i have this code snippets
const int col= 5;const int row= 5;
int a[row][col] = {0};
int (*p)[col] ;
p = a;
And these statements print the same address
cout <<p;
cout << endl;
cout << *p;
in my opinion since p points to an array of 5 ints, dereferencing it
should give the first value which doesn't seem to be the case.
help!
since p points to an array of 5 ints
That much is correct.
dereferencing it should give the first value
No, it's type is "pointer to array"; dereferencing that gives "array", which decays to an int* pointer when you do just about anything with it - including printing it.
If you had a pointer to int
int * p = a;
then *p would indeed give the first array element.
Related
I writing a script which receives a structure (Evt) which holds an array of pointers uint16_t *DataChannel[Evt->ChSize[ch]]. I can then loop over the data and print out of the value as so:
for(uint16_t ch=0; ch<sizeof(Evt->DataChannel)/sizeof(Evt->DataChannel[0]); ++ch){
for(int adc_it=0; adc_it<(Evt->ChSize[ch]); ++adc_it){ for(int adc_it=0; adc_it<(Evt->ChSize[ch]); ++adc_it){
std::cout << (Evt->DataChannel)[ch][adc_it] << " ";
}
}
I don't understand where the second bracket comes from ([adc_it]), what it is doing and how it works. I think (Evt->DataChannel) should be a uint16_t which is not an array so why the second bracket?
I tried to replicate this in a quick piece of code:
#include <iostream>
int main()
{
int* test[10];
*(test[0]) = 5;
std::cout << test[0][0] << std::endl; //Gives 5
std::cout << test[0][1] << std::endl; //Seg faults
return 0;
}
Again can some explain what test[0][0] is doing because I have no idea and why it runs but test[0][1] fails?
Cheers
Dom
C++ permits the use of array notation [ ] when dereferencing a pointer.
In the simplest case, you can say something like:
char *ptr = new char[10];
That creates a pointer to char, which points to the first character in an array of ten characters allocated with new[].
After doing this:
char ch = *ptr;
and
char ch = ptr[0];
do the exact same thing. However since there are 10 characters in the array, you can also say:
char ch5 = ptr[5];
to access an element further up the array.
This leads to an eqivalence in C++ (and C, where it originated) that:
ptr[x];
is identical to:
*(ptr + x);
for any pointer / array referenced by ptr and any index x.
Already the first part
int* test[10];
*(test[0]) = 5;
is undefined behaviour, because you dereference an uninitialized pointer value.
Explanation:
int* test[10] is an array of 10 pointers to ints, each pointer
being not initialized and definitely not pointing to a memory that
you reserved to store ints.
test[0] gives an uninitialized pointer; already this statement is
UB
*(test[0]) = 5 dereferences an arbitrary pointer value, definitely
UB
Try:
int* test[10];
int firstLine[5] = { 1,2,3,4,5 };
test[0] = firstLine;
cout << test[0][0]; // gives 1, firstLine[0]
cout << test[0][1]; // gives 2, is the same as firstLine[1]
// cout << test[1][0]; // undefined behaviour, the second pointer in array test is not initialized
I know that the following is not correct:
int arr[2][3] = {}; //some array initialization here
int** ptr;
ptr = arr;
But I am quite surprised that the following lines actually work
int arr[2][3] = {}; //some array initialization here
auto ptr = arr;
int another_arr[2][3] = {}; //some array initialization here
ptr = another_arr;
Can anyone possibly explain what is the type assigned to ptr in the second block of code, and what happened underneath?
Well, arrays decay to pointers when used practically everywhere. So naturally there's decay going on in your code snippet too.
But it's only the "outer-most" array dimension that decays to a pointer. Since arrays are row-major, you end up with int (*)[3] as the pointer type, which is a pointer to a one-dimensional array, not a two dimensional array. It points to the first "row".
If you want ptr's deduction to be a pointer to the array instead, then use the address-of operator:
auto ptr = &arr;
Now ptr is int(*)[2][3].
In
auto ptr = arr;
arr decays into a pointer to its first element in the normal way; it's equivalent to
auto ptr = &arr[0];
Since arr[0] is an array of three ints, that makes ptr a int (*)[3] - a pointer to int[3].
another_arr decays in exactly the same way, so in
ptr = another_arr;
both sides of the assignment have the type int (*)[3], and you can assign a T* to a T* for any type T.
