Related
Before anyone closes this question because there's another related one, hear me out. I've already looked at the other question and I didn't get it. I would like to create a pointer to a multi dimensional array but I don't know how. I thought I was supposed to do it like this:
int test_arr[2][4];
int *ptr_array = test_arr;
But when I do that, I get a warning saying:
incompatible pointer types initializing 'int *' with an expression of type 'int [2][4]'
I have no idea what I'm doing wrong. Can someone help me please?
In order to do pointer arithmetic the pointer has to know the size of what it is pointing to.
int test_arr[2][4]; is equivalent to 2 elements of type int[4]. So whenever you add 1 to the pointer, it will jump the size of 4 integers. If you had int* it would increment the size of a single integer only.
Like said in the comments, what you want is: int (*ptr_array)[4] = test_arr;
I know, the syntax is a little weird, but that's how you do it.
You need to understand the basics.
[Below in post, read 1D as 1 dimension and 2D as 2 dimension]
First, lets take an example of pointer to 1D array:
Take an int type and a pointer pointing to it:
int x = 1;
int *px = &x;
The type of x is int and the type of &x is int * and type of px is int * i.e. it can hold the address of an int type variable. Hence the assignment is valid.
Take a 1D int array and a pointer pointing to it:
int arr[2] = {1,2};
int *parr = arr;
Note that the type of px and parr is same i.e. int * and assigning arr to parr is completely valid although the arr is a 1D array.
This is because of the fact that, in C, an array of type converted to pointer of type that points to the initial element of the array object but there are few exceptions to this rule and one of them is when used unary & operator.
The type of arr is int [2] but in the above statment (parr declare and initialisation statement) it is converted to pointer pointing to initial element of array which is an integer i.e. arr is converted to int * type in the int *parr = arr; statement. Hence, it is valid to assign arr to parr or to any variable of type int *.
As, I have mentioned that there are few exceptions to the rule (array of type converted to pointer of type....) and one of the exception is when used unary & operator. So type of &arr will be int (*)[2]. That means you can assign &arr to any variable of type int (*)[2].
int (*parr2)[2] = &arr;
i.e. parr2 is a pointer to an array of 2 integer.
The difference between this
int *parr = arr;
and
int (*parr2)[2] = &arr;
is that parr can hold address of any int pointer but parr2 can hold the address of any array of 2 integers. Both are pointing to same address but their types are different.
If you understood the above mentioned things completely then read below otherwise read the above part again.
Now lets come to 2D array.
Take a 2D array and a pointer pointing to it:
int arr[2][2];
int (*ptr)[2] = arr;
In the above ptr declare and initialisation statement, the type of arr is int (*)[2]. Why??
Recall the rule mentioned above - array of type converted to pointer of type that points to the initial element of the array object.
arr is array of 2 1D arrays of size 2 (2 int elements). So, it's first element is array of 2 int and, by rule, it will be converted to pointer to the initial element i.e. pointer to
1D array of 2 int which is int (*)[2].
Hence, you can assign arr to a pointer variable of type int (*)[2].
The type of &arr is int (*)[2][2], so this is also perfectly fine:
int (*ptr2)[2][2] = &arr;
So, there are two ways you can have pointer to 2D arr but remember that the types of ptr and ptr2 is different. This matters, for e.g., if you add 1 to pointer ptr it will be incremented by size of int [2] but if you add 1 to ptr2 it will be incremented by size of int [2][2].
How to access the array members using pointer:
In case of 1D array:
int arr[2] = {1,2};
int *parr = arr;
int (*parr2)[2] = &arr;
// access arr members via parr pointer
printf ("%d\n", parr[0]); // this will give value of first member of array arr
// access arr members via parr2 pointer
printf ("%d\n", (*parr2)[0]); // this will give value of first member of array arr
Similarly, in case of 2D array:
int test_arr[2][4];
int (*ptr_array)[4] = test_arr;
int (*ptr_array2)[2][4] = &test_arr;
// access test_arr members via ptr_array pointer
printf ("%d\n", ptr_array[0][0]); // this will give value of first member of array test_arr
// access test_arr members via ptr_array2 pointer
printf ("%d\n", (*ptr_array2)[0][0]); // this will give value of first member of array test_arr
Based on above explanation, try yourself creating multidimensional array (> 2D array) pointers and play with them.
