int (*a)[5];
How can we Initialize a pointer to an array of 5 integers shown above.
Is the below expression correct ?
int (*a)[3]={11,2,3,5,6};
4 Answers
Suppose you have an array of int of length 5 e.g.
int x[5];
Then you can do a = &x;
int x[5] = {1};
int (*a)[5] = &x;
To access elements of array you: (*a)[i] (== (*(&x))[i]== (*&x)[i] == x[i]) parenthesis needed because precedence of [] operator is higher then *. (one common mistake can be doing *a[i] to access elements of array).
Understand what you asked in question is an compilation time error:
int (*a)[3] = {11, 2, 3, 5, 6};
It is not correct and a type mismatch too, because {11,2,3,5,6} can be assigned to int a[5]; and you are assigning to int (*a)[3].
Additionally,
You can do something like for one dimensional:
int *why = (int p[2]) {1,2};
Similarly, for two dimensional try this(thanks @caf):
int (*a)[5] = (int p[][5]){ { 1, 2, 3, 4, 5 } , { 6, 7, 8, 9, 10 } };
12 Comments
Grijesh Chauhan
@Flow second not correct try ,as
int (*a)[5]= (int(*)[5])&({11,2,3,5,6}); with new compiler
caf
int (*a)[5] = (int [][5]){ { 1, 2, 3, 4, 5 } , { 6, 7, 8, 9, 10 } };Grijesh Chauhan
@caf got it Thanks its similar to:
int *why = (int[2]) {1,2}Grijesh Chauhan
@caf got it!! my commented expression is not correct because I am using
& before constantGrijesh Chauhan
@caf your expression is
c99 ? |
{11,2,3,5,6} is an initializer list, it is not an array, so you can't point at it. An array pointer needs to point at an array, that has a valid memory location. If the array is a named variable or just a chunk of allocated memory doesn't matter.
It all boils down to the type of array you need. There are various ways to declare arrays in C, depending on purpose:
// plain array, fixed size, can be allocated in any scope
int array[5] = {11,2,3,5,6};
int (*a)[5] = &array;
// compound literal, fixed size, can be allocated in any scope
int (*b)[5] = &(int[5]){11,2,3,5,6};
// dynamically allocated array, variable size possible
int (*c)[n] = malloc( sizeof(int[n]) );
// variable-length array, variable size
int n = 5;
int vla[n];
memcpy( vla, something, sizeof(int[n]) ); // always initialized in run-time
int (*d)[n] = &vla;
1 Comment
Lundin
@M.M Indeed. Not sure why I wrote that. Fixed. I suppose I was trying to warn about passing a pointer to a compound literal outside the scope where it was declared.
int vals[] = {1, 2};
int (*arr)[sizeof(vals)/sizeof(vals[0])] = &vals;
and then you access the content of the array as in:
(*arr)[0] = ...
int (*a)[5] = malloc(nrows * sizeof *a);)size_t cols; ...; int (*a)[cols] = malloc( nrows * sizeof *a );. It's also what a 2D array expression decays to when passed to a function. The OP's usage is incorrect, but that doesn't make them useless.