I am looking at some old code of mine that manipulates data buffers. I have many places that have:
char *ptr1 ;
char *ptr2 ;
And then I need to find the number of bytes between the two.
int distance = ptr2 - ptr1 ;
I am getting a lot of warnings about truncation. What is the type of
ptr2 - ptr1
I have found a number of answers dealing with pointer arithmetic but, oddly not an answer to this particular question.
Pointer Arithmetic
The result of subtracting two pointers† is a std::ptrdiff_t. It is an implementation-defined signed integer; it could be larger than what an int could store.
See http://en.cppreference.com/w/cpp/types/ptrdiff_t for more information.
†You can only subtract pointers if they point to elements of the same array, otherwise it's UB.
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;
I casted the memory address from double to an integer .
Even though they point to the same address why the values are different ?
#include<iostream>
using namespace std;
int main()
{
double d = 2.5;
auto p = (int*)&d;
auto q = &d;
cout<<p<<endl; // prints memory address 0x7fff5fbff660
cout<<q<<endl; // print memory address 0x7fff5fbff660
cout<<*p<<endl; //prints 0
cout<<*q<<endl; // prints 2.5
return 0;
}
But why the value are different
0x7fff5fbff660
0x7fff5fbff660
0
2.5
Program ended with exit code: 0
Suppose you have "11" written on a piece of paper. That is eleven if it's decimal digits. That is two if there's one mark for each value. That's three if it's binary. How you interpret stored information affects the value you understand it to be storing.
double d = 2.5;
auto p = (int*)&d;
auto q = &d;
p and q are created pointing to the same memory location. The memory holds a double (usually 8 bytes)
When you create
auto p = (int*)&d;
you are telling the compiler ( reintepret_cast< int*> ( &d) ) that the value in d was an integer.
So the values of the pointers are the same, but the types are not.
When you print out
cout<<*q<<endl; // prints 2.5
You are displaying the correct value - as it came in and out through that.
when you print out
cout<<*p<<endl; //prints 0
You are looking at 4 (typically) bytes of the 8 byte memory, and interpreting them as an integer.
These happen to be 0x00, 0x00, 0x00, 0x00
It's because you've violated the strict aliasing rule, giving you undefined behavior. You cannot acesss type A through a pointer of type B and just pretend it works.
TL;DR:
if you have an int* pointing to some memory containing an int and then
you point a float* to that memory and use it as a float you break the
rule. If your code does not respect this, then the compiler's
optimizer will most likely break your code.
The memory addresses are the same, and they both point to a double-precision floating point number in memory. However, you've asked the compiler to treat one as an integer and another as a double. (A pointer might just be a memory address, but at compile-time the compiler has information about the type as well.) It just so happens that the in-memory representation of this particular double-precision number looks like a 0 when treated as an integer.
Because you have casted them to different types yourself.
When you do auto p = (int*)&d; you are asking the compiler to store a double value in a memory area allocated for an integer. An integer and a double are represented in different formats in a computer's memory. A double is stored using a floating point representation in memory, while an int is not. This is a classic example of undefined behaviour.
I'm trying to store a couple of ints in memory using void* & then retrieve them but it keeps throwing "pointer of type ‘void *’ used in arithmetic" warning.
void *a = new char[4];
memset(a, 0 , 4);
unsigned short d = 7;
memcpy(a, (void *)&d, 2);
d=8;
memcpy(a+2, (void *)&d, 2); //pointer of type ‘void *’ used in arithmetic
/*Retrieving*/
unsigned int *data = new unsigned int();
memcpy(data, a, 2);
cout << (unsigned int)(*data);
memcpy(data, a+2, 2); //pointer of type ‘void *’ used in arithmetic
cout << (unsigned int)(*data);
The results are as per expectation but I fear that these warnings might turn into errors on some compiler. Is there another way to do this that I'm not aware of?
I know this is perhaps a bad practice in normal scenario but the problem statement requires that unsigned integers be stored and sent in 2 byte packets. Please correct me if I'm wrong but as per my understanding, using a char* instead of a void* would have taken up 3 bytes for 3-digit numbers.
a+2, with a being a pointer, means that the pointer is increased to allow space for two items of the pointer type. V.g., if a was int32 *, a + 2 would mean "a position plus 8 bytes".
Since void * has no type, it can only try to guess what do you mean by a + 2, since it does not know the size of the type being referred.
The problem is that the compiler doesn't know what to do with
a+2
This instruction means "Move pointer 'a' forward by 2 * (sizeof-what-is-pointed-to-by-'a')".
If a is void *, the compiler doesn't know the size of the target object (there isn't one!), so it gives an error.
You need to do:
memcpy(data, ((char *)a)+2, 2);
This way, the compiler knows how to add 2 - it knows the sizeof(char).
Please correct me if I'm wrong but as per my understanding, using a char* instead of a void* would have taken up 3 bytes for 3-digit numbers.
Yes, you are wrong, that would be the case if you were transmitting the numbers as chars. 'char*' is just a convenient way of referring to 8-bit values - and since you are receiving pairs of bytes, you could treat the destination memory are char's to do simple math. But it is fairly common for people to use 'char' arrays for network data streams.
I prefer to use something like BYTE or uint8_t to indicate clearly 'I'm working with bytes' as opposed to char or other values.
void* is a pointer to an unknown, more importantly, 0 sized type (void). Because the size is zero, offset math is going to result in zeros, so the compiler tells you it's invalid.
