To make a pointer to a whole array we proceed like that:
int arr[3] = {1,2,3};
int (*p)[3] = &arr;
How come i get an incompatibility error when trying to do the same with a 2D array?
int arr[3][3] = {{12,10,45}, {44,55,66}, {79,85,91}};
int (*p)[3] = &arr;
The problem here is &.
I'm sure this is a simple question that might have already been answered but i don't find any answer to this specific issue concerning the use of &.
In C, "pointer" is a category of types. To get a particular type, you have to specify what type of object the pointer points to.
Accordingly, the unary & does not generically compute a pointer. It creates a pointer to the type of its operand.
Similarly, in C, "array" is a category of types. To get a particular type, you have to specify the element type, the number of dimensions, and the sizes of at least the last n-1 dimensions.
Thus, with
int arr[3] = {1,2,3};
int (*p)[3] = &arr;
arr is defined as an array of 3 int, therefore
&arr has type pointer to array of 3 int (spelled int (*)[3] as a C type name), and
p is declared with that type
so everything is consistent.
How come i get an incompatibility error when trying to do the same with a 2D array?
Because your "the same" is not analogous.
With
int arr[3][3] = {{12,10,45}, {44,55,66}, {79,85,91}};
int (*p)[3] = &arr;
, arr has type array of 3 array of 3 int, and &arr has type pointer to array of 3 array of three int (a.k.a. int (*)[3][3]). This is not the same type that p has. The appropriate declaration of p for this version of arr would be
int (*p)[3][3] = &arr;
You've got p in the second example as a pointer to a 1D array, not a pointer to a 2D array. For that, you need the following:
int (*p)[3][3] = &arr;
arr has type int [3][3], so &arr has type int (*)[3][3]. So you can use:
int arr[3][3] = {{12,10,45}, {44,55,66}, {79,85,91}};
int (*p)[3][3] = &arr;
You can then access p as (*p)[i][j].
But it would be more typical to use:
int arr[3][3] = {{12,10,45}, {44,55,66}, {79,85,91}};
int (*p)[3] = arr;
In this case, int [3][3] is compatible with an int (*)[3] pointer, and you can use p[i][j] when accessing it.
An analogous situation exists with one-dimensional arrays. If you have int x[3], then the usual pointer type to use is int *p, in which case you would do p = x and access it as p[i].
But if you really want the extra level of indirection, you could use:
int (*p)[3] = &x;
Then instead of p[i], you would now need to use (*p)[i] when accessing it.
Related
I have an array of int pointers
int * arr[3];
If I want to define a pointer to arr, I can do the following:
int * (*p)[] = &arr;
However, in my code, I need to first declare that pointer:
int *(*p)[];
My question is, how to assign &arr to it after it has been declared. I have tried (*p)[] = &arr; but that didn't work.
You can simply do p = &arr.
Try here.
If you want to assign p to arr you simply have to do p = &arr;
this is because when you write p you basically are saying to C: this is the pointer to a array of pointers
So when you &arr you get the same type as & gives you the pointer to the variable next to it
Also beware that if you want the value of a variable inside a struct that its pointed to( ie struct something *a, where inside there int i) you put a -> (making it a a->integer;)
You should update the answer to the full extent
I have the following code for a one dimensional array:
#include <stdio.h>
int main()
{
static int b[4] = {11, 12, 13, 14};
int (*p)[4];
p = b;
printf("%d \n", *(p + 1));
return 0;
}
Even though I consider "b (the array name)" as a pointer pointing to a one dimensional array, I got a compiling error as
'=': cannot convert from 'int [4]' to 'int (*)[4]'
However, if I change b array into a two dimensional array "a (the array name)", everything works fine. Does this mean that, in the usage of "int (*p)[4];", "*p" has to represent a[] as in the following:
static int a[3][4] = { {1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12} };
int (*p)[4];
p = a;
As a result, "int (*p)[4]" only provides the flexibility on the number of rows of a two dimensional array.
Any insights on this problem?
Arrays naturally decay to pointers to their first elements, depending on context. That is, when such a decay happen then plain b is the same as &b[0], which have the type int *. Since the types of p and b (or &b[0]) are different you get an error.
As for a it's the same thing here, it decays to a pointer to its first element, i.e. a is the same as &a[0]. But since a[0] is an array of 4 elements, then &a[0] is a pointer to an array of four elements, or int (*)[4]. Which is also the type of p in the second example.