Hope, above explanation resolves your all confusions related to pointer to a 2D array and you understood now why compiler is giving incompatible pointer types .... warning on statement int *ptr_array = test_arr;. It's all about type compatibility during initialisation/assignment.
How can you interpret the following line of code?
int (*arrayABC)[10];
In my textbook, it says that we have a pointer to a pointer to the 0th element of an integer array.
However, I don't quite understand this.
My interpretation: We have some variable, which gets as its value some address. This address is then the address of the 0th element of an UNNAMED integer array. Basically we have a pointer to the 0th element.
Why then to have a pointer TO A POINTER?
This is a pointer to an array. It is not a pointer to a pointer. Arrays and pointers are different. An array has an address, but an array is not an address. An array is a series of contiguous elements.
This pointer points to the whole array and not just the first element, in the same way that a float * points to the whole float and not just the first byte.
If you have for example:
int foo[10];
int (*arrayABC)[10] = &foo;
then the expressions (*arrayABC) and foo are identical. E.g. foo[3] is the same as (*arrayABC)[3].
If you have an object of a type T then a pointer to the object is declared like
T obj;
T *ptr = &obj;
Now let's the type T is defined the following way
typedef int T[10];
Thus the code above
T obj;
T *ptr = &obj;
can be rewritten using the typedef definition like
int obj[10];
int (*ptr)[10] = &obj;
In the both cases, when T is some abstract type and when T is an alias for int[10], ptr is a pointer to an object. In the last case ptr is a pointer to an array of 10 elements of type int. That is the object pointed to by ptr has array type. Arrays and poiters are different types.
Try the following simple demonstrative program
#include <stdio.h>
int main( void )
{
typedef int T[10];
T a;
T *pa = &a;
printf( "%zu\n", sizeof( *pa ) );
int b[10];
int ( *pb )[10] = &b;
printf( "%zu\n", sizeof( *pb ) );
}
Its output is
40
40
As you can see the both values are equal to the size of an integer array of 10 elements. So the pointers point to arrays.
If you write
int *arrayABC[10];
you will get an array of 10 pointers of type int *.
If you write
int (*arrayABC)[10];
you will get a pointer to an array of 10 integers.
In my textbook, it says that we have a pointer to a pointer to the 0th element of an integer array.
Your textbook is ... badly worded, let's leave it at that. It's a pointer to a 10-element array of int.
I can kind of see where the book is coming from. Given the declaration
int (*a)[10];
the expression *a will have type int [10] (10-element array of int). However, unless that expression is the operand of the sizeof or unary & operators, it will be converted ("decay") from type int [10] to int *, and the value of the expression will be the address of the first element of the array:
Expression Type "Decays" to Value
---------- ---- ----------- -----
*a int [10] int * Address of first element of the array;
equivalent to `&(*a)[0]`.
Thus, if I call a function like
foo( *a );
what the function actually receives will be a pointer to the first element of the pointed-to array (its type will be int *, not int **).
So in effect, yes, a often winds up behaving as a pointer to a pointer to the first element of the array; however, its type is not "pointer to pointer to int", and type matters. Given:
int (*a)[10];
int **b;
then you'll get different behavior for the following:
sizeof *a vs. sizeof *b
a + i vs b + i (also applies to a++ vs b++)
I know, that static arrays are laid out contiguously in memory.
So, for example, int T[10][10] is basically stored the same way as int T[100].
