C - function returns an array - c

Question

I'm writing a method that receives a number l and returns a vector of size l with random numbers. I have this code, but does not work
#include <time.h>
int makea (int z) {
int a1[z];
int i;
for (i = 0; i < tam; i++) {
a1[i]=srand(time(0));
}
return a1;
}
These are the errors that the compiler returns me
arrays1.c: In function 'makea':
arrays1.c:12: error: void value not ignored as it ought to be
arrays1.c:14: warning: return makes integer from pointer without a cast
arrays1.c:14: warning: function returns address of local variable
I think is a problem of pointers... but I'm not really sure

Answer 1

A few problems:
Your array is allocated on the stack, meaning that when your function exits, the memory you return will be invalid
In C, you cannot return an array from a function, it must first decay into a pointer.
So, to fix, use malloc and a pointer:
int *makea (int z) {
int *a1 = malloc(sizeof(int) * z);
int i;
srand(time(NULL));
for (i = 0; i < tam; i++) {
a1[i]= rand();
}
// remember to free a1 when you are done!
return a1;
}
Also note that using malloc can sometimes basically grant you the 'random number' scenario for free, negating the need to loop through the elements as the value returned from malloc is garbage (and thus random numbers).
However, also note that malloc is implementation-specific, meaning that an implementation could theoretically clear the memory for you before returning it.

Answer 2

Your best bet is:
Declare the array outside of the routine, and pass it in to initialize it:
void init_array (int a[], nelms)
Plan B is pass a pointer to a pointer, and have the routine allocate and initialize it
Like this:
void alloc_and_init_array (int **a_pp, int nelms)
{
*a_pp = malloc (sizeof (int) * nelms);
...
... or, equivalently ...
int *
alloc_and_init_array (int nelms)
{
int *a_p = malloc (sizeof (int) * nelms);
...
return a_p;

Answer 3

A local variable like your array is allocated on the stack. At function return it is removed from the stack, so the pointer you return points to an unallocated memory location.
You have to allocate the array with malloc() or pass an already existing array to the function.

Answer 4

#include <time.h>
int makea (int z) {
int *a1 = (int*)malloc(z*sizeof(int));
int i;
for (i = 0; i < tam; i++) {
a1[i]=srand(time(0));
}
return a1;
}
IMPORTANT: remember to free memory allocated somewhere outside, when you do not need it anymore.

Answer 5

Well, first off your function says that it returns an int, yet you want to return an array, so that is wrong. Of course, you can't return an array in C either...
Second, you will have to return a pointer. You cannot copy arrays via assignment or assign a new value to an array at all in C, so your function won't be very useful. Either return an int* or take an int** as an output argument and initialize it in your function.
Also, your array is locally allocated, so even if the compiler didn't complain you would be returning invalid memory.
int makea (int size, int **out_array) {
int *temp, i;
if(!out_array)
return 0;
temp = malloc(sizeof(int) * size);
if(!temp)
return 0;
srand(time(0));
for (i = 0; i < size; ++i)
temp[i] = rand();
*out_array = temp;
return 1;
}
int main() {
int *arr;
if(!makea(10, &arr)) {
printf("Failed to allocate array");
return -1;
}
return 0
}
Another note:
temp[i] = srand(time(0));
That is wrong. srand seeds the random number generator, but does not return a random number. You call srand to input the seed and then call rand to get a random number.