how to return float array from a function?

Question

The output shows the values of k as 0.000 but it should contain actual values that are returned from funcTest().

#include <stdio.h>
#include <stdlib.h>

float *funcTest(int *a, int size)
{
    float p[size];
    int i;
    for(i=0; i< size; i++){
        p[i] = *a;
        p[i]=p[i]/2;
        a++;
    }
    for(i =0; i<size; i++){
        printf("%f\n",p[i]);
    }
    return (float *)p;
}

int main()
{
    int a[4] = {1,5,3,7};
    int size = 4;
    int i,j;
    float *k;
    k = funcTest(&a[0],size);
    for(i =0; i<size; i++){
        printf("%f\n",*k);
        k++;
    }
    return 0;
}

OUTPUT :

0.5
2.5
1.5
3.5
1.5
0.000
0.000
0.000

Answer 1

You return a local variable. so, at end of function, local variables are cleaned and do not exist any more. You should use malloc

float *funcTest(int *a, int size)
{
    float *p = malloc(sizeof(float) * size);

And don't forget to free memory when you don't use it anymore with free. If you forget, you will have a memory leak

Answer 2

In your code, float p[size]; is local to the function funcTest(), so you cannot return the address of it from the function and expect it to be valid in the caller. Once the funcTest() function ends, p will cease to exist. Then, any attempt to make use of the return value will cause undefined behavior, because of invalid memory access.

Instead, you can make p a pointer, allocate memory dynamically using malloc() or family and then, return that pointer to the caller. Dynamically allocated memory will have a lifetime equal to the entire execution of the program (unless deallocated manually), so, even after returning from the function, in the caller, the returned pointer will be valid.