How to dynamically fill an array with a fibonachi sequence

Question

My question is about dynamic memory allocation and pointer arithmetic.I have a function that is called in the main function and it should fill the array with n longs and populate them with a Fibonacci sequence dynamically. my function is below but it doesn't appear to give me the correct output.

long* make_fib_array(long n){

int i;
long *fib;  
fib = (long*)malloc(sizeof(long)*n);
if (fib == NULL){
    printf("ERROR: Out of memory\n");
return NULL;}


*fib = 0;
*(fib +1) = 1;
for (i=2;i<n;i++);{
    *(fib +i) = *(fib + (i-1)) + *(fib + (i-2));}     

for(i=0;i<n;i++){
    printf("%ld\n",*(fib + i));}


return fib;}

using a value of the values of 0 and 1 print fine but the values after this are very large and not expected. I suspect it is something to do woth my pointer arithmetic but im not sure what it is.

`*(fib + (i+1)) = 1;` uses `i` before it is assigned a value -> undefined behavior — UnholySheep, Dec 01 '18 at 18:12
I have corrected this but the error still remains, I will edit the post for this though, thank you — tribo32, Dec 01 '18 at 18:18
@tribo32: You must initialize the `i` like `int i = 0` before using it. Otherwise `i` contains ****junk*** value — eapetcho, Dec 01 '18 at 18:18

score 1 · Answer 1 · answered Dec 01 '18 at 18:31

1

Just remove ; after for loop brackets:

From

for (i=2;i<n;i++);{

to

for (i=2;i<n;i++) {

answered Dec 01 '18 at 18:31

Morpheus

377
2
9

much appreciated! – tribo32 Dec 01 '18 at 18:35

How to dynamically fill an array with a fibonachi sequence

1 Answers1