#include<stdio.h>
void main()
{
char s[10][10];
int i;
for(i=0;i<4;i++)
scanf("%s",s[i]);
printf("%s",s);
printf("%s",s+1);
printf("%s",s[1]+1);
}
When I type the above line of code first printf statement will print the first string and second printf will print the second string since s[1] is equivalent to s+1. But the third printf will print the second string starting from the second character.
If s[1] is equivalent to s+1 why s[1]+1 does not give the result of s+2?
I do not get the idea of address calculation for 2D string array.
The way pointer arithmetic works, s[i] is equal to *(s + i). So s[1]+1 is actually *(s + 1) + 1, which is not the same as either s + 2 or *(s + 2).
So, let's talk about pointer arithmetic for a second. Given a pointer to any type T:
T *p;
the expression p+1 will evaluate to the address of the next object of type T. So if T is int, then p+1 will give us the address of the next int object after p. If p is 0x8000 and sizeof (int) is 4, then p+1 evaluates to 0x8004.
Where things get fun is if we're working with array expressions. Assume the following:
T a[N];
Except when it is the operand of the sizeof or unary & operators, an expression of type "N-element array of T" will be converted ("decay") to an expression of type "pointer to T" and its value will be the address of the first element of the array.
So if we write
int a[10];
int *p = a + 1;
the expression a is converted from "10-element array of int" to "pointer to int", and the value of a is the address of the first element (i.e., &a[0]). So the expression a + 1 gives us the address of the next integer object following a, which just happens to be &a[1]1.
Now assume we're working with a 2-D array:
int a[2][3];
int (*p)[3] = a + 1;
The expression a has type "2-element array of 3-element array of int". In this case, a "decays" to type "pointer to 3-element array of int", or int (*)[3]. So a + 1 gives us the address of the next *3-element array of int". Again, assuming a starts at 0x8000 and sizeof (int) is 4, then a + 1 evaluates to 0x800c
1. The expression a[i] is evaluated as *(a+i); that is, we're offsetting i elements from the address specified by a and dereferencing the result. Note that this treats a as a pointer, not an array. In the B language (from which C is derived), the array object a would have been a pointer object that contained the address of the first element (a[0]). Ritchie changed that in C so that the array expression would be converted to a pointer expression as necessary.
Related
This question already has answers here:
How come an array's address is equal to its value in C?
(6 answers)
Closed 7 years ago.
The following program prints that a and array share the same address.
How should I understand this behavior?
Is it &arr the address for the pointer arr, which contains the beginning address the 10 chars?
#include <stdio.h>
int main()
{
char arr[10] = {0};
char* a = (char*)(&arr);
*a = 1;
printf("a=%p,arr=%p.\n", a, arr);
printf("%d\n", arr[0]);
return 0;
}
When you allocate an array in C, what you get is something like the following:
+---+
arr[0]: | |
+---+
arr[1]: | |
+---+
...
+---+
arr[N-1]: | |
+---+
That's it. There's no separate memory location set aside for an object named arr to store the address of the first element of the array. Thus, the address of the first element of the array (&arr[0]) is the same value as the address of the array itself (&arr).
Except when it is the operand of the sizeof 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 of the array.
So the type of the expression arr in the first printf call is char [10]; by the rule above, the expression "decays" to type char *, and the value is the address of arr[0].
In the expression &arr, arr is the operand of the unary & operator, so the conversion isn't applied; instead of getting an expression of type char **, you get an expression of type char (*)[10] (pointer to 10-element array of char). Again, since the address of the first element of the array is the same as the address of whole array, the expressions arr and &arr have the same value.
In idiomatic C, you should write char *a = arr; or char *a = &(arr[0]);. &arr is normally a char **. Even if modern (C++) compilers fixe it automatically, it is not correct C.
As arr is an array of char, arr[0] is a char and arr is the same as &(arr[0]) so it is a char *. It may be strange if you are used to other languages, but it is how C works. And it would be the same if arr was an array of any other type including struct.