I can access element of index i, j many ways, for example:
int T[10][10];
/*filling array*/
int i=3, j=7;
int x = T[i][j];
//EDIT - old and wrong: int * ptr = T;
int * ptr = &T[0][0];
int y = *(ptr + 10* i + j);
On the other hand, when I create dynamicaly allocated 2-dimensional array by myself:
int ** T;
T = malloc(10 * sizeof(int *));
for(i = 0; i < N; i++)
T[i] = malloc(10 * sizeof(int));
My array contains pointers to
It is obvious, that I can access element of this array by:
int i=3, j=7;
int x = *(*(T+i)+j);
And now my question: why and how does it work for static arrays?
Why does
int T[10][10];
/*filling array*/
int i=3, j=7;
int x = *(*(T+i)+j);
return good value to x, when this table doesn't contain pointers to arrays? *(*(T+i)) shouldn't have sense there in my opinion, end even if, it should return T[0][i], as T points at the first element of array. How does compiler interpret this, is * something other than dereference here? Enlighten me.
For starters:
int * ptr = T;
That's not going to actually work, at least without your compiler yelling at you. Very loudly. The correct way to do this is:
int * ptr = &t[0][0];
This point is actually very relevant to your question.
As you know, when used in an expression, an array gets decayed to a pointer. For example:
char foo[10];
bar(foo);
When used in an expression, like a parameter to a function, the array decays to a pointer to the underlying type. foo gets you a char *, here.
However, and this is the key point: the array only decays one level. If the array is a two-dimensional array, the array does not get decayed to the underlying value, an int in this case. The two dimensional array reference decays to a pointer to a one-dimensional array:
int T[10][10];
/*filling array*/
int i=3, j=7;
int x = *(*(T+i)+j);
The sequence of steps that occurs here:
T decays to a pointer to an array of 10 integers, or int (*)[10]
The addition of i advances the pointer by the given value. The pointer is advanced by the size of the element being pointed to. Since the pointer points to array of 10 integers, the pointer is advanced accordingly. If i was 2, the pointer is advanced by "two arrays of 10 integers", loosely speaking.
The * operator takes a "pointer to an array of 10 integers" and gives you "an array of 10 integers" as a result. In other words: from int (*)[10] to int [10].
Since the result is used in an expression, namely the left operand of + j, and the left operand is an array type, the array type decays to a "pointer to int".
j is added to the result, and dereferenced.
Why does
int T[10][10];
/*filling array*/
int i=3, j=7;
int x = *(*(T+i)+j);
return good value to x
The magic is all in *(*(T+3)+7) (I have converted to the literal values).
T is an array (of size 10) of arrays (of size 10) of int.
When T is used in an an expression it decays into a pointer to its first element, so it decays to "pointer to arrays (of size 10) of int".
Adding an integer to that pointer will advance to the fourth element of the array.
So T+3 is a pointer to an array of 10 ints, and specifically the fourth such array in T.
*(T+3) indirects through that pointer to give an l-value of type "array of 10 ints".
Ah-ha! That is another array being used in an expression - so it decays into a pointer to it's first element! (It wouldn't decay in sizeof, so sizeof(*(T+3)) would be typically 40.)
(*(T+3) + 7) just points at the eight element in the array, and ...
*(*(T+3) + 7) is an l-value of type int!
I wrote the following code in C:
#include<stdio.h>
int main()
{
int a[10][10]={1};
//------------------------
printf("%d\n",&a);
printf("%d\n",a);
printf("%d\n",*a);
//-------------------------
printf("%d",**a);
return 0;
}
With the above 3 printf statements I got the same value. On my machine it's 2686384. But with the last statement I got 1.
Isn't it something going wrong? These statements mean:
The address of a is 2686384
The value stored in a is 2686384
the value that is stored at address of variable pointed by a (i.e. at 2686384) is 2686384.
This means a must be something like a variable pointing towards itself...
Then why is the output of *(*a) 1? Why isn't it evaluated as *(*a)=*(2686384)=2686384?