It is possible that your solution could be as simple as to receive the bytes from the network into a byte-based array. An int is 32 bits, 4 bytes. They're not "char" values, but quads of a 4-byte integer.
#include <cstdint>
uint8_t buffer[4];
Once you know you've filled the buffer, you can simply say
uint32_t integer = *(static_cast<uint32*>(buffer));
Whether this is correct will depend on whether the bytes are in network or host order. I'm guessing you'll probably need:
uint32_t integer = ntohl(*(static_cast<uint32*>(buffer)));
If I declare
int x = 5 ;
int* p = &x;
unsigned int y = 10 ;
cout << p+y ;
Is this a valid thing to do in C++, and if not, why?
It has no practical use, but is it possible?
The math is valid; the resulting pointer isn't.
When you say ptr + i (where ptr is an int*), that evaluates to the address of an int that's i * sizeof(int) bytes past ptr. In this case, since your pointer points to a single int rather than an array of them, you have no idea (and C++ doesn't say) what's at p+10.
If, however, you had something like
int ii[20] = { 0 };
int *p = ii;
unsigned int y = 10;
cout << p + y;
Then you'd have a pointer you could actually use, because it still points to some location within the array it originally pointed into.
What you are doing in your code snippet is not converting unsigned int to pointer. Instead you are incrementing a pointer by an integer offset, which is a perfectly valid thing to do. When you access the index of an array, you basically take the pointer to the first element and increase it by the integer index value. The result of this operation is another pointer.
If p is a pointer/array, the following two lines are equivalent and valid (supposing the pointed-to-array is large enough)
p[5] = 1;
*(p + 5) = 1;
To convert unsigned int to pointer, you must use a cast
unsigned int i = 5;
char *p = reinterpret_cast<char *>(i);
However this is dangerous. How do you know 5 is a valid address?
A pointer is represented in memory as an unsigned integer type, the address. You CAN store a pointer in an integer. However you must be careful that the integer data type is large enough to hold all the bits in a pointer. If unsigned int is 32-bits and pointers are 64-bits, some of the address information will be lost.
C++11 introduces a new type uintptr_t which is guaranteed to be big enough to hold a pointer. Thus it is safe to cast a pointer to uintptr_t and back again.
It is very rare (should be never in run-of-the-mill programming) that you need to store pointers in integers.
However, modifying pointers by integer offsets is totally valid and common.
Is this a valid thing to do in c++, and if not why?
Yes. cout << p+y; is valid as you can see trying to compile it. Actually p+y is so valid that *(p+y) can be translated to p[y] which is used in C-style arrays (not that I'm suggesting its use in C++).
Valid doesn't mean it actually make sense or that the resulting pointer is valid. Since p points to an int the resulting pointer will be an offset of sizeof(int) * 10 from the location of x. And you are not certain about what's in there.
A variable of type int is a variable capable of containing an integer value. A variable of type int* is a pointer to a variable copable of containing an integer value.
Every pointer type has the same size and contains the same stuff: A memory address, which the size is 4 bytes for 32-bit arquitectures and 8 bytes for 64-bit arquitectures. What distinguish them is the type of the variable they are poiting to.
Pointers are useful to address buffers and structures allocated dynamically at run time or any sort of variable that is to be used but is stored somewhere else and you have to tell where.
Arithmetic operations with pointers are possible, but they won't do what you think. For instance, summing + 1 to a pointer of type int will increase its value by sizeof(int), not by literally 1, because its a pointer, and the logic here is that you want the next object of this array.
For instance:
int a[] = { 10, 20, 30, 40 };
int *b = a;
printf("%d\n", *b);
b = b + 1;
printf("%d\n", *b);
It will output:
10
20
Because b is pointing to the integer value 10, and when you sum 1 to it, or any variable containing an integer, its then poiting to the next value, 20.
If you want to perform operations with the variable stored at b, you can use:
*b = *b + 3;
Now b is the same pointer, the address has not changed. But the array 10, 20, 30, 40 now contains the values 13, 20, 30, 40, because you increased the element b was poiting to by 3.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Pointer Arithmetic
The given code
int arr[12];
int * cur = arr;
cur++;
cout<<"cur-arr = "<<cur-arr<<endl;
Outputs 1, but I expected sizeof(int). Can someone please explain the nature of this behavior?
It's a defined behavior of C pointer arithmetic. It uses the size of pointed type as a unit. If you change subtraction in the last line to
(char *)cur - (char *)arr
you get 4 in the output.
This is the number of elements (ints here) between arr and cur (which is arr+1 at the time of subtraction). Compiler takes note that cur is a pointer to an integer and arr is an integer array. To get total number of bytes, try this:
(cur - arr) * sizeof(arr[0]);
cur is a pointer to int, initialized to some value (arr - the semantics of array-to-pointer conversion are irrelevant here), incremented (cur++) and compared to its old value. Unsurprisingly, it grew by one through the increment operation.
Pointer arithmetic with a given type works just like regular arithmetic. While the pointer is advanced by sizeof(int) bytes in this example, the difference between pointers is also calculated in units of sizeof(int), so you see plain simple arithmetics.
Addition and substraction for pointers works in accordance to the pointer type.