In C, you can declare an char array either by
char []array;
or
char *array;
The later one is a pointer, why can it be an array?
Pointers and arrays are two completely different animals; a pointer cannot be an array and an array cannot be a pointer.
The confusion comes from two concepts that aren't explained very well in most introductory C texts.
The first is that the array subscript operator [] can be applied to both pointer and array expressions. The expression a[i] is defined as *(a + i); you offset i elements from the address stored in a and dereference the result.
So if you declare a pointer
T *p;
and assign it to point to some memory, like so
p = malloc( N * sizeof *p );
you'll get something like the following:
+---+
p: | | ---+
+---+ |
... |
+---+ |
p[0]: | |<---+
+---+
p[1]: | |
+---+
...
+---+
p[N-1]: | |
+---+
p stores the base address of the array, so *(p + i) gives you the value stored in the i'th element (not byte) following that address.
However, when you declare an array, such as
T a[N];
what you get in memory is the following:
+---+
a[0]: | |
+---+
a[1]: | |
+---+
...
+---+
a[N-1]: | |
+---+
Storage has only been set aside for the array elements themselves; there's no separate storage set aside for a variable named a to store the base address of the array. So how can the *(a+i) mechanism possibly work?
This brings us to the second concept: except when it is the operand of the sizeof or unary & operators, or is a string literal being used to initialize another array ijn a declaration, an expression of type "N-element array of T" will be converted ("decay") to an expression of type "pointer to T", and the value of the expression will be the address of the first element of the array.
In other words, when the compiler sees the expression a in the code, it will replace that expression with a pointer to the first element of a, unless a is the operand of sizeof or unary &. So a evaluates to the address of the first element of the array, meaning *(a + i) will work as expected.
Thus, the subscript operator works exactly the same way for both pointer and array expressions. However, this does not mean that pointer objects are the same thing as array objects; they are not, and anyone who claims otherwise is confused.
Related
This program works in C:
#include <stdio.h>
int main(void) {
char a[10] = "Hello";
char *b = a;
printf("%s",b);
}
There are two things I would expect to be different. One is that we in the second line in the main write: "char *b = &a", then the program is like this:
#include <stdio.h>
int main(void) {
char a[10] = "Hello";
char *b = &a;
printf("%s",b);
}
But this does not work. Why is that? Isn't this the correct way to initialize a pointer with an adress?
The second problem I have is in the last line we should have: printf("%s",*b) so the program is like this:
#include <stdio.h>
int main(void) {
char a[10] = "Hello";
char *b = a;
printf("%s",*b);
}
But this gives a segmentation fault. Why does this not work? Aren't we supposed to write "*" in front of a pointer to get its value?
There is a special rule in C. When you write
char *b = a;
you get the same effect as if you had written
char *b = &a[0];
That is, you automatically get a pointer to the array's first element. This happens any time you try to take the "value" of an array.
Aren't we supposed to write "*" in front of a pointer to get its value?
Yes, and if you wanted to get the single character pointed to by b, you would therefore need the *. This code
printf("first char: %c\n", *b);
would print the first character of the string. But when you write
printf("whole string: %s\n", b);
you get the whole string. %s prints multiple characters, and it expects a pointer. Down inside printf, when you use %s, it loops over and prints all the characters in the string.
Expanding on Steve's answer (which is the correct one to accept)...
This is the special rule he's talking about:
6.3.2.1 Lvalues, arrays, and function designators
...
3 Except when it is the operand of the sizeof operator, the _Alignof operator, or the
unary & operator, or is a string literal used to initialize an array, an expression that has
type ‘‘array of type’’ is converted to an expression with type ‘‘pointer to type’’ that points
to the initial element of the array object and is not an lvalue. If the array object has
register storage class, the behavior is undefined.
C 2011 Prepublication Draft
Arrays are weird and don't behave like other types. You don't get this "decay to a pointer to the first element" behavior in other aggregate types like struct types. You can't assign the contents of an entire array with the = operator like you can with struct types; for example, you can't do something like
int a[5] = {1, 2, 3, 4, 5};
int b[5];
...
b = a; // not allowed; that's what "is not an lvalue" means
Why are arrays weird?