If you have an object of some type T like
T a;
then declaration of a pointer to the object will look like
T *p = &a;
Your array b has the type int[4]. So a pointer to the array will look like
int ( *p )[4] = &b;
To output the second element of the array using the pointer you should write
printf("%d \n", *( *p + 1 ) );
Thus your compiler issued the error message
cannot convert from 'int [4]' to 'int (*)[4]
because instead of writing at least
int ( *p )[4] = &b;
you wrote
int ( *p )[4] = b;
On the other hand, an array designator used in expressions with rare exceptions is implicitly converted to pointer to its first element. For example in this declaration
int *p = b;
the array b used as an initializer is converted to pointer to its firs element. The above declaration is equivalent to
int *p = &b[0];
or that is the same
int *p = b + 0;
Using this pointer you can call the function printf like
printf("%d \n", *(p + 1));
If you have a two-dimensional array as
int a[3][4];
then used in expressions it is converted to pointer to its first element that has the type int[4]. So you may write
int ( *p )[4] = a;
If you want to declare a pointer to the whole array as a single object you can write
int ( *p )[3][4] = &a;
a pointer pointing to a one dimensional array,
No, it points directly to the first element. Likewise:
int *p = b;
is enough.
The number 4 is not really part of any type here;
static int b[] = {11, 12, 13, 14};
It can be left out in the declaration. (Because it is the first dimension unless you make it 2D)
This (from AA)
int (*p)[4] = &b;
...
printf("%d \n", *( *p + 1 ) );
is just a obfuscated and overtyped version of:
int (*p)[] = &b;
...
printf("%d \n", (*p)[1] );
This replaces b with (*p), normally not what you want.
I was reading through some lecture notes that in order for a pointer to reference a 2D array, it has to be given the address of the first element.
int a[10][10];
int *p = &a[0][0];
I've never tried this, so I was curious why isn't it enough to assign the array itself to the pointer, just as we do in a 1D case.
int a[10][10];
int *p = a;
The array is kept in an uninterrupted 'line' of memory anyway, and 2D arrays only have a different type, but the same structure as 1D arrays.
By doing this
int *p = &a[0][0];
I don't see how we give the pointer any more information than by doing this
int *p = a;
Or maybe all arrays regardless of their number of dimensions have the same type, the only difference being that multidimensional arrays store their extra dimensions before their first element and we need to jump over those memory spaces which remember sizes of an array's dimensions?
First, some background:
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.
Given the declaration
int a[10][10];
the expression a has type "10-element array of 10-element array of int". Unless this expression is the operand of the sizeof or unary & operators, it will be converted to an expression of type "pointer to 10-element array of int", or int (*)[10].
Given that declaration, all of the following are true:
Expression Type Decays to
---------- ---- ---------
a int [10][10] int (*)[10]
&a int (*)[10][10]
*a int [10] int *
a[i] int [10] int *
&a[i] int (*)[10]
*a[i] int
a[i][j] int
&a[i][j] int *
Also,
sizeof a == sizeof (int) * 10 * 10
sizeof &a == sizeof (int (*)[10][10])
sizeof *a == sizeof (int) * 10
sizeof a[i] == sizeof (int) * 10
sizeof &a[i] == sizeof (int (*)[10] )
sizeof *a[i] == sizeif (int)
sizeof a[i][j] == sizeof (int)
sizeof &a[i][j] == sizeof (int *)
Note that the different pointer types int (*)[10][10], int (*)[10], and int * don't have to be the same size or have the same representation, although on the platforms I'm familiar with they do.
The address of the first element of the array is the same as the address of the array itself; thus, all of a, &a, a[0], &a[0], and &a[0][0] will yield the same value, but the types will be different (as shown in the table above).
So, assume we add the following declarations:
int *p0 = &a[0][0]; // equivalent to int *p0 = a[0];
int (*p1)[10] = &a[0]; // equivalent to int (*p1)[10] = a;
int (*p2)[10][10] = &a;
All of p0, p1, and p2 initially have the same value, which is the address of the first element in a; however, because of the different pointer types, the results operations involving pointer arithmetic will be different. The expression p0 + 1 will yield the address of the next int object (&a[0][1]). The expression p1 + 1 will yield the address of the next 10-element array of int (&a[1][0]). And finally, the expression p2 + 1 will yield the address of the next 10-element array of 10-element array of int (effectively, &a[11][0]).
Note the types of p1 and p2; neither is a simple int *, because the expressions being used to initialize them are not that type (refer to the first table).
