure*)point);
But on top of that is the fact that long ints aren't the same as pointers. There's probably also a signedness issue here.
point = (long int)&structure;
This takes the address of structure which is a DummyStructure* and assign it to point. So point should be a double pointer (pointer to pointer). And when you assign structure2, it should be properly type casted.
typedef struct
{
int rate;
int duration;
} DummyStructure;
DummyStructure* structure;
DummyStructure* structure2;
long int **point;
structure = (DummyStructure*)malloc(sizeof(DummyStructure));
structure->rate = 19;
structure->duration = 92;
point = (long int **)&structure;
structure2 = (DummyStructure*)*point;
If your intention is to make structure2 point to the same memory location as structure, why don't you directly assign it rather than having an intermediate long int **.
The bug is that point is the address of structure, which is itself a pointer to a DummyStructure. In order for structure2 to point to the same thing as structure, you need to dereference point. Ignoring for a second all length, signedness, and similar issues,
structure2 = *(DummyStructure**)point;
would fix your code. But why not just:
structure2 = structure;
If you really want to hold a pointer in something generic, hold it in a void*. At least that's the right size.
Today I've a weird question.
The Code(C++)
#include <iostream>
union name
{
int num;
float num2;
}oblong;
int main(void)
{
oblong.num2 = 27.881;
std::cout << oblong.num << std::endl;
return 0;
}
The Code(C)
#include <stdio.h>
int main(void)
{
float num = 27.881;
printf("%d\n" , num);
return 0;
}
The Question
As we know, C++ unions can hold more than one type of data element but only one type at a time. So basically the name oblong will only reserve one portion of memory which is 32-bit (because the biggest type in the union is 32-bit, int and float) and this portion could either hold a integer or float.
So I just assign a value of 27.881 into oblong.num2 (as you can see on the above code). But out of curiosity, I access the memory using oblong.num which is pointing to the same memory location.
As expected, it gave me a value which is not 27 because the way float and integer represented inside a memory is different, that's why when I use oblong.num to access the memory portion it'll treat that portion of memory value as integer and interpret it using integer representation way.
I know this phenomena also will happen in C , that's why I initialize a float type variable with a value and later on read it using the %d.So I just try it out by using the same value 27.881 which you can see above. But when I run it, something weird happens, that is the value of the one I get in C is different from C++.
Why does this happen? From what I know the two values I get from the two codes in the end are not garbage values, but why do I get different values? I also use the sizeof to verified both C and C++ integer and float size and both are 32-bit. So memory size isn't the one that causes this to happen, so what prompt this difference in values?
First of all, having the wrong printf() format string is undefined behavior. Now that said, here is what is actually happening in your case:
In vararg functions such as printf(), integers smaller than int are promoted to int and floats smaller than double are promoted to double.
The result is that your 27.881 is being converted to an 8-byte double as it is passed into printf(). Therefore, the binary representation is no longer the same as a float.
Format string %d expects a 4-byte integer. So in effect, you will be printing the lower 4-bytes of the double-precision representation of 27.881. (assuming little-endian)
*Actually (assuming strict-FP), you are seeing the bottom 4-bytes of 27.881 after it is cast to float, and then promoted to double.
In both cases you are encountering undefined behaviour. Your implementation just happens to do something strange.