c++
p is pointing to specific place
int * p
When I try p=p[1] it says can't convert int to int (using devcpp).
While p=&p[1] works fine
Why do I need to do the second method?
p[1] is an address. So the first method should work?
Can you explain me about this error?
p[1] is the same as *(p + 1).
You want the address of this element, which is simply (p + 1). C++ also allows &p[1], as you noticed.
p[1] is equivalent to *(p + 1) so it's a value, not an address. p = p + 1 or just p++ would be what you want.
While p is anint*, p[1] is an element from that array, therefore p[1] is int.
You can do p = &p[1] in other ways, for instance, p = p + 1, or p++. Both will set p to the same final value.
Notice when doing such arithmetic operations with pointers, it will not not increment the address by 1, its incrementing it by 1 times the size of one element, so its really the same thing.
Related
#include <iostream>
using namespace std;
int main ()
{
int numbers[5];
int * p;
p = numbers; *p = 10;
p++; *p = 20;
p = &numbers[2]; *p = 30;
p = numbers + 3; *p = 40;
p = numbers; *(p+4) = 50;
for (int n=0; n<5; n++)
cout << numbers[n] << ", ";
return 0;
}
Fairly new coder here. This is an example taken from the pointers page on cplusplus.com. They talk about dereferencing, the address of operator, and then they talk about the relationship between arrays and pointers. I noticed if I simply printed out a declared array it spits out the address of the array. I'm really trying to understand exactly what some of these lines of code are doing and am having a hard time.
Everything makes sense up until *p=10. I read this as, "the value being pointed to by pointer p is equal to 10" but an array isn't just one value. In this case it's 5... So, whats the deal? Will the compiler just assume we are talking about the first element of the array?
Even more confusing for me is p++;. p right now is the address of the array. How do you do p=p+1 when p is an address?
I understand p= &numbers[2];. As well as the following *p = 30;.
p = numbers + 3; confuses me. Why are we able to take an address and just add 3 to it? And, what does that even mean?
Everything within the for loop makes sense. It's just printing out the array correct?
Thanks for your help in advance!
As Pointed out this due to pointer arithmetic.
So both array numbers and pointer p are int, so after assigning numbers's base address to pointer p and the *p=10 does numbers[0]=10
Now when you do p++ the address of p is incremented but not to the next numeric value as in normal arithmetic operation of ++ but to the address of the next index of numbers.
Let's assume a int is 4byte of memory and numbers's initial address is 1000.
So p++ makes address it to be incremented to 1004 which is numbers[1]
Basically it points to the next int variable position in memory.
So when *p = 20 is encountered the value 20 will be assigned to the next adjacent memory i.e. 1004 or numbers[1]=20.
After that the p directly assigned address of 3rd index numbers[2] i.e 1008 (next int var in memory). Now p is pointing to the 3rd index of numbers so *p=30 is again numbers[2]=30
p = numbers + 3; *p = 40; does similar thing that is assigns base address of numbers incremented by 3 indexes (1000 + 3*sizeof(int) =1012 ==> 1000 + 3*4 =1012) to p and assigns 40 to that. so it becomes equivalent to numbers[3]=40.
Then p = numbers; *(p+4) = 50; is assigning 1000 to p and the adding 4 to the address of p by using the () and assigns 50 to the value by using *() which is value of the address(1000+ 4*4 = 1016) inside the braces.
Using a pointer like this in C++: p = numbers makes p point to the first element of the array. So when you do p++ you are making p point to the second element.
Now, doing *p = 10 assigns 10 to whatever p is pointing to, as in your case p is pointing to the first element of the array, then *p = 10 is the same as numbers[0] = 10.
In case you see this later: *(p + i) = 20, this is pointer arithmetic and is the same as p[i] = 20. That's why p[0] = 10 is equivalent to *(p + 0) = 10 and equivalent to *p = 10.
And about your question: How do you do p=p+1 when p is an address?: as arrays in C++ are stored sequentially in memory, if the element p is pointing at memory address 6400010 then p++ points at 6400014 assuming each element pointed by p occupies 4 bytes. If you wonder why p++ is 6400014 instead of 6400011, the reason is a little magic C++ does: when increasing a pointer, it doesn't increase by 1 byte but by 1 element so if you are pointing to integers, each one occupying 4 bytes, then p will be increased by 4 instead of 1.