C was derived from an earlier language named B, and when you declared an array in B:
auto arr[5];
the compiler set aside an extra word to point to the first element of the array:
+---+
arr: | | ----------+
+---+ |
... |
+---+ |
| | arr[0] <--+
+---+
| | arr[1]
+---+
| | arr[2]
+---+
| | arr[3]
+---+
| | arr[4]
+---+
The array subscript operation arr[i] was defined as *(arr + i) - given the starting address stored in arr, offset i elements from that address and dereference the result. This also meant that &arr would yield a different value from &arr[0].
When he was designing C, Ritchie wanted to keep B's array subscripting behavior, but he didn't want to set aside storage for the separate pointer that behavior required. So instead of storing a separate pointer, he created the "decay" rule. When you declare an array in C:
int arr[5];
the only storage set aside is for the array elements themselves:
+---+
arr: | | arr[0]
+---+
| | arr[1]
+---+
| | arr[2]
+---+
| | arr[3]
+---+
| | arr[4]
+---+
The subscript operation arr[i] is still defined as *(arr + i), but instead of storing a pointer value in arr, a pointer value is computed from the expression arr. This means &arr and &arr[0] will yield the same address value, but the types of the expressions will be different (int (*)[5] vs int *, respectively).
One practical effect of this rule is that you can use the [] operator on pointer expressions as well as array expressions - given your code you can write b[i] and it will behave exactly like a[i].
Another practical effect is that when you pass an array expression as an argument to a function, what the function actually receives is a pointer to the first element. This is why you often have to pass the array size as a separate parameter, because a pointer only points to a single object of the specified type; there's no way to know from the pointer value itself whether you're pointing to the first element of an array, how many elements are in the array, etc.
Arrays carry no metadata around, so there's no way to query an array for its size, or type, or anything else at runtime. The sizeof operator is computed at compile time, not runtime.
#include<stdio.h>
int main(){
int a[] = {1,2,3};
int b[] = {4,5,6};
b = a;
return 0;
}
Result in this error:
array type 'int [3]' is not assignable
I know arrays are lvalues and are not assignable but in this case, all the compiler has to do is
reassign a pointer. b should just point to the address of a. Why isn't this doable?
"I know arrays are lvalues and are not assignable but in this case, all the compiler has to do is reassign a pointer."
"b should just point to the address of a. Why isn't this doable?"
You seem to confuse here something. b isn't a pointer. It is an array of three int elements.
b = a;
Since b is used here as lvalue in the assignment, it is taken as of type int [3], not int *. The pointer to decay rule takes no place in here for b, only for a as rvalue.
You cannot assign an array (here b) by a pointer to the first element of another array (here a) by using b = a; in C.
The syntax doesn't allow that.
That's what the error
"array type 'int [3]' is not assignable"
is saying to you for b.
Also you seem to be under the misunderstanding that the pointer to decay rule means that an array is anyhow converted to a pointer object, which can in any manner store addresses of locations of different objects.
This is not true. This conversion is only happening in a very implicit kind of way and is subject of this SO question:
Is the array to pointer decay changed to a pointer object?
If you want to assign the values from array a to the array b, you can use memcpy():
memcpy(b, a, sizeof(a));
I know arrays are lvalues and are not assignable but in this case, all the compiler has to do is reassign a pointer. b should just point to the address of a. Why isn't this doable?
Because b isn't a pointer. When you declare and allocate a and b, this is what you get:
+---+
| 1 | a[0]
+---+
| 2 | a[1]
+---+
| 3 | a[2]
+---+
...
+---+
| 4 | b[0]
+---+
| 5 | b[1]
+---+
| 6 | b[2]
+---+
No space is set aside for any pointers. There is no pointer object a or b separate from the array elements themselves.
C was derived from an earlier language called B, and in B there was a separate pointer to the first element:
+---+
b: | +-+--+
+---+ |
... |
+----+
|
V
+---+
| | b[0]
+---+
| | b[1]
+---+
...
