Why does my loop terminate even with a condition? - c

Question

I am trying to get user input and store them in an array Fib[i]. After that print the Fib[i] array. When the user enters -1 the loop will quit and the program will end. But my code is not printing or terminating.
#include <stdio.h>
double Fib[50];//globally declared
int fib(int n)
{
int i;
for(i=0; i<50; i++)
{
scanf("%d", &Fib[i]);
if(i==-1)
break;
//printf("numbers entered %d\n", Fib[i]); // <- doesn't terminate if printf is here
}
printf("numbers entered %d\n", Fib[i]); //doesn't print anything??
}
int main()
{
int i, n;
//calling the function
fib(n);
return 0;
}
user input:
4
5
-1
Expected output:
Numbers entered
4
5

Answer 1

First issue: you declare Fib as an array of double:
double Fib[50];
But you use %d to read the values, which is for reading an int:
scanf("%d", &Fib[i]);
Using the wrong format specifier invokes undefined behavior. You presumably want to store integers, so change the array to int:
int Fib[50];
Next is your array breakout condition:
if(i==-1)
i is your array index, which ranges from 0 to 49, so this will never be true. You want to stop when the user enters -1, and that value will be in Fib[i]:
if(Fib[i]==-1)
Finally, printing the array:
printf("numbers entered %d\n", Fib[i]);
This doesn't print the array. It just prints the element at the last index of i, and the value at that index will always be -1. You need a separate loop to print the values:
int j;
printf("numbers entered:\n");
for (j=0; j<i; j++) {
printf("%d\n", Fib[j]);
}

Answer 2

This code has many code writing standard issues but it seems you are new so I am making minimal changes just for your understanding
#include <stdio.h>
double Fib[50];//globally declared
int fib(int n)
{
int i,j;
for(i=0; i<50; i++)
{
scanf("%lf", &Fib[i]);
if(Fib[i]==-1)
break;
}
printf("numbers entered \n");
for(int j=0;j<i;j++)
{
printf("%lf\n",Fib[j]);
}
}
int main()
{
int i, n;
fib(n);
return 0;
}