Note the pattern; for an array type, the simpler the expression, the more complicated the corresponding type will be. The expression a does not refer to a single int object; it refers to a 10x10 array of int objects, so when it appears in an expression, it is treated as a pointer to an array of integers, not a pointer to a single integer.
The compiler knows that "a" is a pointer to ten integers. If you don't declare the dimensions, then the compiler sees the new pointer as a pointer to an unknown number of integers. This will work in your case, but it will generate a compiler warning because the compiler sees them as incompatible pointers. The syntax for what you are trying to do (without generating a compiler warning) is:
int a[10][10];
int *p1 = &a[0][0];
int (*p2)[10] = a;
printf("p1: %p p2: %p\n", p1, p2);
One reason this is important is pointer arithmetic:
p1++; //move forward sizeof(int) bytes
p2++; //move forward sizeof(int) * 10 bytes
You understanding is close, the difference is the type information. Pointer does has its type. For example int* p, the pointer type is int*, as int a[10][10], the corresponding pointer type is int *[10][10].
In your example, p and a do point to the same address, but they're different type, which matters when perform arithmetic operation on them.
Here's an example from this URL
Suppose now that we define three pointers :
char *mychar;
short *myshort;
long *mylong;
and that we know that they point to the memory locations 1000, 2000, and 3000, respectively.
Therefore, if we write:
++mychar;
++myshort;
++mylong;
mychar, as one would expect, would contain the value 1001. But not so obviously, myshort would contain the value 2002, and mylong would contain 3004, even though they have each been incremented only once. The reason is that, when adding one to a pointer, the pointer is made to point to the following element of the same type, and, therefore, the size in bytes of the type it points to is added to the pointer.
You are right, you can assign the array itself to the pointer:
int a[10][10] = {[0][0]=6,[0][1]=1,[1][0]=10,[1][1]=11};
int b[10][10][10] = {[0][0][0]=8,[0][0][1]=1,[0][1][0]=10,[1][0][0]=100};
int *p, *q, *r, *s;
p = &a[0][0];
q = a; // what you are saying
r = &b[0][0][0];
s = b; // what you are saying
printf("p= %p,*p= %d\n",p,*p);
printf("q= %p,*q= %d\n",q,*q);
printf("r= %p,*r= %d\n",r,*r);
printf("s= %p,*s= %d\n",s,*s);
And the output is:
p= 0xbfdd2eb0,*p= 6
q= 0xbfdd2eb0,*q= 6
r= 0xbfdd3040,*r= 8
s= 0xbfdd3040,*s= 8
They point to the same address, regardless of the dimension of the matrix. So, what you are saying is right.
Well in 2D array, the outcome of *a and a is the same, they all point to the first address of this 2D array!
But if you want to define a pointer to point to this array, you could use int (*ptr)[10] for example.
You are right, 1D and 2D share the same structure, but 2D has some additional manipulation on pointers like above.
So all in all, in 2D array, a, *a and &a[0][0] prints the same address, but their usages may vary.
Like this:
#include<stdio.h>
int main() {
int a[10][10];
int *pa1 = &a[0][0];
int *pa2 = *a;
printf("pa1 is %p\n", pa1);
printf("pa2 is %p\n", pa2);
printf("Address of a is %p\n", a);
// pointer to array
int (*pa3)[10];
pa3 = a;
printf("pa3 is %p\n", pa3);
return 0;
}
They print the same address.
Im relatively knew to C, i am used to program in Java so i find C a little bit difficult in what concerns arrays. I still cofuse myself with this cases:
int a [];
int* a;
int *a;
In java, i would do something like this to return an array in a function:
int [] returnArr(int [] a){
... modify a ...
return a;
}
int [] a = {...};
int [] b = returnArr(a); ##
How can i do the same in C, specially the parts with ##.
EDITED:
I have this function:
float *normalizeValues(float *v, float maxY){
int size = sizeof(v) / sizeof(float);
float max = findMax(v);
float ratio = maxY / max;
int i;
for(i = 0; i < size ; ++i){
v[i] = v[i] * ratio;
}
return v;
}
And im doing the following:
float vert [] = {306, 319, 360, 357, 375, 374, 387, 391, 391, 70, 82, 94, 91, 108, 114, 125, 127, 131};
int i = 0;
float *vert2;
vert2 = normalizeValues(vert, 0.7);
for(i = 0; i < sizeof(vert2) / sizeof(float); ++i){
fprintf(stdout,": %f\n",vert2[i]);
}
And the output is only 1 element.