+---+
| | b[N-1]
+---+
When Dennis Ritchie was developing C, he wanted to keep B's array semantics (specifically, a[i] == *(a + i)), but he didn't want to store that separate pointer anywhere. So instead he created the following rule - unless it is the operand of the sizeof or unary & operators, or is a string literal used to initialize a character array in a declaration, an expression of type "N-element array of T" will be converted ("decay") to an expression of type "pointer to T" and the value of the expression will be the address of the first element of the array, and that value is not an lvalue.
This has several practical effects, the most relevant here being that an array expression may not be the target of an assignment. Array expressions lose their "array-ness" under most circumstances, and simply are not treated like other types.
Edit
Actually, that misstates the case - array expression may not be the target of an assignment because an array expression is not a modifiable lvalue. The decay rule doesn't come into play. But the statement "arrays are not treated like other types" still holds.
End Edit
The upshot is that you cannot copy the contents of one array to the other using just the = operator. You must either use a library function like memcpy or copy each element individually.
Others already explained what you got wrong. I'm writing that answer to explain that actually the compiler could assign an array to another, and you can achieve the same effect with minimal change to your sample code.
Just wrap your array in a structure.
#include <stdio.h>
int main(){
struct Array3 {
int t[3];
};
struct Array3 a = {{1,2,3}};
struct Array3 b = {{4,5,6}};
a = b;
printf("%d %d %d", a.t[0], a.t[1], a.t[2]);
return 0;
}
Once the array is wrapped in a structure copying the array member of the structure works exactly as copying any other member. In other words you are copying an array. This trick is usefull in some cases like when you really want to pass an array to a function by copying it. It's slightly cleaner and safer than using memcopy for that purpose, which obviously would also work.
Henceforth the reason why it is not allowed for top level arrays is not because the compiler can't do it, but merely because that's not what most programmers usually wants to do.
Usually they just want to decay the array to a pointer. Obviously that is what you thought it should do, and direct copy of array is likely forbiden to avoid specifically that misunderstanding.
From The C Programming Language:
The array name is the address of the zeroth element.
There is one difference between an array name and a pointer that must be kept in mind. A pointer is a variable. But an array name is not a variable.
My understanding is that the array name is a constant, so it can't be assigned.
The variable b in your code is allocated on the stack as 3 consecutive ints. You can take the address of b and store it in a variable of type int*.
You could assign a value to it if you allocate the array on the heap and store only the pointer to it on the stack, in this case you could, in fact, be able to change the value of the pointer to be the same as a.
Say I have the following code:
// x = whatever number
int *arr_of_ptr[x];
int （*ptr_to_arr)[x]
int **p1 = arr_of_ptr;
int **p2 = ptr_to_arr;
My understanding of arr_of_ptr is that "dereferencing an element of arr_of_ptr results in an int" - therefore the elements of arr_of_ptr are pointers to integers. On the other hand, dereferencing ptr_to_arr results in an array that I can then nab integers from, hence ptr_to_arr points to an array.
I also have a rough understanding that arrays themselves are pointers, and that arr[p] evaluates to (arr + p * sizeof(data_type_of_arr)) where the name arr decays to the pointer to the first element of arr.
So that's all well and good, but is there any way for me to tell whether p1 and p2 are pointers to arrays or arrays of pointers without prior information?
My confusion mostly stems from the fact that (I think) we can evaluate int **p two ways:
*(p + n * size) is what's giving me an int
(*p + n * size) is what's giving me an int
In hindsight this question might be poorly worded because I'm confusing myself a bit just looking back on it, but I really don't know how to articulate myself better. Sorry.
The main difference is that this is legal:
int **p1 = arr_of_ptr;
While this is not:
int **p2 = ptr_to_arr;
Because arr_of_ptr is an array, it can (in most contexts) decay to a pointer to its first element. So because the elements of arr_of_ptr are of type int *, a pointer to an element has type int ** so you can assign it to p1.
ptr_to_arr however is not an array but a pointer, so there's no decaying happening. You're attempting to assign an expression of type int (*)[x] to an expression of type int **. Those types are incompatible, and if you attempt to use p2 you won't get what you expect.
First,
I also have a rough understanding that arrays themselves are pointers, and that arr[p] evaluates to (arr + p * sizeof(data_type_of_arr)) where the name arr decays to the pointer to the first element of arr.