Let's break it down:
int numbers[5];
int * p;
Assigning the array variable to a pointer means that the pointer will be pointing to the first element:
p = numbers; // p points to the 1st element
*p = 10; // 1st element will be 10
p++; // go to next location i.e. 2nd element
*p = 20; // 2nd element will be 20
You can do pointer arithmetic and it works according to the type. Incrementing an int pointer on a 32-bit machine will do a jump of 4 bytes in memory. On a 64-bit machine, that will be 8 bytes. Keep in mind that you cannot do this with a void pointer.
p = &numbers[2]; // assign the address of 3rd element
*p = 30; // 3rd element will be 30
p = numbers + 3; // numbers + 3 means the address of 4th element
*p = 40; // 4th element will be 40
p = numbers; // p points to the 1st element
*(p + 4) = 50; // p + 4 means the address of 5th element
// dereference 5th location and assign 50
// 5th element will be 50
There are a lot of pages on the web on how pointers work. But there are a couple things to quickly point out.
int* p;
The above line mean p is a pointer to an int. Assigning something to it doesn't change what it is. It is a pointer to an int, not an array of int.
int number[5];
p = number;
This assignment takes the address where the array is stored, and assigns it to p. p now points to the beginning of that array.
p++;
p now points to the 2nd int in the array. Pointers are incremented by the size of the object they point to. So the actual address may increase by 4 or 8, but it goes to the next address of an int.
*p = 30;
This is an assignment, not a comparison. It doesn't say that what p points to is 30, it copies 30 into the address pointed to by p. Saying "equals" is generally confusing in programming terms, so use either "assign" or "compare" to be clear.
1> p=numbers;
pointer p now points to the start of the array, i.e first element.
2> *p=10;
first element now becomes 10, as p is pointing to the first element.
3> p++;
this moves the pointer to the next element of the array, i.e second element
So, again *p=20; will make second element=20
4> p = &numbers[2];
this statement means that p now points to 3rd element of the array.
So, *p = 30; will make 3rd element=30
5> p = numbers + 3;
Wen we do p=numbers+1; it points to second element, so p=numbers+3; will now point to 4th element of array. *p = 40; make the fourth element=40
6> p = numbers;
Again p points to first element of array
*(p+4) = 50; will now make 5th element=50
I am studying C++ and came across this Code
int a[] = {9,8,−5};
int∗ p = &a[2] ;
and was wondering what exactly that means? So a is an Array and p is a pointer. But what does the & before a[] mean?
and what does mean then?
std::cout << (p − a) << "\n";
edit:
so i have read some answeres but what i still dont understand is what the & is really for. Is there a difference between
int∗ p = &a[2] ;
and
int∗ p = a[2] ;
?
Thanks for your help.
This code
int∗ p = &a[2];
is equivalent to this:
int∗ p = a + 2;
so even mathematically you can see that p - a is equal to a + 2 - a so I think result should be obvious that it is equal to 2.
Note name of array can decay to a pointer to the first element but is not that pointer, as many may mistakenly say so. There is significant difference.
About your note, yes there is difference, first expression assigns address of third element to p, second is syntax error as you try to assign int to int * which are different types:
int a[] = { 1, 2, 3 };
int var = a[2]; // type of a[2] is int and value is 3
int *pointer = &a[2]; // type of &a[2] is int * and value is address of element that holds value 3
Subject of your question says that you think & means reference in this context. It only means reference when applied to types, when used in expression it means different things:
int i, j;
int &r = i; // refence when applied to types
&i; // address of in this context
i&j; // binary AND in this
In your case & is used to get the address to a value. So what will happen there is that p will contain the address of a[2], thus p will be a pointer to the third element of the array.
EDIT: Demo (with p-a included)
I have some problem understanding the pointers syntax usage in context with two dimensional arrays, though I am comfortable with 1-D array notation and pointers, below is one of the syntax and I am not able to understand how the following expression is evaluated.
To access the element stored at third row second column of array a we will use the subscripted notation as a[2][1] other way to access the same element is
*(a[2]+1)
and if we want to use it as pointers we will do like this
*(*(a+2)+1)
Though I am able to understand the replacement of *(a[2]+1)
as *(*(a+2)+1) but I don't know how this is getting evaluated.
Please explain with example if possible.
Assume array is stored in row wise order and contain the following elements
int a[5][2]={
21,22,
31,32
41,42,
51,52,
61,62
};
and base address of array is 100(just assume) so the address of a[2] is 108 (size of int =2(another assumption)) So the expression *(*(a+2)+1). How does it gets evaluated does it start from the inside bracket and if it does then after the first bracket we have the value to which 1 is being added rather than the address... :/
To start of with
a[i] = *(a+i);
So
a[i][j] = *(a[i] +j)
and
a[i][j] = *(*(a+i) + j)
How a[i] = *(a+i):
If a is a array then the starting address of the array is given by &a[0] or just a
So when you specify
a[i] this will decay to a pointer operation *(a+i) which is, start at the location a and dereference the pointer to get the value stored in the location.
If the memory location is a then the value stored in it is given by *a.