The address printed out by arr and a is the memory address of the first element of the array arr. (Remember, the name of an array is always a pointer to the first element of that array.) This is because after you have defined the array arr, you define a as a pointer to the same address in memory.
According to the C Standard (6.3.2.1 Lvalues, arrays, and function designators)
3 Except when it is the operand of the sizeof 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.
The address of an array is the address of its first element. So though these types are different
char ( * )[10] ( that corresponds to &arr ) and char *(that is used in the casting in statement
char* a = (char*)(&arr); ) they will have the same value that is the address of the first element of the array.
If you would not use the casting then the correct definition of the pointer initialized by expression &arr would be
char ( *a )[10] = &arr;
The difference is seen then the pointer is incremented. For your definition of pointer a the value of expression ++a will be greater sizeof( char) than the initial value . For the pointer I showed the value of expression ++a will be greater 10 * sizeof( char ) than the initial value.
In your case the type of expression *a is char and you may write *a = 1; while in my case the type of expression *a will be char[10] and you may not write *a = 1;
This question already has answers here:
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Possible Duplicate:
C: How come an array’s address is equal to its value?
C pointer : array variable
Considering a multidimensional Array:
int c[1][1];
Why all of the following expression points to the same address??
printf("%x", (int *) c); // 0x00000500
printf("%x", *c); // 0x00000500
printf("%x", c); // 0x00000500
How would a pointer's actual value and it's derefernced value can be the same?
Under most circumstances1, an expression of type "N-element array of T" will be converted ("decay") to 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 c has type int [1][1]; by the rule above, the expression will decay to type int (*)[1], or "pointer to 1-element array of int", and its value will be the same as &c[0]. If we dereference this pointer (as in the expression *c), we get an expression of type "1-element array of int", which, by the rule above again, decays to an expression of type int *, and its value will be the same as &c[0][0].
The address of the first element of the array is the same as the address of the array itself, so &c == &c[0] == &c[0][0] == c == *c == c[0]. All of those expressions will resolve to the same address, even though they don't have the same types (int (*)[1][1], int (*)[1], int *, int (*)[1], int *, and int *, respectively).
1 - the exceptions to this rule are when the array expression is an operand of the sizeof, _Alignof, or unary & operators, or is a string literal being used to initialize another array in a declaration
You just have to think: where is the first position on this array?
Suppose it's on 0x00000050 in your memory space. What is the first item in your array? It's c[0][0], and its address is 0x00000050. Sure enough, the address of the first position is the same of the array. Even if you do c[0] only, it still points to the same address, as long as you cast it to the right type.
But you should not confuse pointers to arrays.
How would a pointer's actual value and it's derefernced value can be the same
It's not a pointer, it's an array.
See Q&A #3 & #4
c is the address of the array. c is also the address of the first element.
could you tell my why the value of a referenced
array and the value of the array itself has the same value?
i know a is a type of int* but with &a it should be int** or am i wrong??
so the value should be a pointer to the a int pointer.
example code:
#include <stdio.h>
int main()
{
int a[10];
printf("a without ref.: 0x%x\n",a);
printf("a with ref.: 0x%x\n",&a);
return 0;
}
http://ideone.com/KClQJ
Name of the array decays to an pointer to its first element in this case.
Name of the array will implicit convert to a pointer ,except for two situation ,the one situations is "&array",the other is "sizeof(array)".In both cases,name of the array is a array ,not a pointer .
For example:
int a[10];
int *p;
p = a; //a is a pointer
p = &a; //a is a array,&a is a constant pointer
sizeof(a); //a is array
Given an array declaration T a[N], the expression &a has type "pointer to N-element array of T (T (*)[N]) and its value is the base address of the array. In that respect, the unary & operator behaves the same for arrays as it does for any other data type.
What's hinky is how C treats the array expression a. Except when it is the operand of the sizeof or unary & operators, or is a string literal being used to initialize another array in a declaration, an array expression of type "N-element array of T" (T [N]) will be replaced with ("decay to") a pointer expression of type "pointer to T" (T *) and its value will be the address of the first element of the array. IOW, a == &a[0].