EDIT: To directly answer your updated question: you have to pass in the size of the array. C has no mechanism to store the size of arrays like Java does. If the compiler knows about the size of the array because the array is a global or local variable, not dynamically allocated, then you can use the sizeof() operator. Otherwise, you have to know the size separately, or use sentinel values in your array (such as a 0.0 at the end, or a NULL).
As for arrays, pointers and arguments in general, see below:
You will be returning a pointer to the array, which is indicated with the '*' syntax:
int *returnArr(int[] a) {
// modify a...
return a;
}
int a[] = { ... };
int *b;
b = returnArr(a);
A few things to note:
You can't do assignments in variable declarations that involve non-constant expressions (e.g., function calls). This might have changed in C99, though.
The brackets go after the variable name, unlike in Java where they are part of the type. Even though Java's syntax is more consistent, it doesn't quite make sense in C where you often give the array size in the brackets in the variable declaration:
int a[3] = { ... };
There's no way to specify that a function returns an array as opposed to a plain pointer. In C, array references decay to pointers (though pointers and arrays are NOT the same thing, as is commonly claimed). That means that whenever you pass an array around, C only provides a means to a pass a pointer to the array. The whole array isn't actually copied. As it happens, the name of the array is also a pointer to the first element of the array.
Please also take note of what user268396 says in their answer. If you are planning to create a new array and return it, you'll need to either allocate the array dynamically, or have a pointer to an already allocated array be passed in (which is what it seems like you are kind of doing anyway).
You can't. When the function returns the stack frame will be wiped out (typically) and your generated array will be clobbered by that. You can however edit the function prototype to accept a pointer to the array to modify. That kind of function argument is known as an "output parameter". Example:
void function func(int a, int b, int[2] to_modify)
{
to_modify[0] = a;
to_modify[1] = b;
}
int main()
{
int foo[2];
func(1, 2, foo);
printf("Result: foo[0] = %d, foo[1] = %d\n", foo[0], foo[1]);
return 0;
}
This will print "Result: foo[0] = 1, foo[1] = 2".
Hope this helps
#include<stdio.h>
void change(int *c)/*Pointer c now has the first location of the array a[]*/
{
*(c+0) = 0;/*assign values to the array by adding step-size to the first array position*/
*(c+1) = 1;
*(c+2) = 2;
*(c+3) = 3;
*(c+4) = 4;
}
main()
{
int a[5]={10,20,30,40,50}; /* Declare and Assign an array a[] of size 5.*/
int *b = a; /*Declare and assign a Pointer to the location of the array.*/
change(b); /*pass the pointer(which is now pointing to first position of array) to the change() function.*/
printf("%d,%d,%d,%d,%d,",a[0],a[1],a[2],a[3],a[4]);/*Print the changed value.*/
}
Output: 0,1,2,3,4,
From Java point of view, Pointers are simply like(not exactly) Object references.
Object O;
O = New SomeClassName();
Like Object Reference O is pointing to some Actual Object of type SomeClassName, so does pointers in C:
int *b;
b = &a;
Variable b is simply pointing to the address location to a.
Taking a deep dive into array concepts:
int a[5];
int *b = a;
Here we are just saying like Mr.*b point to the first location of group a i.e. a[0].
Now the power pointer in C is that from now on, here after:
*b means a[0]
*(b+1) means a[1]
*(b+2) means a[2]
*(b+3) means a[3]
*(b+4) means a[4]
This means you change in *(b+4), you're changing a[4].
int* returnArr(int a[]){
//modify a
return a;
}
One need mention is when you use an array in the parameter list of a function, it will be converted into a pointer. So in main(...)'s declaration, char *argv[] and char **argv are actually same. In this case, int a[] and int* a are same. But array and pointer is not the same thing.
Take the following code as an example:
int a[10];
int main(){
int *p = a;
p[5] = 4;
return p[5];
}
p is a pointer, when we access p[i], note that the address of p is not the address of a, the content of p is the address of a. Then the code will:
access the memory to get the content of p, i.e. the address of a.
compute the offset based on i and type of the pointer(int).
access the memory to get the result.
a is an array of int, if we access a[i], the address of a is just the address of a[0], the code will:
Compute the offset based on i and the type int.
Access the memory.
Pointer and array are different types. So if you declare int *p in one file and use it in that file, but define the p as an array in another file, that will cause problem.