#include<stdio.h>
int main()
{
// a[row][col]
int a[2][2]={ {9, 2}, {3, 4} };
// in C, multidimensional arrays are really one dimensional, but
// syntax alows us to access it as a two dimensional (like here).
//------------------------
printf("&a = %d\n",&a);
printf("a = %d\n",a);
printf("*a = %d\n",*a);
//-------------------------
// Thing to have in mind here, that may be confusing is:
// since we can access array values through 2 dimensions,
// we need 2 stars(asterisk), right? Right.
// So as a consistency in this aproach,
// even if we are asking for first value,
// we have to use 2 dimensional (we have a 2D array)
// access syntax - 2 stars.
printf("**a = %d\n", **a ); // this says a[0][0] or *(*(a+0)+0)
printf("**(a+1) = %d\n", **(a+1) ); // a[1][0] or *(*(a+1)+0)
printf("*(*(a+1)+1) = %d\n", *(*(a+1)+1) ); // a[1][1] or *(*(a+1)+1)
// a[1] gives us the value on that position,
// since that value is pointer, &a[i] returns a pointer value
printf("&a[1] = %d\n", &a[1]);
// When we add int to a pointer (eg. a+1),
// really we are adding the lenth of a type
// to which pointer is directing - here we go to the next element in an array.
// In C, you can manipulate array variables practically like pointers.
// Example: littleFunction(int [] arr) accepts pointers to int, and it works vice versa,
// littleFunction(int* arr) accepts array of int.
int b = 8;
printf("b = %d\n", *&b);
return 0;
}
An expression consisting the the name of an array can decay to a pointer to the first element of the array. So even though a has type int[10][10], it can decay to int(*)[10].
Now, this decay happens in the expression *a. Consequently the expression has type int[10]. Repeating the same logic, this again decays to int*, and so **a is an int, which is moreover the first element of the first element of the array a, i.e. 1.
The other three print statements print out the address of, respectively, the array, the first element of the array, and the first element of the first element of the array (which are of course all the same address, just different types).
First, a word on arrays...
Except when it is the operand0 of the sizeof, _Alignof, or unary & operators, or is a string literal being used to initialize another 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 in the array.
The expression &a has type "pointer to 10-element array of 10-element array of int", or int (*)[10][10]. The expression a has type "10-element array of 10-element array of int", which by the rule above decays to "pointer to 10-element array of int", or int (*)[10]. And finally, the expression *a (which is equivalent to a[0]) has type "10-element array of int", which again by the rule above decays to "pointer to int".
All three expressions have the same value because the address of an array and the address of its first element are the same: &a[0][0] == a[0] == *a == a == &a. However, the types of the expressions are different, which matters when doing pointer arithmetic. For example, if I have the following declarations:
int (*ap0)[10][10] = &a;
int (*ap1)[10] = a;
int *ip = *a;
then ap0++ would advance ap0 to point to the next 10x10 array of int, ap1++ would advance ap1 to pointer to the next 10-element array of int (or a[1]), and ip++ would advance ip to point to the next int (&a[0][1]).
**a is equivalent to *a[0] which is equivalent to a[0][0]. which is the value of the first element of a and has type int and the value 1 (note that only a[0][0] is initialized to 1; all remaining elements are initialized to 0).
Note that you should use %p to print out pointer values:
printf("&a = %p\n", &a);
printf(" a = %p\n", a);
printf("*a = %p\n", *a);
First of all, if you want to print out pointer values, use %p - if you're on a 64 bit machine int almost certainly is smaller than a pointer.
**a is double dereferencing what's effectively a int**, so you end up with what the first element of the first sub-array is: 1.
If you define a as T a[10] (where T is some typedef), then a simple unadorned a means the address of the start of the array, the same as &a[0]. They both have type T*.
&a is also the address of the start of the array, but it has type T**.