A pointer to arr itself has type int(*)[2][3].
If you want a pointer to the array rather than a pointer to the array's first element, you need to use &:
auto ptr = &arr;
First, let's look at why you can't assign int arr[2][3] to int **. To make it easier to visualise, we'll initialise your array with a sequence, and consider what it looks like in memory:
int arr[2][3] = {{1,2,3},{4,5,6}};
In memory, the array data is stored as a single block, just like a regular, 1D array:
arr: [ 1, 2, 3, 4, 5, 6 ]
The variable arr contains the address of the start of this block, and from its type (int[2][3]) the compiler knows to interpret an index like arr[1][0] as meaning "take the value that is at position (1*2 + 0) in the array".
However for a pointer-to-pointer (int**), it is expected that the pointer-to-pointer contains either a single memory address or an array of memory addresses, and this/these adress(es) point to (an)other single int value or array of ints. Let's say we copied the array arr into int **ptrptr. In memory, it would look like this:
ptrptr: [0x203F0B20, 0x203F17D4]
0x203F0B20: [ 1, 2, 3 ]
0x203F17D4: [ 4, 5, 6 ]
So in addition to the actual int data, an extra pointer must be stored for each row of the array. Rather than converting the two indexes into a single array lookup, access must be performed by making a first array lookup ("take the second value in ptrptr to get an int*"), then nother array lookup ("take the first value in the array at the address held by the previously obtained int*").
Here's a program that illustrates this:
#include <iostream>
int main()
{
int arr[2][3] = {{1,2,3},{4,5,6}};
std::cout << "Memory addresses for int arr[2][3]:" << std::endl;
for (int i=0; i<2; i++)
{
for (int j=0; j<3; j++)
{
std::cout << reinterpret_cast<void*>(&arr[i][j]) << ": " << arr[i][j] << std::endl;
}
}
std::cout << std::endl << "Memory addresses for int **ptrptr:" << std::endl;
int **ptrptr = new int*[2];
for (int i=0; i<2; i++)
{
ptrptr[i] = new int[3];
for (int j=0; j<3; j++)
{
ptrptr[i][j] = arr[i][j];
std::cout << reinterpret_cast<void*>(&ptrptr[i][j]) << ": " << ptrptr[i][j] << std::endl;
}
}
// Cleanup
for (int i=0; i<2; i++)
{
delete[] ptrptr[i];
ptrptr[i] = nullptr;
}
delete[] ptrptr;
ptrptr = nullptr;
return 0;
}
Output:
Memory addresses for int arr[2][3]:
0x7ecd3ccc0260: 1
0x7ecd3ccc0264: 2
0x7ecd3ccc0268: 3
0x7ecd3ccc026c: 4
0x7ecd3ccc0270: 5
0x7ecd3ccc0274: 6
Memory addresses for int **ptrptr:
0x38a1a70: 1
0x38a1a74: 2
0x38a1a78: 3
0x38a1a90: 4
0x38a1a94: 5
0x38a1a98: 6
Notice how the memory addresses always increase by 4 bytes for arr, but for ptrptr there is a jump of 24 bytes between values 3 and 4.
A simple assignment can't create the pointer-to-pointer structure needed for type int **, which is why the loops were necessary in the above program. The best it can do is to decay the int[2][3] type into a pointer to a row of that array, i.e. int (*)[3]. That's what your auto ptr = arr; ends up as.
What is the type of [...]
Did you already try to ask the compiler to tell you the type of an expression?
int main()
{
int arr[2][3] = {{0,1,2}, {3,4,5}}; // <-- direct complete initialized here
auto ptr = arr; // <-- address assignment only
cout << "arr: " << typeid(arr).name() << endl;
cout << "ptr: " << typeid(ptr).name() << endl;
return 0;
}
I've to confess that the output
arr: A2_A3_i
ptr: PA3_i
seems to be not very readable at first glance (compared to some other languages), but when in doubt it may help. It's very compact, but one may get used to it soon. The encoding is compiler-dependent, in case you are using gcc, you may read Chapter 29. Demangling to understand how.
