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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

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3
  • 2
    You can't return arrays from functions in C, you need to return a pointer to an array. Try using malloc to create your array. Commented Mar 15, 2012 at 22:50
  • 1
    first of all, this is very basic question and answer is in any C book (K&R, c-faq). 2) when you provide error output, make sure you mark lines (12,14) in code (copy and paste from IDE or do it manually). Commented Mar 16, 2012 at 1:32
  • Btw, C does not use "methods", it uses functions. Commented Mar 16, 2012 at 4:20

5 Answers 5

Reset to default
7

A few problems:

  1. Your array is allocated on the stack, meaning that when your function exits, the memory you return will be invalid
  2. 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.

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7 Comments

Malloc doesn't necessarily return a pointer to garbage. On some systems, unallocated heap memory is set to a specific value on startup to aid debugging. Depending on the reason for populating the array with random data, it may be wholly necessary to do so.
@ChrisBrowne correct, I had updated my post before you commented, obviously it hadn't updated server-side.
srand is used to seed the generator and does not return a value. I realize that you are just using the OP's code, but just to be correct...
Uninitialized memory usually has the same value on each execution of a program, so it is not very random at all. The OS zeroes out memory pages that it gives to the program, so any non-zero data in the memory came from the program itself.
As far as I know, it's even worse - reading from allocated memory before writing to it is undefined behavior (and may crash your program)
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1

Your best bet is:

  1. Declare the array outside of the routine, and pass it in to initialize it:

    void init_array (int a[], nelms)

  2. 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;
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1 Comment

@patz (2) solves the question as posed, but seriously look to see if (1) works for you. With (2), use the return value to flag successful creation of the array.
0

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.

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Comments

0 
#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.

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Comments

0

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.

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