You may also wonder about int *p = a, in ANSI, if you use an array(its name) as an expression, the compiler will convert it into a pointer, pointing to the very first element of the array.
Update based on Jim Balter's comments:
If you use an array(its name) as an expression, the compiler will not always convert it into a pointer, pointing to the very first element of the array. For instance, in sizeof(p->q->a), p->q->a is an expression but if a is an array it isn't converted into a pointer.
"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.
In C, you can only return a pointer of an array in a function.
For example, if you want to return a string(array of char) in a function, you can return a pointer to a null-ended string. If you want to return an array of some other type(int, user-defined struct, etc), you can alloc some memory to store the array, and return the pointer of the array, return the size of the array in the parameter.
example:
int *function(int *size)
{
*size = 10;
int *intP = (int *)malloc((*size)*sizeof(int));
return intP;
}
Please help me understand the programs below.
#include<stdio.h>
int main()
{
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);
printf("value at memory location %p is %d", &a[1], a[1]);
return 0;
}
&a[1] and &a+1. Are they same or different?
#include <stdio.h>
int main()
{
int v[10];
int **p;
int *a[5];
v[0] = 1234;
v[1] = 5678;
a[0] = v;
a[1] = v+1;
printf("%d\t%d\t%d\t%d\n", *a[0],*a[1],a[0][0],**a);
printf("%d\n", sizeof(v));
return 0;
}
I wanted to know how *a[5] is represented in memory. Is *a a base pointer that points to a[0],a[1],a[2],a[3],a[4]?
#include<stdio.h>
int main()
{
int v[10];
int **p;
int (*a)[10];
a=&v;
printf("%d\n",*a);
return 0;
}
a=v; // gives error why? does v here decay into *v. Then does &v get decayed into (*)[]v? & means const pointer. Here, how is it possible to set a const pointer to a non-const pointer without a typecast?
Where does the array get stored in the memory. Does it get stored onto the data segment of the memory.
#include<stdio.h>
int main()
{
int carray[5]={1,2,3,4,5};
printf("%d\n",carray[0]);
printf("%d\t%d\t%d\n",sizeof(carray),sizeof(&carray),sizeof(&carray[0]));
return 0;
}
EDITED:
I have gone through some of the articles which stated that the only two possible situations where an array name cannot be decyed into pointer is the sizeof and &. But in the above program sizeof(&carray) gives the size as 4. and &carray decays into (*)[]carray as its an rvalue.
Then the statement that array name cannot get decayed into pointers on two conditions sizeof and & becomes false here.
&a[1] and &a+1. Are they same or different?
Different. &a[1] is the same as (a+1). In general, x[y] is by definition equivalent to *(x+y).
I wanted to know how *a[5] is represented in memory. Does *a is a base
pointer that points to a[0],a[1],a[2],a[3],a[4].
In your second example, a is an array of pointers. *a[i] is the value of the object, the address of which is stored as the ith element in your array. *a in this case is the same as a[0], which is the first element in your array (which is a pointer).
a=v //why this gives error
Because a (in your last example) is a pointer to an array. You want to assign to a, then you need to assign the address of the array v (or any other array with correct dimensions);
a = &v;
This is very good that you've commited to understanding things, but nothing will help you better than a good C book.
Hope this helps.
Stuff you are gonna need to know when dealing with pointers is that:
int *a and int a[]
is a declaration of an Array, the only diffrence is that in a[] youre gonna have to declare its constant size, *a gives you flexability, it can point at an array size 1 to infinity
int *a[] and int **a
is a declaration of an Array of Array,sometimes called Matrix, the only diffrence is that in *a[] youre gonna have to declare how many Arrays a[] gonna contain pointers of, **a gives you flexability, it can point at any Array of arrays that you want it to be assigned to.
IN GENERAL:
When adding & to a variable, your adding a * to its Type definition:
int a;
&a -> &(int)=int*
when adding * to a variable, you decrase a * from its Type definition
int *a;
*a -> * (int * )=int
int *a;
&a - the Address given to the pointer a by the system(pointer of pointer = **a)
&a+1 - the Address to the beginning of the array + 1 byte
&a[1] == &(a+1) - the Address to the beginning of the array + 1 size of int
int **a;
*a == a[0] - the Address of the first Array in the array of arrays a
*a[0]==a[0][0] - the first int of first array
int *a, b[5];
*a=*b - ERROR because a points at garbage to begin with
a=b - a points at array b
ask me what else you want to know and ill edit this answer.