A function cannot return an array, only a pointer to an array. So I tried doing this:
int *arr = {-1, -1};
Is that valid syntax? I got a warning from the compiler for that. If not, is there a better way of setting a pointer variable to an array without looping?
No. This is not a valid syntax. A valid syntax which could be a pointer to a compound literal is
int *arr = (int []){-1, -1}; // Valid only in C99 and latter
If you wanna return an array then return a pointer to it;
int *arr = malloc(sizeof(*arr)*2);
....
return arr;
int *arr = (int*){-1, -1};memcpy(arr, (int []){ -1, -1}, sizeof(int)*2); But still you have to use malloc to allocate memory if you wants to return from function.
int * arr = malloc(2 * sizeof(*arr)); would be saver.You can set a pointer to an array created on the fly with a compound literal:
int *p = (int []){2, 4};
However, the array lives only as long as the block it is in. You cannot return it from a function. To return an array (by pointer), you should allocate it dynamically, as with malloc.
int* f(){ int* p = malloc(sizeof(int)*2); memcopy(p, (int []){2, 4};, sizeof(int)*2); return p;}
Is there a better way of setting a pointer variable to an array without looping?
There is a option called pointer to an array.
sample:
int a[10];
int (*x)[10] = &a;
arr is pointer to integer. So you can not have multiple inits like this.
You can have below:
int *arr = NULL;
arr = malloc(..)
...
...
arr[0] = some value..
arr[1] = some value..
Regarding
int *arr = {-1, -1};
Is that valid syntax?
No, but this is:
int *arr = {0}; //initialize pointer to 0;
arr = malloc(100*sizof(int));// allocate memory (can also do this at creation of arr)
arr[99]=1; //assign value to elements as needed
free(arr);// free memory when done with arr
And - in this form (pointer), arr is returnable by a function:
int * somefunc(int *arr)
{
//manipulate arr[0] - arr[n]
....
return arr;
}
void* f(){ int size = sizeof(int)*2; return memcopy(malloc(size), (int []){2, 4};, size); }int *ptr = &arr[0];or evenint *ptr = arr;.