Things become trickier in the presence of multi-dimensional arrays. To see what is happening, it is easier to break things down into smaller chunks using typedefs. So, you effectively wrote
typedef int array10[10];
array10 a[10];
[Exercise to reader: What is the type of a? (it is not int**)]
**a correctly evaluates to the first int in the array a.
From C99 Std
Consider the array object defined by the declaration
int x[3][5];
Here x is a 3 × 5 array of ints; more precisely, x is an array of three element objects, each of which is an array of five ints. In the expression x[i], which is equivalent to (*((x)+(i))), x is first converted to a pointer to the initial array of five ints. Then i is adjusted according to the type of x, which conceptually entails multiplying i by the size of the object to which the pointer points, namely an array of five int objects. The results are added and indirection is applied to yield an array of five ints. When used in the expression x[i][j], that array is in turn converted to a pointer to the first of the ints, so x[i][j] yields an int.
so,
Initial array will be x[0][0] only.
all x, &x and *x will be pointing to x[0][0].
No, there's nothing wrong with your code. Just they way you are thinking about it... The more I think about it the harder I realize this is to explain, so before I go in to this, keep these points in mind:
arrays are not pointers, don't think of them that way, they are different types.
the [] is an operator. It's a shift and deference operator, so when I write printf("%d",array[3]); I am shifting and deferencing
So an array (lets think about 1 dimension to start) is somewhere in memory:
int arr[10] = {1};
//Some where in memory---> 0x80001f23
[1][1][1][1][1][1][1][1][1][1]
So if I say:
*arr; //this gives the value 1
Why? because it's the same as arr[0] it gives us the value at the address which is the start of the array. This implies that:
arr; // this is the address of the start of the array
So what does this give us?
&arr; //this will give us the address of the array.
//which IS the address of the start of the array
//this is where arrays and pointers really show some difference
So arr == &arr;. The "job" of an array is to hold data, the array will not "point" to anything else, because it's holding its own data. Period. A pointer on the other hand has the job to point to something else:
int *z; //the pointer holds the address of someone else's values
z = arr; //the pointer holds the address of the array
z != &z; //the pointer's address is a unique value telling us where the pointer resides
//the pointer's value is the address of the array
EDIT:
One more way to think about this:
int b; //this is integer type
&b; //this is the address of the int b, right?
int c[]; //this is the array of ints
&c; //this would be the address of the array, right?
So that's pretty understandable how about this:
*c; //that's the first element in the array
What does that line of code tell you? if I deference c, then I get an int. That means just plain c is an address. Since it's the start of the array it's the address of the array, thus:
c == &c;
Is an array's name a pointer in C?
If not, what is the difference between an array's name and a pointer variable?
An array is an array and a pointer is a pointer, but in most cases array names are converted to pointers. A term often used is that they decay to pointers.
Here is an array:
int a[7];
a contains space for seven integers, and you can put a value in one of them with an assignment, like this:
a[3] = 9;
Here is a pointer:
int *p;
p doesn't contain any spaces for integers, but it can point to a space for an integer. We can, for example, set it to point to one of the places in the array a, such as the first one:
p = &a[0];
What can be confusing is that you can also write this:
p = a;
This does not copy the contents of the array a into the pointer p (whatever that would mean). Instead, the array name a is converted to a pointer to its first element. So that assignment does the same as the previous one.
Now you can use p in a similar way to an array:
p[3] = 17;
The reason that this works is that the array dereferencing operator in C, [ ], is defined in terms of pointers. x[y] means: start with the pointer x, step y elements forward after what the pointer points to, and then take whatever is there. Using pointer arithmetic syntax, x[y] can also be written as *(x+y).
For this to work with a normal array, such as our a, the name a in a[3] must first be converted to a pointer (to the first element in a). Then we step 3 elements forward, and take whatever is there. In other words: take the element at position 3 in the array. (Which is the fourth element in the array, since the first one is numbered 0.)