Since the address of the first element of the array is the same as the base address of the entire array, the expressions a and &a yield the same value, but the types are different (T * as opposed to T (*)[N]).
if there is int *p which point a, then,
a+0 == &a[0] == p+0 == &p[0] : address
*(a+0) == *(&a[0]) == *(p+0) == *(&p[0]) : data
a == &a == &a[0]
Here,I have some Doubt with the output.
Why the Output is same ?
int (*r)[10];
printf("r=%p *r=%p\n",r,*r);
return 0;
Platform- GCC UBUNTU 10.04
Because Name of the array decays to an pointer to its first element.
int (*r)[10];
Is an pointer to an array of 10 integers.
r gives you the pointer itself.
This pointer to the array must be dereferenced to access the value of each element.
So contrary to what you think **r and not *r gives you access to the first element in the array.
*r gives you address of the first element in the array of integers, which is same as r
Important to note here that:
Arrays are not pointers
But expressions involving array name sometimes behave as pointer when those being used as name of the array would not make sense.
You would better understand if you look at the following program.
#include <stdio.h>
int main()
{
int a[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
int (*r)[10] = &a;
printf("r=%p *r=%p *(r+0)=%p *(r+1)=%p\n", r, *r, *(r+0), *(r+1));
printf("sizeof(int)=%d \n", sizeof(int));
return 0;
}
The output is as follows:
r=0xbfeaa4b4 *r=0xbfeaa4b4 *(r+0)=0xbfeaa4b4 *(r+1)=0xbfeaa4dc
sizeof(int)=4
Observations / Point(s)-to-note:
_DO_NOT_ de-reference a pointer which has not yet made to point to an address. So in your program int (*r)[10]; was de-referenced without being assigned to a memory area. This is not acceptable.
If you see the output - *r is same as *(r+0) which is same as r (only w.r.t this case)
If you see the output for *(r+0) and *(r+1) it is 40 bytes (0xbfeaa4dc - 0xbfeaa4b4 = sizeof(int) * size of the array (which is 10 in this case). So when you increment a pointer to a particular type, it gets incremented to sizeof(type) bytes!
the other worth-notable points about a pointer-to-an-array-of-integers are explained here
Hope this helps!
Remember that when an expression of type "N-element array of T" appears in most contexts, it will be converted to an expression of type "pointer to T" and its value will be the address of the first element in the array. The exceptions to this rule are when the array expression is an operand of either the sizeof or unary & (address-of) operands, or if the array expression is a string literal being used as an initializer in an array declaration.
Your situation is a mirror image of the following:
int a[10] = {0};
printf("a = %p, &a = %p\n", (void *) a, (void *) &a);
In the printf call, the expression a has its type converted from "10-element array of int" to "pointer to int" based on the rule above, and its value will be the address of the first element (&a[0]). The expression &a has type "pointer to 10-element array of int", and its value will be the same as a (the address of the first element in the array is the same as the address of the array itself).
Your code has a bit of undefined behavior in that you're dereferencing r before it has been assigned to point anywhere meaningful, so you can't trust that the output is at all accurate. We can fix that like so:
int a[10] = {0};
int (*r)[10] = &a;
printf("r = %p, *r = %p\n", (void *) r, (void *) *r);
In this case, r == &a and *r == a.
The expression r has type "pointer to 10-element array of int", and its value is the address of a. The expression *r has type "10-element array of int, which is converted to "pointer to int", and its value is set to the address of the first element, which in this case is a[0]. Again, the values of the two expressions are the same.
My understanding was that arrays were simply constant pointers to a sequence of values, and when you declared an array in C, you were declaring a pointer and allocating space for the sequence it points to.
