Declaring Array in C with variable length [closed] - arrays

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This block of code gives me the fibonacci numbers
#include <stdio.h>
int main()
{
int n; //integer overflow error for n > 47
printf("How many Fibonacci numbers?\n");
scanf("%d", &n);
int fibs[n];
fibs[0] = 0;
fibs[1] = 1;
printf("%d ", fibs[0]);
printf("%d ", fibs[1]);
for(int i = 2; i < n; i++)
{
fibs[i] = fibs[i - 2] + fibs[i - 1];
printf("%d ", fibs[i]);
}
return 0;
//gives 0 1 1 2 3 5 8 13 21 34 for n = 10
}
But this gives me the wrong output but no errors
#include <stdio.h>
int main()
{
int n, fibs[n];//change
printf("How many Fibonacci numbers?\n");
scanf("%d", &n);
fibs[0] = 0;
fibs[1] = 1;
printf("%d ", fibs[0]);
printf("%d ", fibs[1]);
for(int i = 2; i < n; i++)
{
fibs[i] = fibs[i - 2] + fibs[i - 1];
printf("%d ", fibs[i]);
}
return 0;
//gives 0 1 for n = 10
}
I know it definitely has something to do with the array and its size not being defined but I'm having trouble understanding what exactly is the problem.
Could someone explain what is going on here?

Answer 1

int n, fibs[n]; attempts to define an array using n for the length, but n has not been initialized, so its value is not determined. Common consequences include:
The definition behaves as if n has some small value, possibly zero, and then the following code attempts to store values in the array but overruns the memory reserved for it and thus destroys other data needed by the program.
The definition behaves as if n has some large value, causing the stack to overflow and the program to be terminated.
For example, storing 0 to to fibs[0] or 1 to fibs[1] might write to the memory reserved for n. Then the for loop terminates without executing any iterations because the test i < n is false.

Answer 2

The one big thing that I see in your code is the line int n, fibs[n];. The variable n is located on the stack since it's a local variable. That means that it's value can literally be anything before it's initialized. And since you are declaring an array using that value, the array has a random, unknown length. If it works, then that is purely coincidence. This is why your first code version works because the array is declared AFTER the scanf which initializes n. I think a better way of creating an array with a variable number of array elements is to use malloc instead...
int n, *fibs;
printf("How many Fibinocci numbers?\n");
scanf("%d", &n);
fibs = malloc(sizeof(int) * n);
if (fibs == NULL)
{
fprintf(stderr, "Unable to allocate sufficient memory for operation.\n");
exit(1);
}
Then you can use array indices fibs[0], fibs[1], etc... to access different locations in the block of memory.
Why does this work? Because int fibs[n] is LIKE a pointer to a block of memory. Technically, they are not the same, but you can generally use a pointer to a block of memory as an array. This will only work with single dimensional arrays because the compiler has no idea how many columns there are. But to work around that, you can compute that manually like this (i is the row, j is the column):
array[i * columns + j];