Similarly the address of the next element in the array is given by
&a[1] = (a+1); /* Array decays to a pointer */
Now the location where the element is stored in given by &a[1] or (a+1) so the value stored in that location is given by *(&a[1]) or *(a+1)
For Eg:
int a[3];
They are stored in the memory as shown below:
a a+1 a+2
------------------
| 100 | 102 | 104|
------------------
&a[0] &a[1] &a[2]
Now a is pointing to the first element of the array. If you know the pointer operations a+1 will give you the next location and so on.
In 2D arrays the below is what the access is like :
int arr[m][n];
arr:
will be pointer to first sub array, not the first element of first sub
array, according to relationship of array & pointer, it also represent
the array itself,
arr+1:
will be pointer to second sub array, not the second element of first sub
array,
*(arr+1):
will be pointer to first element of second sub array,
according to relationship of array & pointer, it also represent second
sub array, same as arr[1],
*(arr+1)+2:
will be pointer to third element of second sub array,
*(*(arr+1)+2):
will get value of third element of second sub array,
same as arr[1][2],
A 2D array is actually a consecutive piece of memory. Let me take a example : int a[3][4] is representend in memory by a unique sequence of 12 integer :
a00 a01 a02 a03 a04 a10 a11 a12 a13 a14 a20 a21 a22 a23 a24
| | |
first row second row third row
(of course it can be extended to any muti-dimensional array)
a is an array of int[4] : it decays to a pointer to int[4] (in fact, it decays to &(a[0]))
a[1] is second row. It decays to a int * pointing to beginning of first row.
Even if arrays are not pointer, the fact that they decay to pointers allows to use them in pointer arithmetic : a + 1 is a pointer to second element of array a.
All that explains why a[1] == *(a + 1)
The same reasoning can be applied to a[i] == *(a+i), and from there to all the expressions in your question.
Let's look specifically to *(*(a+2)+1). As said above, a + 2 is a pointer to the third element of an array of int 4. So *(a + 2) is third row, is an array of int[4] and decays to a int * : &(a[2][0]).
As *(a + 2) decays to an int *, we can still use it as a base for pointer arithmetic and *(a + 2) + 1 is a pointer to the second element of third row : *(a + 2) + 1 == &(a[2][1]). Just dereference all that and we get
*(*(a + 2) + 1) == a[2][1]
In the following code, can anyone please explain to my what the line in bold is doing.
struct southParkRec {
int stan[4];
int *kyle[4];
int **kenny;
string cartman;
};
int main()
{
southParkRec cartoon;
cartoon.stan[1] = 4;
cartoon.kyle[0] = cartoon.stan + 1;
cartoon.kenny = &cartoon.kyle[2];
*(cartoon.kenny + 1) = cartoon.stan; //What does this line do?
return 0;
}
Think of it as
cartoon.kenny[1] = cartoon.stan;
They are basically the same thing
If we bring the whole thing to one common style of using the subscript operator [] (possibly with &) instead of a * and + combination, it will look as follows
cartoon.stan[1] = 4;
cartoon.kyle[0] = &cartoon.stan[1];
cartoon.kenny = &cartoon.kyle[2];
cartoon.kenny[1] = &cartoon.stan[0];
After the
cartoon.kenny = &cartoon.kyle[2];
you can think of kenny as an "array" of int * elements embedded into the kyle array with 2 element offset: kenny[0] is equivalent to kyle[2], kenny[1] is equivalent to kyle[3], kenny[2] is equivalent to kyle[4] and so on.
So, when we do
cartoon.kenny[1] = &cartoon.stan[0];
it is equivalent to doing
cartoon.kyle[3] = &cartoon.stan[0];
That's basically what that last line does.
In other words, if we eliminate kenny from the consideration ("kill Kenny"), assuming that the rest of the code (if any) doesn't depend on it, your entire code will be equivalent to
cartoon.stan[1] = 4;
cartoon.kyle[0] = &cartoon.stan[1];
cartoon.kyle[3] = &cartoon.stan[0];
As for what is the point of all this... I have no idea.
In cartoon you have:
- stan, an array of 4 ints.
- kyle, an array of 4 pointers to int.
- kenny, a pointer to a pointer to int, that is, let's say, a pointer to an "array of ints".
cartoon.stan[1] = 4; sets second element of stan array (an int) to 1.
cartoon.kyle[0] = cartoon.stan + 1; sets first element of kyle array (a pointer to int) to point to the second element of stan array (which we have just set to 4).
cartoon.kenny = &cartoon.kyle[2]; sets kenny pointer to point to the third element of kyle array.
*(cartoon.kenny + 1) = cartoon.stan; sets fourth element of kyle array (a pointer to int) to point to the first element of stan array (which hasn't been initialised yet). More in detail:
cartoon.kenny gets kenny pointer's address (third element of kyle array),
cartoon.kenny+1 gets the next int after that address (fourth element of kyle array, which happens to be a pointer to int),
*(cartoon.kenny + 1) dereferences that pointer, so we can set it, and
= cartoon.stan sets it to point to the first element of stan array.