This isn't strictly correct. Arrays are not pointers. Under most circumstances, expressions of array type will be converted ("decay") to expressions of pointer type and the value of the expression will be the address of the first element of the array. That pointer value is computed as necessary and isn't stored anywhere.
Exceptions to the decay rule occur when the array expression is the operand of the sizeof, _Alignof, or unary & operators, or is a string literal used to initialize a character array in a declaration.
Having said all that, ptr_to_arr has pointer type, not array type - it will not "decay" to int **.
Given the declaration
T arr[N];
the following are true:
Expression Type Decays to Equivalent expression
---------- ---- --------- ---------------------
arr T [N] T * &arr[0]
*arr T n/a arr[0]
arr[i] T n/a n/a
&arr T (*)[N] n/a n/a
The expressions arr, &arr[0], and &arr all yield the same value (modulo any differences in representation between types). arr and &arr[0] have the same type, "pointer to T" (T *), while &arr has type "pointer to N-element array of T" (T (*)[N]).
If you replace T with pointer type P *, such that the declaration is now
P *arr[N];
you get the following:
Expression Type Decays to Equivalent expression
---------- ---- --------- ---------------------
arr P *[N] P ** &arr[0]
*arr P * n/a arr[0]
arr[i] P * n/a n/a
&arr P *(*)[N] n/a n/a
So given your declarations, it would be more correct to write something like this:
int arr[x];
int *p1 = arr; // the expression arr "decays" to int *
int *arr_of_ptr[x];
int **p2 = arr_of_ptr; // the expression arr_of_ptr "decays" to int **
/**
* In the following declarations, the array expressions are operands
* of the unary & operator, so the decay rule doesn't apply.
*/
int (*ptr_to_arr)[x] = &arr;
int *(*ptr_to_arr_of_ptr)[x] = &arr_of_ptr;
Again, ptr_to_arr and ptr_to_arr_of_ptr are pointers, not arrays, and do not decay to a different pointer type.
EDIT
From the comments:
Can I just hand-wavily explain it as： an array of pointers has a name that can decay to a pointer,
Yeah, -ish, just be aware that it is hand-wavey and not really accurate (which is shown by example below). If you are a first-year student, your institution isn't doing you any favors by making you deal with C this early. While it is the substrate upon which most of the modern computing ecosystem is built, it is an awful teaching language. Awful. Yes, it's a small language, but aspects of it are deeply unintuitive and confusing, and the interplay between arrays and pointers is one of those aspects.
an array of pointers has a name that can decay to a pointer, but a pointer to an array, even when dereferenced, does not give a give me something that decays to a pointer?
Actually...
If ptr_to_arr has type int (*)[x], then the expression *ptr_to_arr would have type int [x], which would decay to int *. The expression *ptr_to_arr_of_ptr would have type int *[x], which would decay to int **. This is why I keep using the term "expression of array type" when talking about the decay rule, rather than just the name of the array.
Something I have left out of my explanations until now - why do array expressions decay to pointers? What's the reason for this incredibly confusing behavior?
C didn't spring fully-formed from the brain of Dennis Ritchie - it was derived from an earlier language named B (which was derived from BCPL, which was derived from CPL, etc.)1. B was a "typeless" language, where data was simply a sequence of words or "cells". Memory was modeled as a linear array of "cells". When you declared an N-element array in B, such as
auto arr[N];
the compiler would set aside all the cells necessary for the array elements, plus an extra cell that would store the numerical offset (basically, a pointer) to the first element of the array, and that cell would be bound to the variable arr:
+---+
arr: | +-+-----------+
+---+ |
... |
+---+ |
| | arr[0] <--+
+---+
| | arr[1]
+---+
...
+---+
| | arr[N-1]
+---+
To index into the array, you'd offset i cells from the location stored in arr and dereference the result. IOW, a[i] was exactly equivalent to *(a + i).
When Ritchie was developing the C language, he wanted to keep B's array semantics (a[i] is still exactly equivalent to *(a + i)), but for various reasons he didn't want to store that pointer to the first element. So, he got rid of it entirely. Now, when you declare an array in C, such as
int arr[N];
the only storage set aside is for the array elements themselves:
+---+
| | arr[0]
+---+
| | arr[1]
+---+
...