So, in summary, array names in a C program are (in most cases) converted to pointers. One exception is when we use the sizeof operator on an array. If a was converted to a pointer in this context, sizeof a would give the size of a pointer and not of the actual array, which would be rather useless, so in that case a means the array itself.
When an array is used as a value, its name represents the address of the first element.
When an array is not used as a value its name represents the whole array.
int arr[7];
/* arr used as value */
foo(arr);
int x = *(arr + 1); /* same as arr[1] */
/* arr not used as value */
size_t bytes = sizeof arr;
void *q = &arr; /* void pointers are compatible with pointers to any object */
If an expression of array type (such as the array name) appears in a larger expression and it isn't the operand of either the & or sizeof operators, then the type of the array expression is converted from "N-element array of T" to "pointer to T", and the value of the expression is the address of the first element in the array.
In short, the array name is not a pointer, but in most contexts it is treated as though it were a pointer.
Edit
Answering the question in the comment:
If I use sizeof, do i count the size of only the elements of the array? Then the array “head” also takes up space with the information about length and a pointer (and this means that it takes more space, than a normal pointer would)?
When you create an array, the only space that's allocated is the space for the elements themselves; no storage is materialized for a separate pointer or any metadata. Given
char a[10];
what you get in memory is
+---+
a: | | a[0]
+---+
| | a[1]
+---+
| | a[2]
+---+
...
+---+
| | a[9]
+---+
The expression a refers to the entire array, but there's no object a separate from the array elements themselves. Thus, sizeof a gives you the size (in bytes) of the entire array. The expression &a gives you the address of the array, which is the same as the address of the first element. The difference between &a and &a[0] is the type of the result1 - char (*)[10] in the first case and char * in the second.
Where things get weird is when you want to access individual elements - the expression a[i] is defined as the result of *(a + i) - given an address value a, offset i elements (not bytes) from that address and dereference the result.
The problem is that a isn't a pointer or an address - it's the entire array object. Thus, the rule in C that whenever the compiler sees an expression of array type (such as a, which has type char [10]) and that expression isn't the operand of the sizeof or unary & operators, the type of that expression is converted ("decays") to a pointer type (char *), and the value of the expression is the address of the first element of the array. Therefore, the expression a has the same type and value as the expression &a[0] (and by extension, the expression *a has the same type and value as the expression a[0]).
C was derived from an earlier language called B, and in B a was a separate pointer object from the array elements a[0], a[1], etc. Ritchie wanted to keep B's array semantics, but he didn't want to mess with storing the separate pointer object. So he got rid of it. Instead, the compiler will convert array expressions to pointer expressions during translation as necessary.
Remember that I said arrays don't store any metadata about their size. As soon as that array expression "decays" to a pointer, all you have is a pointer to a single element. That element may be the first of a sequence of elements, or it may be a single object. There's no way to know based on the pointer itself.
When you pass an array expression to a function, all the function receives is a pointer to the first element - it has no idea how big the array is (this is why the gets function was such a menace and was eventually removed from the library). For the function to know how many elements the array has, you must either use a sentinel value (such as the 0 terminator in C strings) or you must pass the number of elements as a separate parameter.
Which *may* affect how the address value is interpreted - depends on the machine.
An array declared like this
int a[10];
allocates memory for 10 ints. You can't modify a but you can do pointer arithmetic with a.