Edit:
some experimentation with some simple_cpp_name function like this rudimentary hack
#include <typeinfo>
#include <cxxabi.h>
#include <stdlib.h>
#include <string>
std::string simple_cpp_name(const std::type_info& ti)
{
/// simplified code extracted from "Chapter 29. Demangling"
/// https://gcc.gnu.org/onlinedocs/libstdc++/manual/ext_demangling.html
char* realname = abi::__cxa_demangle(ti.name(), 0, 0, 0);
std::string name = realname;
free(realname);
return name;
}
will show you that auto &rfa = arr; makes rfa having the same type as arr which is int [2][3].
I having some issue when it comes to initializing pointers.
void findMM (int *PMM, int *theG)
{
// code I haven't written yet. It will essentially take two variables from //theG and store it in MM
}
int main()
{
int size;
int MM [2] = {1000, 0};
int *theG = NULL;
cout << "\nPlease insert size of array:" << endl;
cin >> size;
theG = new int [size];
findMM(&MM, &theG); //Get error with &MM
delete [] theG;
return 0;
}
The complier says that argument of type int (*)[2] is incompatible with parameter of type int ** So obviously that I have issue with the code in particular my (reference?) of array MM. Or perhaps there is other obvious faults that I am missing?
Edit attempt 2
void findMM (int *PMM, int *theG)
{
PMM [1] = 5;
theG [0] = 7;
}
int main()
{
int size;
int MM [2] = {1000, 0};
int *theG = NULL;
cout << "\nPlease insert size of array:" << endl;
cin >> size;
theG = new int [size];
findMM(MM, theG);
cout << MM [1] << endl << theG[0];
delete [] theG;
return 0;
}
The output would be 5 and 7 correct?
Since MM is an array, &MM is a pointer to an array (that's the type int (*)[2] that you see in the error). Instead, you seem to want to pass a pointer to the first element of the array. There are two ways to do that. Firstly, you can explicitly get the first element and then take the address of it: &MM[0]. Secondly, you can rely on array-to-pointer conversion to do it for you and just pass MM. Array-to-pointer conversion converts an array to a pointer to its first element.
I know this question has already been answered but I believe I can contribute to the asker's understanding.
Let's start with the basics:
void main()
{
int a = 2; // a is an int
cout << a << endl; // print 2
int *b; // b is a pointer-to-int
b = &a; // store the address of a in b
cout << *b << endl;// print the value that b points to, which is 2
int my_array = new int[3]; // allocate an array with 3 integers
my_array[0] = 50; // store 50 in the first element of the array
my_array[1] = 51; // store 51 in the second element of the array
my_array[2] = 52; // store 52 in the third element of the array
cout << c[0] << endl; // print 50
some_function(my_array, 3); // explained below
}
Now let's see how to pass arrays into functions. Assume we want to have a function called some_function that receives an array.
void some_function(int *some_array, int size_of_the_array)
{
// use the array however you like here
}
The function some_function receives a pointer to an int (also known as "pointer-to-int"). The name of an array is always the address of its first element, so if a function expects a pointer to an int and you give it the name of an array of ints, you are actually giving it the address of the first element in the array (this is just C++ syntax rules). So the function now has the address of the first element in the array, it can do stuff like *some_array to access the first element in the array, but what if it wants to access the other elements? It adds 1 to the pointer it already has and then applies the * operator to it: *(some_array + 1). Let's say an int is 4 bytes, if you add 1 to a pointer-to-int, the result of this addition is a new pointer that points to a location in memory 4 bytes ahead, so *(some_array + 93) is the value in the 94th element of the array some_array (array elements are stored sequentially in memory). A shorthand notation for this is some_array[93]. So if you have int *some_array = new int[100];, then some_array is a pointer and some_array[93] is the same as *(some_array + 93), which is the 94th element in the array.
The address itself though is not enough, you also need to know the number of entries in the array so that you don't try to access an element past the end of the array. In this example, assume that some_function simply prints the contents of the array, so if you don't provide 3 as the second argument to the function then it will have no way of knowing when to stop adding 1 to the pointer it received in the first argument. Beware, however, that by passing an array to a function this way, you are not passing the function a copy of the array, you are simply telling it where to find its contents in memory.
int x = 4;
int* q = &x; // Is it always equivalent to int *q = &x; ?
cout << "q = " << q << endl; // output: q = 0xbfdded70
int i = *q; // A
int j = *(int*)q; // B, when is this necessary?
cout << "i = " << i << endl; // output: i = 4
cout << "j = " << j << endl; // output: j = 4
My question is what does lines A and B do, and why the outputs are both 4?