It is incrementing the pointer at *cartoon.kenny. 'kenny' is a pointer to a pointer, so the first dereference returns a pointer, which is incremented, and a value assigned. So, *(kenny + 1) now points to the beginning of the array 'stan'.
It sets the last element of Kyle to point to the first element of Stan.
This was a question in our old C++ exam. This code is driving me crazy, could anyone explain what it does and - especially - why?
int arr[3]={10,20,30};
int *arrp = new int;
(*(arr+1)+=3)+=5;
(arrp=&arr[0])++;
std::cout<<*arrp;
This statement writes to the object *(arr+1) twice without an intervening sequence point so has undefined behavior.
(*(arr+1)+=3)+=5;
This statement writes to the object arrp twice without an intervening sequence point so has undefined behavior.
(arrp=&arr[0])++;
The code could result in anything happening.
Reference: ISO/IEC 14882:2003 5 [expr]/4: "Between the previous and next sequence point a scalar object shall have its stored value modified at most once by the evaluation of an expression."
(*(arr+1)+=3)+=5;
arr + 1 - element with index 1
*(arr + 1) - value of this element
(arr + 1) += 3 - increase by 3
((arr+1)+=3)+=5 - increase by 5;
so arr[1] == 28
(arrp=&arr[0])++;
arr[0] - value of element 0
&arr[0] - address of element 0
arrp=&arr[0] - setting arrp to point to elem 0
(arrp=&arr[0])++ - set arr to point to elem 1
result: 28
This line:
(*(arr+1)+=3)+=5;
produces the same result as this (see footnote):
arr[1] += 3;
arr[1] += 5;
This line:
(arrp=&arr[0])++;
produces the same result as this (see footnote):
int* arrp = arr+1;
So this line:
std::cout<<*arrp
prints out 28.
But this code leaks memory because int *arrp = new int; allocates a new int on the heap which will be lost on assignment by (arrp=&arr[0])++;
Footnote: Of course I'm assuming an absence of weirdness.
Edit: Apparently some of the lines in fact lead to undefined behavior, due to C++ Standard 5/4. So this really is a crappy exam question.
int arr[3]={10,20,30}; // obvious?
int *arrp = new int; // allocated memory for an int
(*(arr+1)+=3)+=5; // (1)
(arrp=&arr[0])++; // (2)
std::cout<<*arrp; // (3)
(1)
*(arr+1) is the same as arr[1], which means that *(arr+1)+=3 will increase arr[1] by 3, so arr[1] == 23 now.
(*(arr+1)+=3)+=5 means arr[1] is increased by another 5, so it will be 28 now.
(2)
arrp will pont to the address of the first element of arr (arr[0]). The pointer arrp will then be incremented, thus it will point to the second element after the entire statement is executed.
(3)
Prints what arrp points to: the second element of arr, meaning 28.
Well, remember that arrays can be interpreted as pointers
int arr[3]={10,20,30};
int *arrp = new int;
creates an array arr of three integers and an int pointer that gets assigned with a freshly allocated value.
Since assignment operators return a reference to the value that has been assigned in order to allow multi-assignment,
(*(arr+1)+=3)+=5;
is equivalent to
*(arr+1)+=3;
*(arr+1)+=5;
*(arr + 1) refers to the first element of the array arr, therefore arr[1] is effectively increased by eight.
(arrp=&arr[0])++; assigns the address of the first array element to arrp and afterward increments this pointer which now points to the second element (arr[1] again).
By dereferencing it in std::cout<<*arrp, you output arr[1] which now holds the value 20 + 3 + 5 = 28.
So the code prints 28 (and furthermore creates a memory-leak since the new int initially assigned to arrp never gets deleted)
I'll try to answer you by rewriting the code in a simpler way.
int arr[3]={10,20,30};
int *arrp = new int;
(*(arr+1)+=3)+=5;
(arrp=&arr[0])++;
std::cout<<*arrp;
=== equals ===
int arr[3]={10,20,30};//create array of 3 elements and assign them
int *arrp = new int;//create an integer pointer and allocate an int to it(useless)
//(*(arr+1)+=3)+=5;
arr[1] = arr[1] + 3;//arr[1] == arr+1 because it is incrementing the arr[0] pointer
arr[1] = arr[1] + 5;
//(arrp=&arr[0])++;
arrp = &arr[0];//point the integer pointer to the first element in arr[]
arrp++;//increment the array pointer, so this really is now pointing to arr[1]
std::cout<<*arrp;//just print out the value, which is arr[1]
I am assuming you understand pointers and basic c.