A pointer like this allocates memory for just the pointer p:
int *p;
It doesn't allocate any ints. You can modify it:
p = a;
and use array subscripts as you can with a:
p[2] = 5;
a[2] = 5; // same
*(p+2) = 5; // same effect
*(a+2) = 5; // same effect
The array name by itself yields a memory location, so you can treat the array name like a pointer:
int a[7];
a[0] = 1976;
a[1] = 1984;
printf("memory location of a: %p", a);
printf("value at memory location %p is %d", a, *a);
And other nifty stuff you can do to pointer (e.g. adding/substracting an offset), you can also do to an array:
printf("value at memory location %p is %d", a + 1, *(a + 1));
Language-wise, if C didn't expose the array as just some sort of "pointer"(pedantically it's just a memory location. It cannot point to arbitrary location in memory, nor can be controlled by the programmer). We always need to code this:
printf("value at memory location %p is %d", &a[1], a[1]);
I think this example sheds some light on the issue:
#include <stdio.h>
int main()
{
int a[3] = {9, 10, 11};
int **b = &a;
printf("a == &a: %d\n", a == b);
return 0;
}
It compiles fine (with 2 warnings) in gcc 4.9.2, and prints the following:
a == &a: 1
oops :-)
So, the conclusion is no, the array is not a pointer, it is not stored in memory (not even read-only one) as a pointer, even though it looks like it is, since you can obtain its address with the & operator. But - oops - that operator does not work :-)), either way, you've been warned:
p.c: In function ‘main’:
pp.c:6:12: warning: initialization from incompatible pointer type
int **b = &a;
^
p.c:8:28: warning: comparison of distinct pointer types lacks a cast
printf("a == &a: %d\n", a == b);
C++ refuses any such attempts with errors in compile-time.
Edit:
This is what I meant to demonstrate:
#include <stdio.h>
int main()
{
int a[3] = {9, 10, 11};
void *c = a;
void *b = &a;
void *d = &c;
printf("a == &a: %d\n", a == b);
printf("c == &c: %d\n", c == d);
return 0;
}
Even though c and a "point" to the same memory, you can obtain address of the c pointer, but you cannot obtain the address of the a pointer.
The following example provides a concrete difference between an array name and a pointer. Let say that you want to represent a 1D line with some given maximum dimension, you could do it either with an array or a pointer:
typedef struct {
int length;
int line_as_array[1000];
int* line_as_pointer;
} Line;
Now let's look at the behavior of the following code:
void do_something_with_line(Line line) {
line.line_as_pointer[0] = 0;
line.line_as_array[0] = 0;
}
void main() {
Line my_line;
my_line.length = 20;
my_line.line_as_pointer = (int*) calloc(my_line.length, sizeof(int));
my_line.line_as_pointer[0] = 10;
my_line.line_as_array[0] = 10;
do_something_with_line(my_line);
printf("%d %d\n", my_line.line_as_pointer[0], my_line.line_as_array[0]);
};
This code will output:
0 10
That is because in the function call to do_something_with_line the object was copied so:
The pointer line_as_pointer still contains the same address it was pointing to
The array line_as_array was copied to a new address which does not outlive the scope of the function
So while arrays are not given by values when you directly input them to functions, when you encapsulate them in structs they are given by value (i.e. copied) which outlines here a major difference in behavior compared to the implementation using pointers.
NO. An array name is NOT a pointer. You cannot assign to or modify an array name, but you can for a pointer.
int arr[5];
int *ptr;
/* CAN assign or increment ptr */
ptr = arr;
ptr++;
/* CANNOT assign or increment arr */
arr = ptr;
arr++;
/* These assignments are also illegal */
arr = anotherarray;
arr = 0;
From K&R Book:
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.
sizeof is the other big difference.
sizeof(arr); /* size of the entire array */
sizeof(ptr); /* size of the memory address */
Arrays do behave like or decay into a pointer in some situations (&arr[0]). You can see other answers for more examples of this. To reiterate a few of these cases:
void func(int *arr) { }
void func2(int arr[]) { } /* same as func */
ptr = arr + 1; /* pointer arithmetic */
func(arr); /* passing to function */
Even though you cannot assign or modify the array name, of course can modify the contents of the array
arr[0] = 1;
The array name behaves like a pointer and points to the first element of the array. Example:
int a[]={1,2,3};
printf("%p\n",a); //result is similar to 0x7fff6fe40bc0
printf("%p\n",&a[0]); //result is similar to 0x7fff6fe40bc0
Both the print statements will give exactly same output for a machine. In my system it gave:
0x7fff6fe40bc0