+---+
| | arr[N-1]
+---+
There is no separate object arr which stores a pointer to the first element (which is part of why array expressions cannot be the target of an assignment - there's nothing to assign to). Instead, that pointer value is computed as necessary when you need to subscript into the array.
This same principal holds for multi-dimensional arrays as well. Assume the following:
int a[2][2] = { { 1, 2 }, { 3, 4 } };
What you get in memory is the following:
Viewed as int Viewed as int [2]
+---+ +---+
a: | 1 | a[0][0] a:| 1 | a[0]
+---+ + - +
| 2 | a[0][1] | 2 |
+---+ +---+
| 3 | a[1][0] | 3 | a[1]
+---+ + - +
| 4 | a[1][1] | 4 |
+---+ +---+
On the left we view it as a sequence of int, while on the right we view it as a sequence of int [2].
Each a[i] has type int [2], which decays to int *. The expression a itself decays from type int [2][2] to int (*)[2] (not int **).
The expression a[i][j] is exactly equivalent to *(a[i] + j), which is equivalent to *( *(a + i) + j ).
As detailed in The Development of the C Language
#include <stdio.h>
int main(void) {
// your code goes here
int arr[] = {1,2,3};
int *p1 = &arr[0];
int *p2 = &arr[1];
int *p3 = &arr[2];
int* arr2[3];
arr2[0] = p1;
arr2[1] = p2;
arr2[2] = p3;
int *p4 = &arr;
printf("%d\n", sizeof(p4));
printf("%d\n", sizeof(arr2));
printf("%d\n", *p4); // not **p4
printf("%d\n", **arr2);
return 0;
}
In the above code arr is a normal integer array with 3 elements.
p1, p2, and p3 are normal pointers to these elements.
arr2 is an array of pointers storing p1, p2, and p3.
p4 is a pointer to array pointing to array arr
According to your question, you need to differentiate between p4 and arr2
Since, p4 is a pointer, its size is fixed (8 bytes) while size of arr2 vaires on how many elements it contains (8x3=24).
Also, to print value contained in p4 use use single dereferencing (*p4) not **p4 (illegal), while to print value contained in arr2 use use double dereferencing (**arr2).
The output of above code is :
8
24
1
1
I learnt that there are two ways of declaring an array in C:
int array[] = {1,2,3};
and:
int* arr = malloc(3*sizeof(int));
Why is arr called a free pointer ? And why can't I change the address contained in array while I can do it with array ?
As said in comments, you learned something incorrect, from a bad source.
In the second case, arr is not an array, it's a pointer. A pointer that (if the allocation succeeds) happens to contain the address of a block of memory that can hold three ints, but that's not an array.
This confusion probably comes from the fact that arrays "decay" to pointers in some contexts, but that does not make them equivalent.
Let's look at how the two objects are laid out in memory:
+---+
array: | 1 | array[0]
+---+
| 2 | array[1]
+---+
| 3 | array[2]
+---+
+---+ +---+
arr: | | ---------> | ? | arr[0]
+---+ +---+
| ? | arr[1]
+---+
| ? | arr[2]
+---+
So, one immediate difference - there is no array object that is separate from the array elements themselves, whereas arr is a separate object from the array elements. Only array is an actual array as far as C is concerned - arr is just a pointer to a single object, which may be the first element of a sequence of objects or not.
This is why you can assign a new address value to arr, but not to array - in the second case, there's nothing to assign the new address value to. It's like trying to change the address of a scalar variable - you can't do it, because the operation doesn't make any sense.
It also means that the address of array[0] is the same as the address of array. The expressions &array[0], array, and &array will all yield the same address value, although the types of the expressions will be different (int *, int *, and int (*)[3], respectively). By contrast, the address of arr is not the same as the address of arr[0]; the expressions arr and &arr[0] will yield the same value, but &arr will not, and its type will be int ** instead of int (*)[3].
I know array in C does essentially behaves like a pointer except at some places like (sizeof()). Apart from that pointer and array variables dont differ except in their declaration.