But this confuses me: the following code:
char y[20];
char *z = y;
printf("y size is %lu\n", sizeof(y));
printf("y is %p\n", y);
printf("z size is %lu\n", sizeof(z));
printf("z is %p\n", z);
when compiled with Apple GCC gives the following result:
y size is 20
y is 0x7fff5fbff930
z size is 8
z is 0x7fff5fbff930
(my machine is 64 bit, pointers are 8 bytes long).
If 'y' is a constant pointer, why does it have a size of 20, like the sequence of values it points to? Is the variable name 'y' replaced by a memory address during compilation time whenever it is appropiate? Are arrays, then, some sort of syntactic sugar in C that is just translated to pointer stuff when compiled?
Here's the exact language from the C standard (n1256):
6.3.2.1 Lvalues, arrays, and function designators
...
3 Except when it is the operand of the sizeof 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.
The important thing to remember here is that there is a difference between an object (in C terms, meaning something that takes up memory) and the expression used to refer to that object.
When you declare an array such as
int a[10];
the object designated by the expression a is an array (i.e., a contiguous block of memory large enough to hold 10 int values), and the type of the expression a is "10-element array of int", or int [10]. If the expression a appears in a context other than as the operand of the sizeof or & operators, then its type is implicitly converted to int *, and its value is the address of the first element.
In the case of the sizeof operator, if the operand is an expression of type T [N], then the result is the number of bytes in the array object, not in a pointer to that object: N * sizeof T.
In the case of the & operator, the value is the address of the array, which is the same as the address of the first element of the array, but the type of the expression is different: given the declaration T a[N];, the type of the expression &a is T (*)[N], or pointer to N-element array of T. The value is the same as a or &a[0] (the address of the array is the same as the address of the first element in the array), but the difference in types matters. For example, given the code
int a[10];
int *p = a;
int (*ap)[10] = &a;
printf("p = %p, ap = %p\n", (void *) p, (void *) ap);
p++;
ap++;
printf("p = %p, ap = %p\n", (void *) p, (void *) ap);
you'll see output on the order of
p = 0xbff11e58, ap = 0xbff11e58
p = 0xbff11e5c, ap = 0xbff11e80
IOW, advancing p adds sizeof int (4) to the original value, whereas advancing ap adds 10 * sizeof int (40).
More standard language:
6.5.2.1 Array subscripting
Constraints
1 One of the expressions shall have type ‘‘pointer to object type’’, the other expression shall have integer type, and the result has type ‘‘type’’.
Semantics
2 A postfix expression followed by an expression in square brackets [] is a subscripted designation of an element of an array object. The definition of the subscript operator [] is that E1[E2] is identical to (*((E1)+(E2))). Because of the conversion rules that apply to the binary + operator, if E1 is an array object (equivalently, a pointer to the initial element of an array object) and E2 is an integer, E1[E2] designates the E2-th element of E1 (counting from zero).
Thus, when you subscript an array expression, what happens under the hood is that the offset from the address of the first element in the array is computed and the result is dereferenced. The expression
a[i] = 10;
is equivalent to
*((a)+(i)) = 10;
which is equivalent to
*((i)+(a)) = 10;
which is equivalent to
i[a] = 10;
Yes, array subscripting in C is commutative; for the love of God, never do this in production code.