It is a basic usage of pointers, in A you dereference pointer (access the variable to which a pointer points)":
int i = *q; // A
while B is doing exactly the same but it additionally casts pointer to the same type. You could write it like that:
int j = *q; // B
there is no need for (int*)
int x = 4;
x is 4
int* q = &x;
q is the memory location of x (which holds 4)
cout << "q = " << q << endl; // output: q = 0xbfdded70
There's your memory location.
int i = *q; // A
i is the value at memory location q
int j = *(int*)q; // B
j is the value at memory location q. q is being cast to an int pointer, but that's what it already is.
int i = *q; // A
Dereferences a pointer to get the pointed value
int j = *(int*)q; // B
type casts the pointer to an int * and then dereferences it.
Both are same because the pointer is already pointing to an int. So typecasting to int * in second case is not needed at all.
Further derefenecing yields the pointed integer variable value in both cases.
Lines A and B are equivelent as q is already an int* and therefor (int*)q equals q.
int i = *q; yelds that i becomes the value of the integer pointed to by q. If you want to make i to be equal to the adress itself remove the asterisk.
A: Dereference - takes a pointer to a value (variable or object) and returns the value
B: Cast to int* and than dereference
The result is the same because the pointer is already to int. That's it.
Line A takes the value that q points to and assigns it to i. Line b casts q to the type int* (which is q's type already, so that cast is entirely redundant/pointless), then takes the value that q points to and assigns it to j.
Both give you 4 because that's the value that q points to.
Line A de-reference pointer q typed as int *, i.e. a pointer points to an int value.
Line B cast q as (int *) before de-reference, so line B is the same as int j = *q;.
Given a pointer to int, how can I obtain the actual int?
I don't know if this is possible or not, but can someone please advise me?
Use the * on pointers to get the variable pointed (dereferencing).
int val = 42;
int* pVal = &val;
int k = *pVal; // k == 42
If your pointer points to an array, then dereferencing will give you the first element of the array.
If you want the "value" of the pointer, that is the actual memory address the pointer contains, then cast it (but it's generally not a good idea) :
int pValValue = reinterpret_cast<int>( pVal );
If you need to get the value pointed-to by the pointer, then that's not conversion. You simply dereference the pointer and pull out the data:
int* p = get_int_ptr();
int val = *p;
But if you really need to convert the pointer to an int, then you need to cast. If you think this is what you want, think again. It's probably not. If you wrote code that requires this construct, then you need to think about a redesign, because this is patently unsafe. Nevertheless:
int* p = get_int_ptr();
int val = reinterpret_cast<int>(p);
I'm not 100% sure if I understand what you want:
int a=5; // a holds 5
int* ptr_a = &a; // pointing to variable a (that is holding 5)
int b = *ptr_a; // means: declare an int b and set b's
// value to the value that is held by the cell ptr_a points to
int ptr_v = (int)ptr_a; // means: take the contents of ptr_a (i.e. an adress) and
// interpret it as an integer
Hope this helps.
use the dereference operator * e.g
void do_something(int *j) {
int k = *j; //assign the value j is pointing to , to k
...
}
You should differentiate strictly what you want: cast or dereference?
int x = 5;
int* p = &x; // pointer points to a location.
int a = *p; // dereference, a == 5
int b = (int)p; //cast, b == ...some big number, which is the memory location where x is stored.
You can still assign int directly to a pointer, just don't dereference it unless you really know what you're doing.
int* p = (int*) 5;
int a = *p; // crash/segfault, you are not authorized to read that mem location.
int b = (int)p; // now b==5
You can do without the explicit casts (int), (int*), but you will most likely get compiler warnings.
Use * to dereference the pointer:
int* pointer = ...//initialize the pointer with a valid address
int value = *pointer; //either read the value at that address
*pointer = value;//or write the new value
int Array[10];
int *ptr6 = &Array[6];
int *ptr0 = &Array[0];
uintptr_t int_adress_6 = reinterpret_cast<uintptr_t> (ptr6);
uintptr_t int_adress_0 = reinterpret_cast<uintptr_t> (ptr0);
cout << "difference of casted addrs = " << int_adress_6 - int_adress_0 << endl; //24 bits
cout << "difference in integer = " << ptr6 - ptr0 << endl; //6