For example consider the two declarations:
int arr[] = {11,22,33};
int *arrptr = arr;
Now here is how they behave same:
printf("%d %d", arrptr[0], arr[0]); //11 11
printf("%d %d", *arrptr, *arr); //11 11
But here is one more place I found they differ:
//the outputs will be different on your machine
printf("%d %d", &arrptr, &arr); //2686688 2686692 (obviously different address)
printf("%d %d", arrptr, arr); //2686692 2686692 (both same)
Here the issue is with last line. I understand that arrptr contains the address of arr. Thats why the first address printed in last line is 2686692. I also understand that logically the address (&arr) and contents (arr) of arr should be same unlike arrptr. But then whats exactly that which (internally at implementation level) that makes this happen?
When the unary & operator is applied to an array, it returns a pointer to an array. When applied to a pointer, it returns a pointer to a pointer. This operator together with sizeof represent the few contexts where arrays do not decay to pointers.
In other words, &arrptr is of type int **, whereas &arr is of type int (*)[3]. &arrptr is the address of the pointer itself and &arr is the beginning of the array (like arrptr).
The subtle part: arrptr and &arr have the same value (both point to the beginning of the array), but are of a different type. This difference will show if you do any pointer arithmetic to them – with arrptr the implied offset will be sizeof(int), whereas with &arr it will be sizeof(int) * 3.
Also, you should be using the %p format specifier to print pointers, after casting to void *.
I know array in C does essentially behaves like a pointer except at some places like (sizeof()). Apart from that pointer and array variables dont differ except in their declaration.
This is not quite true. Array expressions are treated as pointer expressions in most circumstances, but arrays and pointers are completely different animals.
When you declare an array as
T a[N];
it's laid out in memory as
+---+
a: | | a[0]
+---+
| | a[1]
+---+
| | a[2]
+---+
...
+---+
| | a[N-1]
+---+
One thing immediately becomes obvious - the address of the first element of the array is the same as the address of the array itself. Thus, &a[0] and &a will yield the same address value, although the types of the two expressions are different (T * vs. T (*)[N]), and the value may possibly adjusted based on type.
Here's where things get a little confusing - except when it is the operand of the sizeof or unary & operator, or is a string literal used to initialize a character array in a declaration, an expression of type "N-element array of T" will be converted ("decay") to an expression of type "pointer to T", and the value of the expression will be the address of the first element of the array.
This means the expression a also yields the same address value as &a[0] and &a, and it has the same type as &a[0]. Putting this all together:
Expression Type Decays to Value
---------- ---- --------- -----
a T [N] T * Address of a[0]
&a T (*)[N] n/a Address of a
*a T n/a Value of a[0]
a[i] T n/a Value of a[i]
&a[i] T * n/a Address of a[i]
sizeof a size_t n/a Number of bytes in a
So why does this conversion rule exist in the first place?
C was derived from an earlier language called B (go figure). B was a typeless
language - everything was treated as basically an unsigned integer. Memory
was seen as a linear array of fixed-length "cells". When you declared an
array in B, an extra cell was set aside to store the offset to the first
element of the array:
+---+
a:| | ----+
+---+ |
... |
+-------+
|
V
+---+
| | a[0]
+---+
| | a[1]
+---+
...
+---+
| | a[N-1]
+---+
The array subscript operation a[i] was defined as *(a + i); that is, take the offset value stored in a, add i, and dereference the result.
When Ritchie was designing C, he wanted to keep B's array semantics, but couldn't figure out what to do with the explicit pointer to the first element, so he got rid of it. Thus, C keeps the array subscripting definition a[i] == *(a + i) (given the address a, offset i elements from that address and dereference the result), but doesn't set aside space for a separate pointer to the first element of the array - instead, it converts the array expression a to a pointer value.
This is why you see the same output when you print the values of arr and arrptr. Note that you should print out pointer values using the %p conversion specifier and cast the argument to void *:
printf( "arr = %p, arrptr = %p\n", (void *) arr, (void *) arrptr );
This is pretty much the only place you need to explicitly cast a pointer value to void * in C.