Since array subscripting is defined in terms of pointer operations, you can apply the subscript operator to expressions of pointer type as well as array type:
int *p = malloc(sizeof *p * 10);
int i;
for (i = 0; i < 10; i++)
p[i] = some_initial_value();
Here's a handy table to remember some of these concepts:
Declaration: T a[N];
Expression Type Converts to Value
---------- ---- ------------ -----
a T [N] T * Address of the first element in a;
identical to writing &a[0]
&a T (*)[N] Address of the array; value is the same
as above, but the type is different
sizeof a size_t Number of bytes contained in the array
object (N * sizeof T)
*a T Value at a[0]
a[i] T Value at a[i]
&a[i] T * Address of a[i]
Declaration: T a[N][M];
Expression Type Converts to Value
---------- ---- ------------ -----
a T [N][M] T (*)[M] Address of the first subarray (&a[0])
&a T (*)[N][M] Address of the array (same value as
above, but different type)
sizeof a size_t Number of bytes contained in the
array object (N * M * sizeof T)
*a T [M] T * Value of a[0], which is the address
of the first element of the first subarray
(same as &a[0][0])
a[i] T [M] T * Value of a[i], which is the address
of the first element of the i'th subarray
&a[i] T (*)[M] Address of the i-th subarray; same value as
above, but different type
sizeof a[i] size_t Number of bytes contained in the i'th subarray
object (M * sizeof T)
*a[i] T Value of the first element of the i'th
subarray (a[i][0])
a[i][j] T Value at a[i][j]
&a[i][j] T * Address of a[i][j]
Declaration: T a[N][M][O];
Expression Type Converts to
---------- ---- -----------
a T [N][M][O] T (*)[M][O]
&a T (*)[N][M][O]
*a T [M][O] T (*)[O]
a[i] T [M][O] T (*)[O]
&a[i] T (*)[M][O]
*a[i] T [O] T *
a[i][j] T [O] T *
&a[i][j] T (*)[O]
*a[i][j] T
a[i][j][k] T
From here, the pattern for higher-dimensional arrays should be clear.
So, in summary: arrays are not pointers. In most contexts, array expressions are converted to pointer types.
Arrays are not pointers, though in most expressions an array name evaluates to a pointer to the first element of the array. So it is very, very easy to use an array name as a pointer. You will often see the term 'decay' used to describe this, as in "the array decayed to a pointer".
One exception is as the operand to the sizeof operator, where the result is the size of the array (in bytes, not elements).
A couple additional of issues related to this:
An array parameter to a function is a fiction - the compiler really passes a plain pointer (this doesn't apply to reference-to-array parameters in C++), so you cannot determine the actual size of an array passed to a function - you must pass that information some other way (maybe using an explicit additional parameter, or using a sentinel element - like C strings do)
Also, a common idiom to get the number of elements in an array is to use a macro like:
#define ARRAY_SIZE(arr) ((sizeof(arr))/sizeof(arr[0]))
This has the problem of accepting either an array name, where it will work, or a pointer, where it will give a nonsense result without warning from the compiler. There exist safer versions of the macro (particularly for C++) that will generate a warning or error when it's used with a pointer instead of an array. See the following SO items:
C++ version
a better (though still not perfectly safe) C version
Note: C99 VLAs (variable length arrays) might not follow all of these rules (in particular, they can be passed as parameters with the array size known by the called function). I have little experience with VLAs, and as far as I know they're not widely used. However, I do want to point out that the above discussion might apply differently to VLAs.
sizeof is evaluated at compile-time, and the compiler knows whether the operand is an array or a pointer. For arrays it gives the number of bytes occupied by the array. Your array is a char[] (and sizeof(char) is 1), thus sizeof happens to give you the number of elements. To get the number of elements in the general case, a common idiom is (here for int):
int y[20];
printf("number of elements in y is %lu\n", sizeof(y) / sizeof(int));
For pointers sizeof gives the number of bytes occupied by the raw pointer type.
In
char hello[] = "hello there"
int i;
and
char* hello = "hello there";
int i;
In the first instance (discounting alignment) 12 bytes will be stored for hello with the allocated space initialised to hello there while in the second hello there is stored elsewhere (possibly static space) and hello is initialised to point to the given string.
hello[2] as well as *(hello + 2) will return 'e' in both instances however.
In addition to what the others said, perhaps this article helps: http://en.wikipedia.org/wiki/C_%28programming_language%29#Array-pointer_interchangeability
If 'y' is a constant pointer, why does it have a size of 20, like the sequence of values it points to?
Because z is the address of the variable, and will always return 8 for your machine. You need to use the dereference pointer (&) in order to get the contents of a variable.
EDIT: A good distinction between the two: http://www.cs.cf.ac.uk/Dave/C/node10.html